The Vector Space Rn
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Linear Combination, Dependence and Span
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Linear Dependence in R^n
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Basis for Rn
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Solution Space of Homogenous SLE
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Subspaces
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Row and Column Spaces
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Change of Basis
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Vector Spaces
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Linear Combinations and Span
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Linear Dependence
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Basis for Known Vector Spaces
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Basis for a Solution Space, Homogeneous SLE
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Basis of a Subspace of a Known Vector Space
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Coordinate Vectors and Change of Basis
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[{"Name":"The Vector Space Rn","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"3D vectors, Motivation","Duration":"8m 12s","ChapterTopicVideoID":25052,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25052.jpeg","UploadDate":"2021-06-27T05:23:42.7170000","DurationForVideoObject":"PT8M12S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.165","Text":"In this clip, we start a new topic."},{"Start":"00:03.165 ","End":"00:10.380","Text":"Which is quite a central topic in linear algebra and it\u0027s vector spaces."},{"Start":"00:10.380 ","End":"00:12.810","Text":"It\u0027s quite theoretical,"},{"Start":"00:12.810 ","End":"00:20.130","Text":"so we\u0027ll take it nice and easy and introduce it slowly."},{"Start":"00:20.130 ","End":"00:25.260","Text":"As you can see, it\u0027s made up of 2 words, vector and spaces."},{"Start":"00:25.260 ","End":"00:30.870","Text":"First of all, we\u0027ll talk about vectors and not just any old vectors,"},{"Start":"00:30.870 ","End":"00:32.070","Text":"that\u0027s still too general,"},{"Start":"00:32.070 ","End":"00:39.420","Text":"let\u0027s narrow it down to 3-dimensional vectors, 3D for short."},{"Start":"00:39.420 ","End":"00:45.850","Text":"Before that, I want to talk about what is 3D space?"},{"Start":"00:45.850 ","End":"00:51.019","Text":"Here\u0027s a picture which is going to help me to explain 3D space."},{"Start":"00:51.019 ","End":"00:56.165","Text":"Notice these blue lines with the arrows, these are axes."},{"Start":"00:56.165 ","End":"00:58.340","Text":"Notice there\u0027s an x-axis,"},{"Start":"00:58.340 ","End":"01:03.635","Text":"a y-axis, and z-axis, that\u0027s 3 axes."},{"Start":"01:03.635 ","End":"01:10.550","Text":"In contrast with high school where you probably just had 2 axes,"},{"Start":"01:10.550 ","End":"01:13.730","Text":"maybe there was an x-axis and"},{"Start":"01:13.730 ","End":"01:19.955","Text":"a y-axis was good for describing the flat world, the 2D world."},{"Start":"01:19.955 ","End":"01:23.180","Text":"But in real life we live in a 3D world,"},{"Start":"01:23.180 ","End":"01:25.175","Text":"and so we need 3 axes,"},{"Start":"01:25.175 ","End":"01:27.140","Text":"x, y, and z,"},{"Start":"01:27.140 ","End":"01:30.890","Text":"if you like, width, length, and height."},{"Start":"01:30.890 ","End":"01:34.100","Text":"When we were in a 2D world and we had a point,"},{"Start":"01:34.100 ","End":"01:36.230","Text":"it was described by 2 numbers."},{"Start":"01:36.230 ","End":"01:39.110","Text":"We dropped a perpendicular here and 1 here,"},{"Start":"01:39.110 ","End":"01:41.245","Text":"and this was the x of the point,"},{"Start":"01:41.245 ","End":"01:44.220","Text":"maybe, this was 3 and this was 4."},{"Start":"01:44.220 ","End":"01:47.010","Text":"When we would say this is the point 3,"},{"Start":"01:47.010 ","End":"01:48.945","Text":"4 and that would do it."},{"Start":"01:48.945 ","End":"01:52.920","Text":"But in 3-dimensional space, well,"},{"Start":"01:52.920 ","End":"01:54.705","Text":"let\u0027s take this point here,"},{"Start":"01:54.705 ","End":"01:56.070","Text":"and we call it P,"},{"Start":"01:56.070 ","End":"01:58.245","Text":"it needs 3 coordinates;"},{"Start":"01:58.245 ","End":"02:00.450","Text":"an x, a y, and a z."},{"Start":"02:00.450 ","End":"02:05.820","Text":"1 way to do this is just to drop a perpendicular into the x,"},{"Start":"02:05.820 ","End":"02:08.130","Text":"y plane, which is this 1 here."},{"Start":"02:08.130 ","End":"02:12.600","Text":"We see that for this point here, for example,"},{"Start":"02:12.600 ","End":"02:15.630","Text":"then x is 2 and y is 3,"},{"Start":"02:15.630 ","End":"02:17.715","Text":"so we start with 2, 3."},{"Start":"02:17.715 ","End":"02:20.220","Text":"But then we also have to go up,"},{"Start":"02:20.220 ","End":"02:21.830","Text":"and that would be 5 units."},{"Start":"02:21.830 ","End":"02:26.480","Text":"This was meant to be parallel with this is the same height as this,"},{"Start":"02:26.480 ","End":"02:28.835","Text":"so we need an x, a y,"},{"Start":"02:28.835 ","End":"02:31.805","Text":"and a z for the 3-dimensional world."},{"Start":"02:31.805 ","End":"02:35.105","Text":"As another example, let\u0027s take this point here."},{"Start":"02:35.105 ","End":"02:41.165","Text":"We can see that the x is 2 the y is 3."},{"Start":"02:41.165 ","End":"02:43.010","Text":"But since we don\u0027t go up at all,"},{"Start":"02:43.010 ","End":"02:44.360","Text":"we stay in the plane,"},{"Start":"02:44.360 ","End":"02:48.425","Text":"the last component is a 0."},{"Start":"02:48.425 ","End":"02:51.755","Text":"Another example, this point here,"},{"Start":"02:51.755 ","End":"02:54.140","Text":"we drop the perpendicular down to the x,"},{"Start":"02:54.140 ","End":"02:57.575","Text":"y plane, we see that it is the origin here."},{"Start":"02:57.575 ","End":"03:01.610","Text":"The x is 0, the y is 0,"},{"Start":"03:01.610 ","End":"03:07.200","Text":"but the z component we go up 5 is 5."},{"Start":"03:07.640 ","End":"03:11.795","Text":"That\u0027s an introduction to 3D points."},{"Start":"03:11.795 ","End":"03:15.080","Text":"But we want 3D vectors."},{"Start":"03:15.080 ","End":"03:17.400","Text":"Now, let\u0027s take this point,"},{"Start":"03:17.400 ","End":"03:19.200","Text":"corresponding to each point.,"},{"Start":"03:19.200 ","End":"03:24.770","Text":"we can take a segment that joins the origin to that point."},{"Start":"03:24.770 ","End":"03:28.985","Text":"Like so and with the little arrow on the end."},{"Start":"03:28.985 ","End":"03:35.770","Text":"This is the line segment with an arrow that takes me from O to P, always the origin."},{"Start":"03:35.770 ","End":"03:38.735","Text":"Associated with this point,"},{"Start":"03:38.735 ","End":"03:41.765","Text":"we have this arrow and this is the vector,"},{"Start":"03:41.765 ","End":"03:44.360","Text":"and we call this the vector OP,"},{"Start":"03:44.360 ","End":"03:48.230","Text":"and we write a little line with an arrow on top."},{"Start":"03:48.230 ","End":"03:51.530","Text":"OP is denoted with"},{"Start":"03:51.530 ","End":"03:59.070","Text":"these brackets, \u003c2,3,5\u003e."},{"Start":"03:59.070 ","End":"04:02.780","Text":"Points are written with round brackets and vectors,"},{"Start":"04:02.780 ","End":"04:06.170","Text":"which are arrows, are written with angular brackets."},{"Start":"04:06.170 ","End":"04:12.770","Text":"Although, I must say that some books use round brackets here also."},{"Start":"04:12.770 ","End":"04:15.085","Text":"It\u0027s not standard,"},{"Start":"04:15.085 ","End":"04:20.415","Text":"in this course we\u0027ll use angular brackets not round brackets."},{"Start":"04:20.415 ","End":"04:24.410","Text":"Sometimes we just want to give a single letter name,"},{"Start":"04:24.410 ","End":"04:27.040","Text":"a variable name to a vector."},{"Start":"04:27.040 ","End":"04:30.724","Text":"Typical letters would be u,"},{"Start":"04:30.724 ","End":"04:34.130","Text":"v, w. Just like variables are x, y,"},{"Start":"04:34.130 ","End":"04:36.260","Text":"z and constants are a, b,"},{"Start":"04:36.260 ","End":"04:44.105","Text":"c. I could say in this case that u is the vector \u003c2, 3, 5\u003e."},{"Start":"04:44.105 ","End":"04:48.390","Text":"It\u0027s customary to put a line underneath."},{"Start":"04:49.000 ","End":"04:51.785","Text":"Let\u0027s take another example,"},{"Start":"04:51.785 ","End":"04:57.350","Text":"draw the line from the origin to that point and put a little arrow at the end."},{"Start":"04:57.350 ","End":"05:01.895","Text":"This is now the corresponding vector."},{"Start":"05:01.895 ","End":"05:05.615","Text":"Another vector, we write it as (2,"},{"Start":"05:05.615 ","End":"05:10.820","Text":"3, 0) the same numbers but with a different brackets."},{"Start":"05:10.820 ","End":"05:13.495","Text":"Also this example,"},{"Start":"05:13.495 ","End":"05:17.345","Text":"here\u0027s the line and with the arrow at the end."},{"Start":"05:17.345 ","End":"05:24.485","Text":"This would be the vector\u003c0,0,5\u003e."},{"Start":"05:24.485 ","End":"05:26.555","Text":"Let\u0027s give them names,"},{"Start":"05:26.555 ","End":"05:28.340","Text":"this 1 will use the letter U,"},{"Start":"05:28.340 ","End":"05:30.965","Text":"let\u0027s call this 1 V,"},{"Start":"05:30.965 ","End":"05:36.835","Text":"and this 1 we\u0027ll call W. Here are 3 examples of vectors,"},{"Start":"05:36.835 ","End":"05:40.400","Text":"3 points and their associated vectors."},{"Start":"05:40.400 ","End":"05:44.330","Text":"It\u0027s sometimes called the position vector of the point,"},{"Start":"05:44.330 ","End":"05:48.560","Text":"because it tells us where it is relative to the origin."},{"Start":"05:48.560 ","End":"05:55.495","Text":"The 2 means, we start from the origin and we go 2 units in the x-direction."},{"Start":"05:55.495 ","End":"06:02.040","Text":"Then the 3 here means we now go 3 units in the y direction,"},{"Start":"06:02.040 ","End":"06:07.980","Text":"and 5 means we now go 5 units in the z direction."},{"Start":"06:07.980 ","End":"06:10.350","Text":"So 2 this way, 3 this way,"},{"Start":"06:10.350 ","End":"06:11.760","Text":"5 this way,"},{"Start":"06:11.760 ","End":"06:13.560","Text":"and we get to the point."},{"Start":"06:13.560 ","End":"06:16.440","Text":"That\u0027s basically what a 3D vector is,"},{"Start":"06:16.440 ","End":"06:23.985","Text":"it\u0027s a vector which joins the origin to a point in 3D space."},{"Start":"06:23.985 ","End":"06:29.700","Text":"Let me just summarize in writing some of the things I said."},{"Start":"06:30.170 ","End":"06:32.400","Text":"Again, in our example,"},{"Start":"06:32.400 ","End":"06:34.710","Text":"this point here (2,3,"},{"Start":"06:34.710 ","End":"06:41.140","Text":"5) in the 3-dimensional world can be viewed as an arrow with a tail at the origin,"},{"Start":"06:41.140 ","End":"06:47.440","Text":"and ahead at the point P. The arrow is called the vector OP,"},{"Start":"06:47.440 ","End":"06:50.590","Text":"and we write it as OP with an arrow on top,"},{"Start":"06:50.590 ","End":"06:53.800","Text":"and equals the \u003c2,3,5\u003e like this."},{"Start":"06:53.800 ","End":"06:55.975","Text":"Or we give it a latter name,"},{"Start":"06:55.975 ","End":"06:59.550","Text":"typically u with an underscore."},{"Start":"06:59.550 ","End":"07:03.310","Text":"Sometimes instead of the underscore,"},{"Start":"07:03.310 ","End":"07:07.060","Text":"we write it in boldface and sometimes just as irregular u,"},{"Start":"07:07.060 ","End":"07:09.820","Text":"you get fed up of writing the underscore all the time,"},{"Start":"07:09.820 ","End":"07:12.275","Text":"so we just call it u."},{"Start":"07:12.275 ","End":"07:20.645","Text":"Now we also showed how the arrow is the set of instructions to get from O to P,"},{"Start":"07:20.645 ","End":"07:23.515","Text":"that we go 2 in the x-direction."},{"Start":"07:23.515 ","End":"07:27.435","Text":"The 3 means that we go 3 steps in the y-direction."},{"Start":"07:27.435 ","End":"07:33.120","Text":"The 5 means that we go 5 steps in the z direction."},{"Start":"07:33.120 ","End":"07:38.210","Text":"I have 1 last diagram that might help a bit more general,"},{"Start":"07:38.210 ","End":"07:40.355","Text":"not just 2, 3, 5."},{"Start":"07:40.355 ","End":"07:45.990","Text":"In general, we have a point in 3-dimensional space with 3 coordinates,"},{"Start":"07:45.990 ","End":"07:47.715","Text":"x, y, and z,"},{"Start":"07:47.715 ","End":"07:53.285","Text":"and the arrow from here to here will be u,"},{"Start":"07:53.285 ","End":"07:55.670","Text":"usually with an underscore,"},{"Start":"07:55.670 ","End":"07:59.375","Text":"which is equal to \u003cx,"},{"Start":"07:59.375 ","End":"08:03.620","Text":"y, z\u003e with angular brackets."},{"Start":"08:03.620 ","End":"08:04.640","Text":"But like I said,"},{"Start":"08:04.640 ","End":"08:08.890","Text":"some people use round brackets for vectors also."},{"Start":"08:08.890 ","End":"08:13.530","Text":"Let\u0027s take a break now continue in the next clip."}],"ID":25767},{"Watched":false,"Name":"3D Vector Space - Definition","Duration":"12m 50s","ChapterTopicVideoID":25051,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25051.jpeg","UploadDate":"2021-06-27T05:21:26.9470000","DurationForVideoObject":"PT12M50S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"We\u0027re now continuing from the previous clip."},{"Start":"00:03.210 ","End":"00:08.385","Text":"There we talk about vectors in 3-dimensional space,"},{"Start":"00:08.385 ","End":"00:13.560","Text":"but we didn\u0027t yet talk about the concept of a vector space,"},{"Start":"00:13.560 ","End":"00:16.710","Text":"which is a definite mathematical concept."},{"Start":"00:16.710 ","End":"00:19.755","Text":"The concept of vector space is quite abstract,"},{"Start":"00:19.755 ","End":"00:22.350","Text":"and we\u0027re going to start with 1 particular case."},{"Start":"00:22.350 ","End":"00:25.435","Text":"That is going to be called R^3,"},{"Start":"00:25.435 ","End":"00:28.910","Text":"which is 3D Euclidean space."},{"Start":"00:28.910 ","End":"00:33.330","Text":"It\u0027s the set of all 3D vectors."},{"Start":"00:33.330 ","End":"00:37.020","Text":"They form what is called a vector space,"},{"Start":"00:37.020 ","End":"00:39.450","Text":"and we write it with this funny R,"},{"Start":"00:39.450 ","End":"00:41.610","Text":"this double R,"},{"Start":"00:41.610 ","End":"00:45.405","Text":"and the 3 for 3 dimensions."},{"Start":"00:45.405 ","End":"00:48.650","Text":"I\u0027m informally writing it with dot, dot, dot."},{"Start":"00:48.650 ","End":"00:52.670","Text":"There\u0027s obviously infinitely number of points in space,"},{"Start":"00:52.670 ","End":"00:59.110","Text":"but I just wrote a few as an example in dot, dot, dot informally."},{"Start":"00:59.110 ","End":"01:04.925","Text":"But more formally to say that we take all such vectors,"},{"Start":"01:04.925 ","End":"01:07.865","Text":"this is mathematical notation,"},{"Start":"01:07.865 ","End":"01:12.195","Text":"and this says we take all the triplets x, y, z."},{"Start":"01:12.195 ","End":"01:15.129","Text":"Like I said, we\u0027re going to use angular brackets,"},{"Start":"01:15.129 ","End":"01:17.280","Text":"but others use round brackets,"},{"Start":"01:17.280 ","End":"01:19.745","Text":"and we really should get used to both."},{"Start":"01:19.745 ","End":"01:25.330","Text":"This vertical line is such that x, y,"},{"Start":"01:25.330 ","End":"01:30.670","Text":"and z belong to R. This is the membership sign in set theory."},{"Start":"01:30.670 ","End":"01:34.390","Text":"In other words, any and all x, y,"},{"Start":"01:34.390 ","End":"01:39.180","Text":"z for real numbers will give us R^3."},{"Start":"01:39.180 ","End":"01:41.160","Text":"All triplets of real numbers,"},{"Start":"01:41.160 ","End":"01:43.885","Text":"if you like, and these are just some examples."},{"Start":"01:43.885 ","End":"01:47.435","Text":"Now, it\u0027s a set of vectors,"},{"Start":"01:47.435 ","End":"01:50.780","Text":"but what makes it a vector space?"},{"Start":"01:50.780 ","End":"01:53.029","Text":"Turns out there were very strict conditions."},{"Start":"01:53.029 ","End":"01:57.830","Text":"You can\u0027t just take any bunch of vectors and say that\u0027s a vector space, no."},{"Start":"01:57.830 ","End":"02:03.250","Text":"It has to satisfy some strict conditions and that\u0027s what I\u0027m going to talk about now."},{"Start":"02:03.250 ","End":"02:05.460","Text":"I\u0027m going to call them demands."},{"Start":"02:05.460 ","End":"02:09.260","Text":"There\u0027s 2 big demands and each demand is made up of sub-conditions."},{"Start":"02:09.260 ","End":"02:13.365","Text":"The first demand is that we have to define some operations."},{"Start":"02:13.365 ","End":"02:16.235","Text":"We\u0027ll call them arithmetic operations."},{"Start":"02:16.235 ","End":"02:21.080","Text":"There\u0027s 2 operations defined on this set of vectors."},{"Start":"02:21.080 ","End":"02:23.960","Text":"The first one, we\u0027ll call it addition of 2 vectors,"},{"Start":"02:23.960 ","End":"02:25.970","Text":"but we have yet to define it."},{"Start":"02:25.970 ","End":"02:32.840","Text":"The other operation will be to multiply a scalar by a vector."},{"Start":"02:32.840 ","End":"02:35.970","Text":"A scalar means just a number."},{"Start":"02:36.170 ","End":"02:40.795","Text":"The addition of 2 vectors in R^3,"},{"Start":"02:40.795 ","End":"02:42.860","Text":"let\u0027s say the first one is a, b,"},{"Start":"02:42.860 ","End":"02:44.870","Text":"c and the second one is A, B,"},{"Start":"02:44.870 ","End":"02:49.250","Text":"C. Then the sum is just component-wise."},{"Start":"02:49.250 ","End":"02:51.260","Text":"We add the first 2 components,"},{"Start":"02:51.260 ","End":"02:53.405","Text":"and that becomes the first component of the sum,"},{"Start":"02:53.405 ","End":"02:57.815","Text":"the second with the B and the third with the C. This is the definition."},{"Start":"02:57.815 ","End":"03:00.875","Text":"That\u0027s the addition of 2 vectors."},{"Start":"03:00.875 ","End":"03:02.195","Text":"Then the secondly,"},{"Start":"03:02.195 ","End":"03:05.295","Text":"we need a scalar times a vector."},{"Start":"03:05.295 ","End":"03:09.050","Text":"We\u0027ll call the scalar k and the vector will be a, b,"},{"Start":"03:09.050 ","End":"03:14.870","Text":"c. We take the scalar and just multiply it by each of the 3 components."},{"Start":"03:14.870 ","End":"03:17.120","Text":"We get ka, kb, kc."},{"Start":"03:17.120 ","End":"03:21.110","Text":"Then I\u0027ll give an example of each of the operations for the addition."},{"Start":"03:21.110 ","End":"03:22.760","Text":"Let\u0027s take 1, 2,"},{"Start":"03:22.760 ","End":"03:24.680","Text":"3 plus 4, 5, 6,"},{"Start":"03:24.680 ","End":"03:27.425","Text":"so 1 plus 4 is 5,"},{"Start":"03:27.425 ","End":"03:29.425","Text":"2 and 5 is 7,"},{"Start":"03:29.425 ","End":"03:31.710","Text":"3 and 6 is 9."},{"Start":"03:31.710 ","End":"03:33.660","Text":"As for scalar multiplication,"},{"Start":"03:33.660 ","End":"03:37.125","Text":"let\u0027s take 4 times the 1, 2, 3."},{"Start":"03:37.125 ","End":"03:39.960","Text":"We would say 4 times 1 is 4,"},{"Start":"03:39.960 ","End":"03:41.370","Text":"4 times 2 is 8,"},{"Start":"03:41.370 ","End":"03:44.010","Text":"and 4 times 3 is 12."},{"Start":"03:44.010 ","End":"03:46.335","Text":"That\u0027s how we do it."},{"Start":"03:46.335 ","End":"03:49.400","Text":"Now we come to the second demand."},{"Start":"03:49.400 ","End":"03:54.605","Text":"The second demand is going to be conditions on those 2 operations."},{"Start":"03:54.605 ","End":"03:56.540","Text":"Actually, we won\u0027t call them conditions,"},{"Start":"03:56.540 ","End":"03:58.580","Text":"they\u0027re going to be called axioms."},{"Start":"03:58.580 ","End":"04:00.165","Text":"There\u0027s going to be 8 of them,"},{"Start":"04:00.165 ","End":"04:03.690","Text":"4 for the addition and 4 for the scalar multiplication."},{"Start":"04:03.690 ","End":"04:05.890","Text":"We\u0027ll start with the addition."},{"Start":"04:05.890 ","End":"04:10.550","Text":"The addition operation is what we call associative."},{"Start":"04:10.550 ","End":"04:15.480","Text":"That just means that you can bunch them in any order."},{"Start":"04:15.480 ","End":"04:17.075","Text":"If you have to add 3 of them,"},{"Start":"04:17.075 ","End":"04:19.760","Text":"doesn\u0027t matter if you add the first 2 and then add"},{"Start":"04:19.760 ","End":"04:24.650","Text":"that with the third or bunch the last 2 together."},{"Start":"04:24.650 ","End":"04:28.990","Text":"But we still maintain the order here."},{"Start":"04:28.990 ","End":"04:31.670","Text":"These are numerical example."},{"Start":"04:31.670 ","End":"04:33.050","Text":"If I have 1, 2,"},{"Start":"04:33.050 ","End":"04:34.250","Text":"3 plus 4, 5,"},{"Start":"04:34.250 ","End":"04:35.300","Text":"6 plus 7, 8,"},{"Start":"04:35.300 ","End":"04:38.540","Text":"9, I can either take the first 2,"},{"Start":"04:38.540 ","End":"04:40.160","Text":"add them together,"},{"Start":"04:40.160 ","End":"04:41.780","Text":"and then I would get,"},{"Start":"04:41.780 ","End":"04:43.365","Text":"for example, for these 2,"},{"Start":"04:43.365 ","End":"04:50.520","Text":"I\u0027d get 5, 7, 9."},{"Start":"04:50.840 ","End":"04:54.090","Text":"If I added that to 7,"},{"Start":"04:54.090 ","End":"04:55.480","Text":"8, 9,"},{"Start":"04:55.480 ","End":"04:59.930","Text":"then we would get 5 and 7 is 12,"},{"Start":"04:59.930 ","End":"05:02.810","Text":"7 and 8 is 15,"},{"Start":"05:02.810 ","End":"05:05.795","Text":"9 and 9 is 18."},{"Start":"05:05.795 ","End":"05:10.605","Text":"On the other hand, if we added the last 2 first,"},{"Start":"05:10.605 ","End":"05:17.585","Text":"then we\u0027d get 4 and 7 is 11, 13, 15."},{"Start":"05:17.585 ","End":"05:20.285","Text":"If we add that to 1, 2, 3,"},{"Start":"05:20.285 ","End":"05:24.600","Text":"then we also get 12,"},{"Start":"05:24.600 ","End":"05:28.725","Text":"2 and 13 is 15,"},{"Start":"05:28.725 ","End":"05:31.515","Text":"and 3 and 15 is 18."},{"Start":"05:31.515 ","End":"05:34.415","Text":"This is the same as this."},{"Start":"05:34.415 ","End":"05:40.805","Text":"The second requirement is called commutativity of the addition."},{"Start":"05:40.805 ","End":"05:44.840","Text":"I\u0027m not writing the words associativity and commutativity, you don\u0027t remember them."},{"Start":"05:44.840 ","End":"05:47.600","Text":"Commutativity says the order doesn\u0027t matter."},{"Start":"05:47.600 ","End":"05:50.730","Text":"If I add 1, 2,"},{"Start":"05:50.730 ","End":"05:52.280","Text":"3 plus 4, 5, 6,"},{"Start":"05:52.280 ","End":"05:53.300","Text":"or the other way around,"},{"Start":"05:53.300 ","End":"05:54.725","Text":"we\u0027ll get the same result."},{"Start":"05:54.725 ","End":"05:56.120","Text":"You can see that in each case,"},{"Start":"05:56.120 ","End":"05:59.620","Text":"we\u0027ll get 1 and 4 is 5."},{"Start":"05:59.620 ","End":"06:01.920","Text":"It doesn\u0027t matter which way we do it,"},{"Start":"06:01.920 ","End":"06:03.120","Text":"2 plus 5 or here,"},{"Start":"06:03.120 ","End":"06:07.380","Text":"5 plus 2, we\u0027ll get 7."},{"Start":"06:07.380 ","End":"06:09.150","Text":"The third one will be 3 plus 6,"},{"Start":"06:09.150 ","End":"06:10.800","Text":"should be the same as 6 plus 3,"},{"Start":"06:10.800 ","End":"06:12.210","Text":"which will be 9."},{"Start":"06:12.210 ","End":"06:15.020","Text":"Either way you do it, you get the same answer."},{"Start":"06:15.020 ","End":"06:20.899","Text":"The third axiom talks about a 0 vector as opposed to a 0 number."},{"Start":"06:20.899 ","End":"06:27.725","Text":"It says it has to be a 0 vector such that when you add it to any vector,"},{"Start":"06:27.725 ","End":"06:30.065","Text":"leaves that vector as is,"},{"Start":"06:30.065 ","End":"06:38.430","Text":"and you can easily guess that the 0 vector will be just 0, 0, 0."},{"Start":"06:38.510 ","End":"06:41.690","Text":"For example, we can take any vector here,"},{"Start":"06:41.690 ","End":"06:43.445","Text":"but let\u0027s say 1, 2, 3."},{"Start":"06:43.445 ","End":"06:45.290","Text":"We add it to 0, 0, 0,"},{"Start":"06:45.290 ","End":"06:47.885","Text":"you just get back your 1, 2, 3."},{"Start":"06:47.885 ","End":"06:50.270","Text":"Start with any v at the 0 vector,"},{"Start":"06:50.270 ","End":"06:51.850","Text":"this is the 0 vector,"},{"Start":"06:51.850 ","End":"06:54.165","Text":"and it just leaves it alone."},{"Start":"06:54.165 ","End":"06:58.110","Text":"The fourth, this axiom."},{"Start":"06:58.110 ","End":"07:03.060","Text":"The fourth one says that for each vector v,"},{"Start":"07:03.060 ","End":"07:06.425","Text":"there\u0027s another vector, which is the inverse vector."},{"Start":"07:06.425 ","End":"07:10.145","Text":"Sometimes it\u0027s called the additive inverse,"},{"Start":"07:10.145 ","End":"07:14.215","Text":"we call it minus v, denote it."},{"Start":"07:14.215 ","End":"07:18.850","Text":"When you add it to the original vector, you get 0."},{"Start":"07:18.850 ","End":"07:22.700","Text":"Probably you need to guess that if we have 1, 2, 3,"},{"Start":"07:22.700 ","End":"07:26.210","Text":"for example, that its inverse will be minus 1,"},{"Start":"07:26.210 ","End":"07:27.950","Text":"minus 2, minus 3."},{"Start":"07:27.950 ","End":"07:29.435","Text":"When you add them together,"},{"Start":"07:29.435 ","End":"07:31.615","Text":"you get the 0 vector."},{"Start":"07:31.615 ","End":"07:35.300","Text":"That\u0027s the inverse or additive inverse."},{"Start":"07:35.300 ","End":"07:38.540","Text":"Now we come to the axioms for the other operations,"},{"Start":"07:38.540 ","End":"07:40.250","Text":"scalar times a vector."},{"Start":"07:40.250 ","End":"07:42.665","Text":"There were 4 axioms for addition,"},{"Start":"07:42.665 ","End":"07:46.070","Text":"there\u0027s going to be 4 axioms here also."},{"Start":"07:46.070 ","End":"07:49.425","Text":"We\u0027re continuing the numbering at 5."},{"Start":"07:49.425 ","End":"07:53.130","Text":"If we have a scalar k and 2 vectors, u and v,"},{"Start":"07:53.130 ","End":"07:57.200","Text":"then we have a distributive law that if I"},{"Start":"07:57.200 ","End":"08:01.430","Text":"first add the vectors and then the scalar times that,"},{"Start":"08:01.430 ","End":"08:07.075","Text":"it\u0027s the same as multiplying the scalar by each of them and then adding."},{"Start":"08:07.075 ","End":"08:09.590","Text":"There\u0027s an example here."},{"Start":"08:09.590 ","End":"08:13.565","Text":"I\u0027m not going to go through every step, you should try it."},{"Start":"08:13.565 ","End":"08:16.970","Text":"Though add these 2 and you\u0027d get 5,"},{"Start":"08:16.970 ","End":"08:18.485","Text":"7, 9,"},{"Start":"08:18.485 ","End":"08:20.890","Text":"then multiply that by 4."},{"Start":"08:20.890 ","End":"08:25.790","Text":"Then the other way you would multiply 4 times this and get 4, 8,"},{"Start":"08:25.790 ","End":"08:28.460","Text":"12, and then 4 times this and then add it,"},{"Start":"08:28.460 ","End":"08:32.250","Text":"and you\u0027ll get the same result."},{"Start":"08:32.270 ","End":"08:34.985","Text":"There\u0027s another distributive law."},{"Start":"08:34.985 ","End":"08:39.425","Text":"This time we have 2 scalars and 1 vector."},{"Start":"08:39.425 ","End":"08:44.580","Text":"If we take the sum of the 2 scalars and then multiply by the vector or"},{"Start":"08:44.580 ","End":"08:46.610","Text":"we take the first scalar times"},{"Start":"08:46.610 ","End":"08:49.340","Text":"the vector and the other scalar times the vector and then add,"},{"Start":"08:49.340 ","End":"08:51.470","Text":"you\u0027ll get the same thing."},{"Start":"08:51.470 ","End":"08:55.240","Text":"There\u0027s a numerical example."},{"Start":"08:55.240 ","End":"08:57.735","Text":"I\u0027ll just do the left side,"},{"Start":"08:57.735 ","End":"08:59.795","Text":"4 plus 10 is 14,"},{"Start":"08:59.795 ","End":"09:01.370","Text":"14 times 1, 2,"},{"Start":"09:01.370 ","End":"09:04.880","Text":"3 would be 14 times 1,"},{"Start":"09:04.880 ","End":"09:08.630","Text":"14 times 2, 14 times 3."},{"Start":"09:08.630 ","End":"09:12.950","Text":"I\u0027ll leave you to check the right-hand side that you get 4, 8,"},{"Start":"09:12.950 ","End":"09:15.365","Text":"12 plus 10, 20,"},{"Start":"09:15.365 ","End":"09:19.260","Text":"30 anyway, easy to check."},{"Start":"09:19.260 ","End":"09:25.040","Text":"The next axiom is another kind of associative law,"},{"Start":"09:25.040 ","End":"09:28.355","Text":"that if I have 2 scalars and a vector,"},{"Start":"09:28.355 ","End":"09:29.600","Text":"and you want to multiply,"},{"Start":"09:29.600 ","End":"09:30.740","Text":"there\u0027s 2 ways of doing it."},{"Start":"09:30.740 ","End":"09:33.575","Text":"You could multiply the 2 scalars and then the vector,"},{"Start":"09:33.575 ","End":"09:39.605","Text":"or multiply the second scalar with the vector and then the first scalar with that,"},{"Start":"09:39.605 ","End":"09:41.800","Text":"and you\u0027ll get the same answer."},{"Start":"09:41.800 ","End":"09:43.845","Text":"There\u0027s a numerical example."},{"Start":"09:43.845 ","End":"09:48.470","Text":"I\u0027ll leave it to you to check that in both cases,"},{"Start":"09:48.470 ","End":"09:53.930","Text":"you will get 40, 80, 120."},{"Start":"09:53.930 ","End":"09:56.450","Text":"On the right-hand side,"},{"Start":"09:56.450 ","End":"09:58.100","Text":"you\u0027d first of all get 10, 20,"},{"Start":"09:58.100 ","End":"10:00.860","Text":"30, and then 4 times that."},{"Start":"10:00.860 ","End":"10:06.665","Text":"But here you\u0027d get 4 times 10 is 40 and multiply each of these by 40 anyway."},{"Start":"10:06.665 ","End":"10:14.060","Text":"The last axiom is that the scalar 1 is a special scalar and we"},{"Start":"10:14.060 ","End":"10:20.780","Text":"require that the scalar 1 multiplied by any vector is that vector itself."},{"Start":"10:20.780 ","End":"10:24.720","Text":"It\u0027s what you\u0027d expect from 1 to do."},{"Start":"10:24.720 ","End":"10:28.320","Text":"But not just 1 times 5 is 5,"},{"Start":"10:28.320 ","End":"10:30.725","Text":"but 1 times a vector is also a vector."},{"Start":"10:30.725 ","End":"10:34.925","Text":"If I take 1 times the vector 1, 2, 3,"},{"Start":"10:34.925 ","End":"10:36.830","Text":"then we really do get 1, 2,"},{"Start":"10:36.830 ","End":"10:38.930","Text":"3 because you get 1 times 1,"},{"Start":"10:38.930 ","End":"10:41.935","Text":"1 times 2, 1 times 3, same thing,"},{"Start":"10:41.935 ","End":"10:45.270","Text":"so 4 axioms we had for addition,"},{"Start":"10:45.270 ","End":"10:48.795","Text":"4 axioms for scalar multiplication,"},{"Start":"10:48.795 ","End":"10:54.095","Text":"and that\u0027s 8 axioms altogether that we require for our operations."},{"Start":"10:54.095 ","End":"10:56.945","Text":"Next, I\u0027m going to summarize."},{"Start":"10:56.945 ","End":"11:02.810","Text":"What we did is we took the set of all 3-dimensional vectors,"},{"Start":"11:02.810 ","End":"11:05.075","Text":"which we called R^3,"},{"Start":"11:05.075 ","End":"11:07.600","Text":"defined it as follows."},{"Start":"11:07.600 ","End":"11:13.085","Text":"Then we defined 2 operations: addition and scalar multiplication."},{"Start":"11:13.085 ","End":"11:14.825","Text":"This is the addition,"},{"Start":"11:14.825 ","End":"11:17.120","Text":"this is the scalar multiplication."},{"Start":"11:17.120 ","End":"11:18.530","Text":"A scalar times a vector,"},{"Start":"11:18.530 ","End":"11:21.350","Text":"you can put a dot here or not."},{"Start":"11:21.350 ","End":"11:27.185","Text":"After that, we showed that these operations satisfied axioms."},{"Start":"11:27.185 ","End":"11:29.540","Text":"We had 8 axioms in all."},{"Start":"11:29.540 ","End":"11:36.540","Text":"There were 4 for addition and 4 for the multiplication."},{"Start":"11:36.540 ","End":"11:43.220","Text":"Here again are the axioms for the addition operation on vectors."},{"Start":"11:43.220 ","End":"11:47.135","Text":"This kind of associative law, commutativity law,"},{"Start":"11:47.135 ","End":"11:50.530","Text":"the existence of a 0 vector,"},{"Start":"11:50.530 ","End":"11:55.535","Text":"and an inverse vector or additive inverse."},{"Start":"11:55.535 ","End":"12:01.150","Text":"We showed that these 4 were satisfied for R^3."},{"Start":"12:01.150 ","End":"12:05.210","Text":"After that, we showed that the 4 axioms for"},{"Start":"12:05.210 ","End":"12:09.290","Text":"scalar multiplication were also satisfied here."},{"Start":"12:09.290 ","End":"12:11.565","Text":"They are 5, 6, 7, and 8,"},{"Start":"12:11.565 ","End":"12:14.540","Text":"2 kinds of distributive law,"},{"Start":"12:14.540 ","End":"12:16.685","Text":"a kind of associative law,"},{"Start":"12:16.685 ","End":"12:20.365","Text":"and the special scalar 1."},{"Start":"12:20.365 ","End":"12:25.950","Text":"We showed that these 4 were good for R^3."},{"Start":"12:25.950 ","End":"12:27.950","Text":"After we showed all this,"},{"Start":"12:27.950 ","End":"12:34.815","Text":"we can then conclude that what we call the R^3 really is a vector space,"},{"Start":"12:34.815 ","End":"12:38.780","Text":"and this is our first example of a vector space."},{"Start":"12:38.780 ","End":"12:45.090","Text":"Set of vectors with 2 operations that satisfy certain axioms."},{"Start":"12:45.470 ","End":"12:51.000","Text":"Let\u0027s take a break here and continue in the next clip."}],"ID":25766},{"Watched":false,"Name":"The Vector Space Rn","Duration":"6m 1s","ChapterTopicVideoID":25053,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25053.jpeg","UploadDate":"2021-06-27T05:25:58.9370000","DurationForVideoObject":"PT6M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.450","Text":"We\u0027re continuing from the previous clip where we did"},{"Start":"00:03.450 ","End":"00:07.935","Text":"the vector space R, 3-dimensional space."},{"Start":"00:07.935 ","End":"00:11.010","Text":"Let\u0027s move on to 4-dimensional space,"},{"Start":"00:11.010 ","End":"00:15.810","Text":"which is really pretty much the same as 3-dimensional space."},{"Start":"00:15.810 ","End":"00:21.915","Text":"We can define R^4 as the set of all vectors in 4D space."},{"Start":"00:21.915 ","End":"00:24.960","Text":"We reached the end of the alphabet with x, y, z,"},{"Start":"00:24.960 ","End":"00:31.275","Text":"so we\u0027ll call the 4th component t. You see we have 4 components instead of 3."},{"Start":"00:31.275 ","End":"00:35.160","Text":"You don\u0027t have to imagine 4-dimensional space."},{"Start":"00:35.160 ","End":"00:38.400","Text":"The real world, we live in a 3-dimensional space."},{"Start":"00:38.400 ","End":"00:44.085","Text":"But in mathematics you can have 4 dimensions or any number,"},{"Start":"00:44.085 ","End":"00:47.530","Text":"we just treat it abstractly."},{"Start":"00:47.600 ","End":"00:51.725","Text":"Here are some examples of 4D vectors,"},{"Start":"00:51.725 ","End":"00:54.835","Text":"1 minus 4, 2, 0,"},{"Start":"00:54.835 ","End":"00:57.405","Text":"just anything, any 4 real numbers,"},{"Start":"00:57.405 ","End":"00:59.700","Text":"e minus 4.1,"},{"Start":"00:59.700 ","End":"01:02.800","Text":"3, and 5 and so on."},{"Start":"01:03.740 ","End":"01:12.240","Text":"An infinite list in there somewhere is also 0,0,0,0,0 and so on."},{"Start":"01:12.240 ","End":"01:14.085","Text":"This is just for illustration."},{"Start":"01:14.085 ","End":"01:18.270","Text":"Now we have to define operations on a vector space."},{"Start":"01:18.270 ","End":"01:23.145","Text":"We need addition of vectors and multiplication of scalar by a vector."},{"Start":"01:23.145 ","End":"01:27.750","Text":"Here they are just as you\u0027d expect if you looked at the 3D case."},{"Start":"01:27.750 ","End":"01:31.080","Text":"It\u0027s just that we have slightly longer vectors,"},{"Start":"01:31.080 ","End":"01:32.925","Text":"4 instead of 3."},{"Start":"01:32.925 ","End":"01:35.640","Text":"We add component-wise,"},{"Start":"01:35.640 ","End":"01:41.925","Text":"the first component we add the second component correspondingly, and so on."},{"Start":"01:41.925 ","End":"01:44.550","Text":"Similarly with multiplication by a scalar,"},{"Start":"01:44.550 ","End":"01:49.215","Text":"we multiply it with each of the 4 components."},{"Start":"01:49.215 ","End":"01:53.610","Text":"Analogously almost the same as we showed that"},{"Start":"01:53.610 ","End":"01:58.530","Text":"the 8 axioms are satisfied in the case of R^3."},{"Start":"01:58.530 ","End":"02:01.890","Text":"They will also be satisfied in the case of R^4,"},{"Start":"02:01.890 ","End":"02:03.720","Text":"you can go and check."},{"Start":"02:03.720 ","End":"02:09.700","Text":"It would just be tedious since it\u0027s almost exactly the same as R^3."},{"Start":"02:09.770 ","End":"02:13.950","Text":"While we\u0027re at it, why not go down a dimension?"},{"Start":"02:13.950 ","End":"02:18.775","Text":"Let\u0027s look at the vector space R^2."},{"Start":"02:18.775 ","End":"02:22.710","Text":"This would be the 2-dimensional space that you\u0027re used to,"},{"Start":"02:22.710 ","End":"02:24.750","Text":"the flat plane,"},{"Start":"02:24.750 ","End":"02:28.605","Text":"Euclidean plane, or Cartesian plane."},{"Start":"02:28.605 ","End":"02:32.840","Text":"This is going to be again, very similar to what we did with R^3 and R^4."},{"Start":"02:32.840 ","End":"02:35.165","Text":"We do this briefly."},{"Start":"02:35.165 ","End":"02:38.690","Text":"First we define the vectors in R^2."},{"Start":"02:38.690 ","End":"02:43.670","Text":"They\u0027re just pairs x and y of real numbers."},{"Start":"02:43.670 ","End":"02:46.735","Text":"Here are a few examples,"},{"Start":"02:46.735 ","End":"02:50.925","Text":"just pairs of numbers 1 and minus 4,"},{"Start":"02:50.925 ","End":"02:52.875","Text":"0, and 0,"},{"Start":"02:52.875 ","End":"02:56.155","Text":"minus root 2 and 99, so on."},{"Start":"02:56.155 ","End":"03:01.445","Text":"The 2 operations, addition component wise,"},{"Start":"03:01.445 ","End":"03:03.950","Text":"scalar times a vector,"},{"Start":"03:03.950 ","End":"03:06.545","Text":"multiply by each of the components,"},{"Start":"03:06.545 ","End":"03:10.310","Text":"just analogously to 3."},{"Start":"03:10.310 ","End":"03:13.280","Text":"The 8 axioms are satisfied,"},{"Start":"03:13.280 ","End":"03:15.500","Text":"it would be too tedious to go through this again,"},{"Start":"03:15.500 ","End":"03:19.700","Text":"it\u0027s almost the same as showing it for R^3."},{"Start":"03:19.700 ","End":"03:25.045","Text":"We now have R^2 is a vector space."},{"Start":"03:25.045 ","End":"03:29.030","Text":"We now have 3 vector spaces that we\u0027re familiar with."},{"Start":"03:29.030 ","End":"03:34.340","Text":"We started out with R^3 and that was a previous clip and"},{"Start":"03:34.340 ","End":"03:41.560","Text":"then R^4 in this clip and we just this moment did R^2."},{"Start":"03:41.560 ","End":"03:43.905","Text":"We have 3, 4, and 2,"},{"Start":"03:43.905 ","End":"03:49.725","Text":"why not 5-dimensional space, 6-dimensional, 100-dimensional?"},{"Start":"03:49.725 ","End":"03:56.560","Text":"Yes, what we\u0027re going to do is generalized for any number dimensions."},{"Start":"03:56.560 ","End":"04:01.655","Text":"We\u0027ll talk generally about the vector space R^n,"},{"Start":"04:01.655 ","End":"04:07.580","Text":"where n can be any natural number, 1, 2, 3, 4,"},{"Start":"04:07.580 ","End":"04:14.310","Text":"5, and so on R^1000, R^243, whatever."},{"Start":"04:14.310 ","End":"04:18.870","Text":"We define n-dimensional space, R^n."},{"Start":"04:18.870 ","End":"04:21.630","Text":"We have to use the ellipsis,"},{"Start":"04:21.630 ","End":"04:24.230","Text":"the dot because we can\u0027t say x,"},{"Start":"04:24.230 ","End":"04:27.620","Text":"y, z, t. We don\u0027t know how many letters we need,"},{"Start":"04:27.620 ","End":"04:30.080","Text":"and there could be more than 26 anyway."},{"Start":"04:30.080 ","End":"04:32.360","Text":"We say x_1,"},{"Start":"04:32.360 ","End":"04:36.020","Text":"x_2 through x_n, we use subscripts."},{"Start":"04:36.020 ","End":"04:38.780","Text":"There are n components."},{"Start":"04:38.780 ","End":"04:41.060","Text":"Each of the components is a real number."},{"Start":"04:41.060 ","End":"04:51.105","Text":"These are the vectors now I have to tell you what the operations are."},{"Start":"04:51.105 ","End":"04:53.865","Text":"As usual, 2 operations,"},{"Start":"04:53.865 ","End":"04:58.400","Text":"addition of 2 vectors and multiplication of a scalar by a vector."},{"Start":"04:58.400 ","End":"05:01.520","Text":"It\u0027s just what you\u0027d expect for the addition."},{"Start":"05:01.520 ","End":"05:03.500","Text":"We add component wise,"},{"Start":"05:03.500 ","End":"05:05.600","Text":"the first component with the first component,"},{"Start":"05:05.600 ","End":"05:09.620","Text":"the second component with the second component, and so on."},{"Start":"05:09.740 ","End":"05:13.415","Text":"Similarly for scalar multiplication,"},{"Start":"05:13.415 ","End":"05:16.355","Text":"the scalar times the vector."},{"Start":"05:16.355 ","End":"05:20.645","Text":"We just multiply that scalar with each of the components from 1"},{"Start":"05:20.645 ","End":"05:25.115","Text":"through n. Of course these,"},{"Start":"05:25.115 ","End":"05:32.150","Text":"we\u0027ll have to satisfy the 8 axioms but I\u0027m not going to check them."},{"Start":"05:32.150 ","End":"05:37.820","Text":"It works completely analogously to R^3."},{"Start":"05:37.820 ","End":"05:41.270","Text":"You can follow it for R^3 and R^4 then,"},{"Start":"05:41.270 ","End":"05:44.470","Text":"we\u0027ve got it for general R^n."},{"Start":"05:44.470 ","End":"05:48.860","Text":"We now have a vector space for any dimension."},{"Start":"05:48.860 ","End":"05:53.090","Text":"There are other vector spaces besides these,"},{"Start":"05:53.090 ","End":"05:58.490","Text":"but they\u0027re beyond the scope of this course or at least this chapter."},{"Start":"05:58.490 ","End":"06:01.770","Text":"For this clip, we\u0027re done."}],"ID":25768},{"Watched":false,"Name":"Exercise 1","Duration":"5m ","ChapterTopicVideoID":9894,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9894.jpeg","UploadDate":"2017-08-07T11:26:04.3030000","DurationForVideoObject":"PT5M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:04.950","Text":"In this exercise, we start with the vector space R^3,"},{"Start":"00:04.950 ","End":"00:08.025","Text":"and then we consider the subset W,"},{"Start":"00:08.025 ","End":"00:11.730","Text":"which is all the triplets a,"},{"Start":"00:11.730 ","End":"00:15.495","Text":"b, c, such that a plus b plus c is 0."},{"Start":"00:15.495 ","End":"00:19.290","Text":"We have to see if this subset is a subspace."},{"Start":"00:19.290 ","End":"00:22.335","Text":"Let\u0027s, first of all, get a feel for what W is."},{"Start":"00:22.335 ","End":"00:30.400","Text":"For example, if I had the vector 1, 1 minus 2,"},{"Start":"00:30.400 ","End":"00:35.909","Text":"that would belong to W because if I add the 3 components,"},{"Start":"00:35.909 ","End":"00:39.035","Text":"1 and 1, and minus 2, it is indeed 0."},{"Start":"00:39.035 ","End":"00:44.440","Text":"Another example would be say, 0, 4,"},{"Start":"00:44.440 ","End":"00:51.320","Text":"minus 4, that would also belong to W because these add up to 0."},{"Start":"00:51.320 ","End":"00:53.330","Text":"Now let\u0027s get started."},{"Start":"00:53.330 ","End":"00:56.705","Text":"There are 3 things to prove for a subspace."},{"Start":"00:56.705 ","End":"01:02.270","Text":"The first condition is that the 0 vector has to belong to W,"},{"Start":"01:02.270 ","End":"01:06.980","Text":"but the 0 vector is just the vector 0, 0, 0."},{"Start":"01:06.980 ","End":"01:09.580","Text":"Is this in W?"},{"Start":"01:09.580 ","End":"01:13.520","Text":"Well, yes, because if we look at the a plus b plus c here,"},{"Start":"01:13.520 ","End":"01:17.990","Text":"at 0 plus 0 plus 0 which is 0 as required,"},{"Start":"01:17.990 ","End":"01:21.379","Text":"so the 0 vector is in W."},{"Start":"01:21.379 ","End":"01:28.115","Text":"The second condition is sometimes called closure under scalar multiplication."},{"Start":"01:28.115 ","End":"01:33.515","Text":"What it means is that if we know that some vector u is in W,"},{"Start":"01:33.515 ","End":"01:37.790","Text":"then we need for all scalar multiples of it to be"},{"Start":"01:37.790 ","End":"01:42.965","Text":"also be in W. Let\u0027s check if this is so always."},{"Start":"01:42.965 ","End":"01:48.550","Text":"Let\u0027s say that the components of u are some a, b, c,"},{"Start":"01:48.550 ","End":"01:53.040","Text":"then ku is just this vector ka,"},{"Start":"01:53.040 ","End":"01:56.780","Text":"kb, kc because that\u0027s how we do scalar multiplication."},{"Start":"01:56.780 ","End":"01:59.110","Text":"We do it component wise."},{"Start":"01:59.110 ","End":"02:05.650","Text":"Now to say that u belongs to W is the same thing as saying that a plus b plus c is 0.,"},{"Start":"02:05.650 ","End":"02:08.660","Text":"and to say that ku is in W would be"},{"Start":"02:08.660 ","End":"02:12.785","Text":"the same thing as saying that this plus this plus this is 0."},{"Start":"02:12.785 ","End":"02:14.570","Text":"We have to show that if this is true,"},{"Start":"02:14.570 ","End":"02:16.330","Text":"then this is true."},{"Start":"02:16.330 ","End":"02:18.170","Text":"That if this is true,"},{"Start":"02:18.170 ","End":"02:20.075","Text":"then this is also true."},{"Start":"02:20.075 ","End":"02:23.690","Text":"But I think that\u0027s fairly clear and"},{"Start":"02:23.690 ","End":"02:27.590","Text":"perhaps I should have written an extra line that this is just"},{"Start":"02:27.590 ","End":"02:35.810","Text":"k times a plus b plus c. Now since this already we know is 0,"},{"Start":"02:35.810 ","End":"02:39.545","Text":"then k times 0 will indeed be 0,"},{"Start":"02:39.545 ","End":"02:46.620","Text":"and so ku will indeed be in W from this here."},{"Start":"02:46.620 ","End":"02:51.125","Text":"That\u0027s Part 2 and there are 3 parts to the proof."},{"Start":"02:51.125 ","End":"02:55.645","Text":"The last part is called closure under addition,"},{"Start":"02:55.645 ","End":"02:58.870","Text":"and what we have to prove here is that if"},{"Start":"02:58.870 ","End":"03:02.845","Text":"a vector u is in W and another vector v is in W,"},{"Start":"03:02.845 ","End":"03:06.835","Text":"then the sum u plus v has to also be"},{"Start":"03:06.835 ","End":"03:11.585","Text":"in W. I put the question mark here because this is what we have to prove."},{"Start":"03:11.585 ","End":"03:15.580","Text":"Let\u0027s describe u and v. U would be some a, b,"},{"Start":"03:15.580 ","End":"03:19.690","Text":"c, and v we\u0027ll use A, B,"},{"Start":"03:19.690 ","End":"03:28.570","Text":"and C. U plus v that we need here is just obtained by adding component-wise,"},{"Start":"03:28.570 ","End":"03:30.860","Text":"so this is what it is."},{"Start":"03:30.860 ","End":"03:37.900","Text":"That we can interpret or re-interpret these statements here."},{"Start":"03:37.900 ","End":"03:46.340","Text":"To say that u and v are in W as 2 of them means that a plus b plus c is 0."},{"Start":"03:46.340 ","End":"03:47.615","Text":"That\u0027s for u."},{"Start":"03:47.615 ","End":"03:50.960","Text":"For v, we need this plus this plus this to be 0."},{"Start":"03:50.960 ","End":"03:53.825","Text":"To say that u plus v is in W,"},{"Start":"03:53.825 ","End":"03:57.740","Text":"is to say that this plus this plus this is 0 like this."},{"Start":"03:57.740 ","End":"04:02.060","Text":"We have to show this arrow that if this line is true,"},{"Start":"04:02.060 ","End":"04:04.105","Text":"then so is this line."},{"Start":"04:04.105 ","End":"04:07.940","Text":"It seems fairly clear to me that if a plus b plus c is 0,"},{"Start":"04:07.940 ","End":"04:11.405","Text":"and A plus B plus C is 0, and so is this."},{"Start":"04:11.405 ","End":"04:17.554","Text":"But perhaps I should have put in an extra step that this is equal"},{"Start":"04:17.554 ","End":"04:25.735","Text":"to a plus b plus c plus A plus B plus C,"},{"Start":"04:25.735 ","End":"04:30.655","Text":"because I can open the brackets and rearrange the order and it won\u0027t matter,"},{"Start":"04:30.655 ","End":"04:33.055","Text":"and this part we know is 0,"},{"Start":"04:33.055 ","End":"04:34.975","Text":"and this part is 0,"},{"Start":"04:34.975 ","End":"04:36.460","Text":"so this is 0."},{"Start":"04:36.460 ","End":"04:38.405","Text":"But this is the same as this."},{"Start":"04:38.405 ","End":"04:40.535","Text":"But you can see this addition,"},{"Start":"04:40.535 ","End":"04:44.125","Text":"the order in opening brackets doesn\u0027t matter."},{"Start":"04:44.125 ","End":"04:48.405","Text":"This says that u plus v is in W,"},{"Start":"04:48.405 ","End":"04:51.870","Text":"and so we\u0027ve proven the third part also,"},{"Start":"04:51.870 ","End":"05:00.840","Text":"so all 3 parts and so indeed W is a subspace of R^3, and we\u0027re done."}],"ID":10095},{"Watched":false,"Name":"Exercise 2","Duration":"3m 47s","ChapterTopicVideoID":9895,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9895.jpeg","UploadDate":"2017-08-07T11:26:11.0270000","DurationForVideoObject":"PT3M47S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.370","Text":"In this exercise, we have to check if W is a subspace"},{"Start":"00:05.370 ","End":"00:11.190","Text":"of R^3 and W is just defined 1st as a subset of R^3."},{"Start":"00:11.190 ","End":"00:14.175","Text":"It\u0027s all the triples a, b, c,"},{"Start":"00:14.175 ","End":"00:20.520","Text":"such that a equals c. Let\u0027s just interpret this what it means with a couple"},{"Start":"00:20.520 ","End":"00:27.720","Text":"of examples the vector 1 for instance,"},{"Start":"00:27.720 ","End":"00:34.200","Text":"would be in W because a equals c means 1st and 3rd components are the same."},{"Start":"00:34.200 ","End":"00:37.650","Text":"Which is true, 1 equals 1 so this is okay."},{"Start":"00:37.650 ","End":"00:43.910","Text":"On the other hand if I took 1,4,4 then the 1st and the"},{"Start":"00:43.910 ","End":"00:50.450","Text":"last are not the same so this is not in W. You get the idea what W is now,"},{"Start":"00:50.450 ","End":"00:52.115","Text":"is it a subspace?"},{"Start":"00:52.115 ","End":"00:53.840","Text":"There are 3 things to check,"},{"Start":"00:53.840 ","End":"00:59.920","Text":"and the 1st of them is whether the 0 vector is in W or not."},{"Start":"00:59.920 ","End":"01:07.575","Text":"The 0 vector in case of R^3 is just the vector 0,0,0 is it in W?"},{"Start":"01:07.575 ","End":"01:11.320","Text":"Well, yes of course it is because the a equals c in"},{"Start":"01:11.320 ","End":"01:14.860","Text":"this case means that this has got to equal this so"},{"Start":"01:14.860 ","End":"01:21.765","Text":"this 0 equals this 0 and so this thing is in W. That\u0027s easy 1."},{"Start":"01:21.765 ","End":"01:23.325","Text":"Now the 2nd 1,"},{"Start":"01:23.325 ","End":"01:26.575","Text":"is called closure under scalar multiplication,"},{"Start":"01:26.575 ","End":"01:30.850","Text":"we have to show that if some vectors are in W then so"},{"Start":"01:30.850 ","End":"01:36.625","Text":"are all the scalars times that vector."},{"Start":"01:36.625 ","End":"01:41.905","Text":"Let\u0027s check if u is a, b, c,"},{"Start":"01:41.905 ","End":"01:50.270","Text":"then the scalar k times u we just multiply by each of the components is ka, kb, kc."},{"Start":"01:50.270 ","End":"01:53.705","Text":"Now let\u0027s interpret this condition the 1st part,"},{"Start":"01:53.705 ","End":"01:58.520","Text":"u belongs to W is the same thing as a equals c and to say"},{"Start":"01:58.520 ","End":"02:04.090","Text":"that ku is in W is the same as saying that ka equals kc."},{"Start":"02:04.090 ","End":"02:06.960","Text":"Does this imply this?"},{"Start":"02:06.960 ","End":"02:10.055","Text":"Well, yes, because if this is true,"},{"Start":"02:10.055 ","End":"02:15.559","Text":"I can multiply both sides of the equation or the equality by k,"},{"Start":"02:15.559 ","End":"02:18.065","Text":"and then this will also be true."},{"Start":"02:18.065 ","End":"02:26.770","Text":"Indeed ku is in W well that\u0027s 2 parts proved now we have to get on to the 3rd part."},{"Start":"02:26.770 ","End":"02:30.140","Text":"This part is called closure under addition,"},{"Start":"02:30.140 ","End":"02:35.180","Text":"that if u is in W and v is in W then also u"},{"Start":"02:35.180 ","End":"02:40.110","Text":"plus v must be in W. Let\u0027s give them some components,"},{"Start":"02:40.110 ","End":"02:42.140","Text":"let u be little a, b,"},{"Start":"02:42.140 ","End":"02:45.390","Text":"c, and v be capital A, B,"},{"Start":"02:45.390 ","End":"02:54.290","Text":"C. Then u plus v. We get from this by component-wise addition so this will be a plus A,"},{"Start":"02:54.290 ","End":"03:00.255","Text":"b plus B, c plus C and now I interpret each of these conditions."},{"Start":"03:00.255 ","End":"03:05.655","Text":"To say that u and v belong to W means, 2 things."},{"Start":"03:05.655 ","End":"03:12.045","Text":"It\u0027s really 2 things a equals c from the 1st 1 and big A equals big C."},{"Start":"03:12.045 ","End":"03:18.945","Text":"To say that u plus v is in W is to say that this equals this A that this equals this."},{"Start":"03:18.945 ","End":"03:21.950","Text":"Now does this imply this?"},{"Start":"03:21.950 ","End":"03:24.260","Text":"Well, certainly because we can add"},{"Start":"03:24.260 ","End":"03:27.875","Text":"inequalities if we have this equality and add it to this equality,"},{"Start":"03:27.875 ","End":"03:33.375","Text":"equals plus equals give equals and so this is true."},{"Start":"03:33.375 ","End":"03:38.280","Text":"Interpreting this means that u plus v is in W as required."},{"Start":"03:38.280 ","End":"03:42.120","Text":"We\u0027ve proven all 3 parts so really I should also write"},{"Start":"03:42.120 ","End":"03:47.650","Text":"the answer yes W is a subspace. We\u0027re done."}],"ID":10096},{"Watched":false,"Name":"Exercise 3","Duration":"4m 4s","ChapterTopicVideoID":9896,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9896.jpeg","UploadDate":"2017-08-07T11:26:17.7230000","DurationForVideoObject":"PT4M4S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.480","Text":"In this exercise, we have to check if W is a subspace of R^3."},{"Start":"00:06.480 ","End":"00:09.945","Text":"W is just defined as a subset of R^3,"},{"Start":"00:09.945 ","End":"00:11.730","Text":"the set of all a, b, c,"},{"Start":"00:11.730 ","End":"00:15.105","Text":"with the condition that a equals 3b."},{"Start":"00:15.105 ","End":"00:18.045","Text":"We have to check if it\u0027s a subspace."},{"Start":"00:18.045 ","End":"00:20.685","Text":"Let me just illustrate the condition."},{"Start":"00:20.685 ","End":"00:26.550","Text":"I\u0027ll give an example of something that is in W. Let\u0027s say 6, 2,"},{"Start":"00:26.550 ","End":"00:31.320","Text":"4 would be in W,"},{"Start":"00:31.320 ","End":"00:34.380","Text":"because a is 3 times b,"},{"Start":"00:34.380 ","End":"00:37.080","Text":"the first component is 3 times the second."},{"Start":"00:37.080 ","End":"00:39.240","Text":"On the other hand if I took,"},{"Start":"00:39.240 ","End":"00:41.070","Text":"5, 4,"},{"Start":"00:41.070 ","End":"00:49.265","Text":"3 that would definitely not be in W because 5 is not equal to 3 times 4."},{"Start":"00:49.265 ","End":"00:52.580","Text":"We\u0027re going to show that this is in fact a subspace."},{"Start":"00:52.580 ","End":"00:54.050","Text":"There\u0027s 3 things to prove."},{"Start":"00:54.050 ","End":"00:57.380","Text":"One, we have to show that the 0 vector is in"},{"Start":"00:57.380 ","End":"01:01.775","Text":"W. The 0 vector written generally maybe like this,"},{"Start":"01:01.775 ","End":"01:03.485","Text":"but in the case of R^3,"},{"Start":"01:03.485 ","End":"01:05.255","Text":"this is the 0 vector,"},{"Start":"01:05.255 ","End":"01:06.710","Text":"0, 0, 0."},{"Start":"01:06.710 ","End":"01:10.520","Text":"Does it belong to W? Of course it does."},{"Start":"01:10.520 ","End":"01:11.845","Text":"A equals 3b."},{"Start":"01:11.845 ","End":"01:16.120","Text":"This 0 is equal to 3 times that 0."},{"Start":"01:16.120 ","End":"01:18.810","Text":"0 belongs."},{"Start":"01:18.810 ","End":"01:23.870","Text":"The second thing we have to show called closure and the scalar multiplication,"},{"Start":"01:23.870 ","End":"01:28.100","Text":"means that if a vector is in W,"},{"Start":"01:28.100 ","End":"01:33.320","Text":"then a scalar times that vector has to always also be"},{"Start":"01:33.320 ","End":"01:38.700","Text":"in W. Let\u0027s say u is the vector a,"},{"Start":"01:38.700 ","End":"01:42.495","Text":"b, c then ku would be the vector ka,"},{"Start":"01:42.495 ","End":"01:47.790","Text":"kb, kc because we multiply scalars component-wise."},{"Start":"01:47.790 ","End":"01:51.890","Text":"I\u0027m going to re-interpret this condition."},{"Start":"01:51.890 ","End":"01:56.210","Text":"This part on the left is equivalent to saying that a is"},{"Start":"01:56.210 ","End":"02:00.740","Text":"3b and this part on the right is equivalent to saying,"},{"Start":"02:00.740 ","End":"02:04.080","Text":"that this is 3 times this meaning that ka,"},{"Start":"02:05.660 ","End":"02:13.695","Text":"it should really say 3 times kb but clearly that\u0027s the same thing."},{"Start":"02:13.695 ","End":"02:20.150","Text":"To say that this implies this or the question is the same as does this imply this?"},{"Start":"02:20.150 ","End":"02:22.325","Text":"The answer is of course yes,"},{"Start":"02:22.325 ","End":"02:27.289","Text":"because you can multiply both sides of an inequality by a constant"},{"Start":"02:27.289 ","End":"02:32.330","Text":"and you\u0027ll still get an equality which shows that ku is in W,"},{"Start":"02:32.330 ","End":"02:34.685","Text":"and this is part 2 proven."},{"Start":"02:34.685 ","End":"02:39.680","Text":"We had part 1 and now we have to proceed to part 3,"},{"Start":"02:39.680 ","End":"02:42.665","Text":"which is called closure under addition."},{"Start":"02:42.665 ","End":"02:47.420","Text":"What we have to show here is that if u and v are both in W,"},{"Start":"02:47.420 ","End":"02:53.510","Text":"then u plus v is also in W with a question mark here because we have to still show it."},{"Start":"02:53.510 ","End":"02:57.530","Text":"Let\u0027s see. Let\u0027s give names to the components."},{"Start":"02:57.530 ","End":"02:58.940","Text":"Let u be little a, b,"},{"Start":"02:58.940 ","End":"03:01.160","Text":"c and v be capital A, B,"},{"Start":"03:01.160 ","End":"03:05.165","Text":"C. We\u0027re concerned with u plus v."},{"Start":"03:05.165 ","End":"03:10.205","Text":"That is just this component wise because that\u0027s how we do vector addition."},{"Start":"03:10.205 ","End":"03:15.890","Text":"Now I can reinterpret these parts in terms of the components."},{"Start":"03:15.890 ","End":"03:19.280","Text":"This part that UV belong to W means that a is"},{"Start":"03:19.280 ","End":"03:24.850","Text":"3b and that big A is 3 times big B, capital B."},{"Start":"03:24.850 ","End":"03:30.590","Text":"This part on the right is equivalent to saying that this is 3 times this,"},{"Start":"03:30.590 ","End":"03:32.360","Text":"which is what we have here."},{"Start":"03:32.360 ","End":"03:34.640","Text":"I\u0027ll show that this implies this in 2 steps."},{"Start":"03:34.640 ","End":"03:38.860","Text":"First of all, from these 2,"},{"Start":"03:38.860 ","End":"03:46.275","Text":"I can add the 2 equalities and get a plus A is 3b plus 3B."},{"Start":"03:46.275 ","End":"03:49.114","Text":"Now I can take 3 outside the brackets."},{"Start":"03:49.114 ","End":"03:53.410","Text":"That gives us this, which is exactly what we wanted here."},{"Start":"03:53.410 ","End":"04:01.760","Text":"Because this part means that u plus v is indeed in W. We\u0027ve done the third part also,"},{"Start":"04:01.760 ","End":"04:04.980","Text":"and we\u0027re altogether done."}],"ID":10097},{"Watched":false,"Name":"Exercise 4","Duration":"1m 28s","ChapterTopicVideoID":9897,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9897.jpeg","UploadDate":"2017-08-07T11:26:22.4300000","DurationForVideoObject":"PT1M28S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"In this exercise, we have to check if W,"},{"Start":"00:04.170 ","End":"00:05.340","Text":"defined in a moment,"},{"Start":"00:05.340 ","End":"00:07.665","Text":"is a subspace of R^3."},{"Start":"00:07.665 ","End":"00:11.250","Text":"W is actually defined as the subset of R^3."},{"Start":"00:11.250 ","End":"00:12.735","Text":"It doesn\u0027t say so explicitly."},{"Start":"00:12.735 ","End":"00:14.250","Text":"It\u0027s all the a, b, c,"},{"Start":"00:14.250 ","End":"00:16.020","Text":"such that a is less than b,"},{"Start":"00:16.020 ","End":"00:18.870","Text":"less than c. Before we go into checking,"},{"Start":"00:18.870 ","End":"00:21.630","Text":"let\u0027s just see we understand what this condition means."},{"Start":"00:21.630 ","End":"00:23.685","Text":"It would mean, for example,"},{"Start":"00:23.685 ","End":"00:26.190","Text":"that the vector 1, 2,"},{"Start":"00:26.190 ","End":"00:31.200","Text":"3 in R^3 is in W because this is less than this,"},{"Start":"00:31.200 ","End":"00:32.730","Text":"is less than this."},{"Start":"00:32.730 ","End":"00:36.450","Text":"For example, would 4, 7,"},{"Start":"00:36.450 ","End":"00:40.620","Text":"10, that would also be in W because this is less than this, less than this,"},{"Start":"00:40.620 ","End":"00:46.950","Text":"but an example of something that is not in W is 1,"},{"Start":"00:46.950 ","End":"00:51.470","Text":"1, 5 because 1 is not strictly less than 1,"},{"Start":"00:51.470 ","End":"00:53.045","Text":"which we would need."},{"Start":"00:53.045 ","End":"00:55.850","Text":"Now, as usual, to be a subspace there\u0027s"},{"Start":"00:55.850 ","End":"00:59.755","Text":"3 things that we have to check and if you fail any 1 then you\u0027re out."},{"Start":"00:59.755 ","End":"01:05.670","Text":"The first thing we need to ask is is the 0 vector in W. Now,"},{"Start":"01:05.670 ","End":"01:08.310","Text":"the 0 vector of R^3 is just this 1,"},{"Start":"01:08.310 ","End":"01:10.910","Text":"0, 0, 0, is it in W?"},{"Start":"01:10.910 ","End":"01:12.725","Text":"Well, the answer is no,"},{"Start":"01:12.725 ","End":"01:14.870","Text":"because this a less than b,"},{"Start":"01:14.870 ","End":"01:17.209","Text":"less than c is not observed."},{"Start":"01:17.209 ","End":"01:19.339","Text":"0 is not less than 0,"},{"Start":"01:19.339 ","End":"01:23.615","Text":"so this is not in W. Once we\u0027ve failed the first condition,"},{"Start":"01:23.615 ","End":"01:29.100","Text":"then we immediately conclude that W is not a subspace and we\u0027re done."}],"ID":10098},{"Watched":false,"Name":"Exercise 5","Duration":"3m 24s","ChapterTopicVideoID":9898,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9898.jpeg","UploadDate":"2017-08-07T11:26:32.5770000","DurationForVideoObject":"PT3M24S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:06.195","Text":"In this exercise, we have to check if W is a subspace of 3."},{"Start":"00:06.195 ","End":"00:08.400","Text":"W is defined as follows."},{"Start":"00:08.400 ","End":"00:12.000","Text":"It\u0027s the set of all vectors with 3 components, a, b,"},{"Start":"00:12.000 ","End":"00:15.855","Text":"c, such that a equals c squared."},{"Start":"00:15.855 ","End":"00:17.430","Text":"Just to get a better feel for this,"},{"Start":"00:17.430 ","End":"00:23.585","Text":"let me give an example of something that\u0027s in W. I would say that 4,"},{"Start":"00:23.585 ","End":"00:31.680","Text":"1 minus 2 would be in W because a is c squared,"},{"Start":"00:31.680 ","End":"00:34.335","Text":"4 is the square of negative 2."},{"Start":"00:34.335 ","End":"00:36.750","Text":"That\u0027s okay. On the other hand,"},{"Start":"00:36.750 ","End":"00:39.840","Text":"if I take 4, 1, 3,"},{"Start":"00:39.840 ","End":"00:46.870","Text":"let say, then that is definitely not in W because 4 is not equal to 3 squared."},{"Start":"00:46.870 ","End":"00:50.690","Text":"Now as usual, there are 3 steps that we need to check."},{"Start":"00:50.690 ","End":"00:54.710","Text":"But if we fail, then the 1 stage then we know it\u0027s not a subspace."},{"Start":"00:54.710 ","End":"00:58.400","Text":"The first part is easy where we check if the 0 vector,"},{"Start":"00:58.400 ","End":"00:59.510","Text":"in this case 0, 0,"},{"Start":"00:59.510 ","End":"01:00.990","Text":"0 is in W,"},{"Start":"01:00.990 ","End":"01:03.545","Text":"and the answer is yes,"},{"Start":"01:03.545 ","End":"01:08.485","Text":"a equals c squared because this 0 is this 0 squared."},{"Start":"01:08.485 ","End":"01:11.880","Text":"We\u0027re okay for step 1."},{"Start":"01:11.880 ","End":"01:13.875","Text":"Now let\u0027s go to step 2."},{"Start":"01:13.875 ","End":"01:19.160","Text":"This is where we check multiplication by a scalar that if u is in the subspace,"},{"Start":"01:19.160 ","End":"01:22.715","Text":"k times u also has to be in."},{"Start":"01:22.715 ","End":"01:25.030","Text":"Does this hold?"},{"Start":"01:25.030 ","End":"01:26.940","Text":"Now you have 3 components."},{"Start":"01:26.940 ","End":"01:28.725","Text":"Let\u0027s call them a, b, c,"},{"Start":"01:28.725 ","End":"01:32.655","Text":"then ku will be just simply ka,"},{"Start":"01:32.655 ","End":"01:37.235","Text":"kb, kc because that\u0027s how we multiply by scalars."},{"Start":"01:37.235 ","End":"01:43.309","Text":"Now the condition that u is in W just says that a equals c squared."},{"Start":"01:43.309 ","End":"01:47.630","Text":"The condition for ku to be in W is that"},{"Start":"01:47.630 ","End":"01:51.710","Text":"this component is equal to the square of this component."},{"Start":"01:51.710 ","End":"01:53.570","Text":"The first is the square of the third,"},{"Start":"01:53.570 ","End":"01:56.665","Text":"which would mean that we\u0027d have to have this to be true."},{"Start":"01:56.665 ","End":"01:58.940","Text":"Now is the delicate part,"},{"Start":"01:58.940 ","End":"02:01.430","Text":"this is the same,"},{"Start":"02:01.430 ","End":"02:03.740","Text":"or this implies this,"},{"Start":"02:03.740 ","End":"02:09.634","Text":"because kc all squared is k squared c squared."},{"Start":"02:09.634 ","End":"02:11.165","Text":"On the other hand,"},{"Start":"02:11.165 ","End":"02:14.139","Text":"from this that I highlighted in blue,"},{"Start":"02:14.139 ","End":"02:19.610","Text":"we get that ka equals kc squared just by multiplying"},{"Start":"02:19.610 ","End":"02:25.175","Text":"both sides by k. Now look at the right-hand sides."},{"Start":"02:25.175 ","End":"02:27.005","Text":"They could be equal,"},{"Start":"02:27.005 ","End":"02:30.000","Text":"but they won\u0027t always be equal."},{"Start":"02:30.250 ","End":"02:37.730","Text":"In general, we would have that ku is not in W. Like I said,"},{"Start":"02:37.730 ","End":"02:40.100","Text":"it\u0027s possible for these 2 to be equal,"},{"Start":"02:40.100 ","End":"02:41.885","Text":"but they won\u0027t always be equal."},{"Start":"02:41.885 ","End":"02:46.220","Text":"Usually it\u0027s best to actually give an example, a counterexample."},{"Start":"02:46.220 ","End":"02:48.950","Text":"A counterexample really clinches it."},{"Start":"02:48.950 ","End":"02:51.980","Text":"For example, if u is 4,"},{"Start":"02:51.980 ","End":"02:55.230","Text":"1, 2, then certainly 4 is 2 squared."},{"Start":"02:55.230 ","End":"02:57.860","Text":"So u is in W, no problem."},{"Start":"02:57.860 ","End":"03:00.875","Text":"Now if I multiply this by the scalar 2,"},{"Start":"03:00.875 ","End":"03:03.865","Text":"then this becomes 8, 2, 4."},{"Start":"03:03.865 ","End":"03:08.100","Text":"You will agree with me that 8 is not equal to 4 squared,"},{"Start":"03:08.100 ","End":"03:10.665","Text":"so it doesn\u0027t satisfy the condition,"},{"Start":"03:10.665 ","End":"03:16.310","Text":"and so it\u0027s not in W. Once we failed on step 2,"},{"Start":"03:16.310 ","End":"03:18.230","Text":"there\u0027s no point in going on to step 3."},{"Start":"03:18.230 ","End":"03:25.170","Text":"We can already declare W is not a subspace of R^3. We\u0027re done."}],"ID":10099},{"Watched":false,"Name":"Exercise 6","Duration":"4m 51s","ChapterTopicVideoID":9899,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9899.jpeg","UploadDate":"2017-08-07T11:26:47.0370000","DurationForVideoObject":"PT4M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.030","Text":"In this exercise, we have the subspace of R^3,"},{"Start":"00:06.030 ","End":"00:08.550","Text":"W, which is the set of all a, b,"},{"Start":"00:08.550 ","End":"00:13.425","Text":"c which satisfy this strange condition that c minus b,"},{"Start":"00:13.425 ","End":"00:17.925","Text":"in other words, this minus this is the same as this minus this."},{"Start":"00:17.925 ","End":"00:23.880","Text":"You might immediately recognize this condition as the condition that a,"},{"Start":"00:23.880 ","End":"00:26.895","Text":"b, c form an arithmetic sequence."},{"Start":"00:26.895 ","End":"00:29.685","Text":"Just 3 members of the sequence."},{"Start":"00:29.685 ","End":"00:34.530","Text":"Because when the differences of the same between consecutive elements,"},{"Start":"00:34.530 ","End":"00:36.150","Text":"then it\u0027s an arithmetic sequence."},{"Start":"00:36.150 ","End":"00:39.825","Text":"The question is, are the arithmetic sequences in R^3,"},{"Start":"00:39.825 ","End":"00:42.105","Text":"are they a subspace?"},{"Start":"00:42.105 ","End":"00:44.730","Text":"As usual, we have 3 things to check."},{"Start":"00:44.730 ","End":"00:49.299","Text":"The first 1 that we ask is the 0 vector,"},{"Start":"00:49.299 ","End":"00:50.830","Text":"in this case, 0, 0, 0,"},{"Start":"00:50.830 ","End":"00:56.010","Text":"is it in the set W?"},{"Start":"00:56.010 ","End":"01:00.640","Text":"The answer, of course, is yes because this 0 minus this"},{"Start":"01:00.640 ","End":"01:05.680","Text":"0 is the same as this 0 minus this 0."},{"Start":"01:05.680 ","End":"01:07.240","Text":"It\u0027s a tongue twister there."},{"Start":"01:07.240 ","End":"01:11.845","Text":"Yeah, anyway, at 0 minus 0 equals 0 minus 0 and so this"},{"Start":"01:11.845 ","End":"01:17.645","Text":"0 vector is indeed in W. The next 1 is multiplication by a scalar,"},{"Start":"01:17.645 ","End":"01:24.900","Text":"i.e., if we have a vector in W and we multiply it by k, is that also in W?"},{"Start":"01:24.900 ","End":"01:29.390","Text":"You might intuitively think that the answer is yes and it is yes."},{"Start":"01:29.390 ","End":"01:35.180","Text":"If you multiply an arithmetic sequence by an enlarging or shrinking factor,"},{"Start":"01:35.180 ","End":"01:37.160","Text":"it\u0027s still going to be an arithmetic sequence,"},{"Start":"01:37.160 ","End":"01:40.080","Text":"but we have to prove it."},{"Start":"01:40.310 ","End":"01:44.295","Text":"Let\u0027s call u the vector a, b, c,"},{"Start":"01:44.295 ","End":"01:49.370","Text":"and then ku will by definition of scalar multiplication be ka,"},{"Start":"01:49.370 ","End":"01:52.290","Text":"kb, and kc."},{"Start":"01:53.170 ","End":"01:56.300","Text":"We want to see if this implies this."},{"Start":"01:56.300 ","End":"01:59.930","Text":"Let me reinterpret both of them to say that u is in W is"},{"Start":"01:59.930 ","End":"02:04.070","Text":"equivalent to saying that c minus b is b minus a by definition."},{"Start":"02:04.070 ","End":"02:09.590","Text":"To say that ku is in W is to say that this minus this equals this minus this,"},{"Start":"02:09.590 ","End":"02:11.660","Text":"which is what I\u0027ve written here."},{"Start":"02:11.660 ","End":"02:14.270","Text":"Well, it\u0027s not hard to get from here to here."},{"Start":"02:14.270 ","End":"02:15.799","Text":"We just take this equation,"},{"Start":"02:15.799 ","End":"02:18.065","Text":"multiply it by k,"},{"Start":"02:18.065 ","End":"02:20.090","Text":"both sides, so we get this."},{"Start":"02:20.090 ","End":"02:22.070","Text":"But if you expand the brackets,"},{"Start":"02:22.070 ","End":"02:26.405","Text":"the distributive law, then this becomes kc minus kb, which is this."},{"Start":"02:26.405 ","End":"02:29.395","Text":"This is kb minus ka which is this."},{"Start":"02:29.395 ","End":"02:37.370","Text":"Which means that ku really is in W. We\u0027ve done the first part and the second part."},{"Start":"02:37.370 ","End":"02:39.800","Text":"Now we have to go to the third part,"},{"Start":"02:39.800 ","End":"02:44.840","Text":"which is the closure under addition,"},{"Start":"02:44.840 ","End":"02:46.610","Text":"it\u0027s called, in other words,"},{"Start":"02:46.610 ","End":"02:50.875","Text":"that if we have u and v both in W,"},{"Start":"02:50.875 ","End":"02:53.610","Text":"the sum also has to be in"},{"Start":"02:53.610 ","End":"02:58.860","Text":"W. Is the sum of 2 arithmetic sequences and arithmetic sequence?"},{"Start":"02:58.860 ","End":"03:01.335","Text":"Well, let\u0027s check."},{"Start":"03:01.335 ","End":"03:05.610","Text":"We\u0027re going to let u be a, b, c,"},{"Start":"03:05.610 ","End":"03:09.220","Text":"that\u0027s the name of the components and v will be capital A,"},{"Start":"03:09.220 ","End":"03:12.890","Text":"B, C, and then u plus v will be by"},{"Start":"03:12.890 ","End":"03:17.650","Text":"definition of vector addition component-wise will be this."},{"Start":"03:17.650 ","End":"03:22.040","Text":"We have to check if this arrow, this implication follows."},{"Start":"03:22.040 ","End":"03:30.660","Text":"The left-hand side, I can reinterpret it as c minus b is b minus a."},{"Start":"03:30.660 ","End":"03:31.980","Text":"That\u0027s the u part."},{"Start":"03:31.980 ","End":"03:35.720","Text":"The v part says that the same is true in capitals."},{"Start":"03:35.720 ","End":"03:38.600","Text":"We have to show that that implies this,"},{"Start":"03:38.600 ","End":"03:43.820","Text":"which is that this minus this equals this minus this,"},{"Start":"03:43.820 ","End":"03:46.145","Text":"which is what we\u0027ve written here,"},{"Start":"03:46.145 ","End":"03:48.310","Text":"and does it follow?"},{"Start":"03:48.310 ","End":"03:50.670","Text":"Answer is yes, you\u0027ll see in a moment."},{"Start":"03:50.670 ","End":"03:52.370","Text":"We just have to do some simple algebra."},{"Start":"03:52.370 ","End":"04:01.530","Text":"I wrote these 2 as a pair of equalities and we can certainly add equalities together."},{"Start":"04:02.390 ","End":"04:07.100","Text":"This plus this equals this plus this,"},{"Start":"04:07.100 ","End":"04:08.660","Text":"which is what I wrote here."},{"Start":"04:08.660 ","End":"04:12.455","Text":"Now we\u0027re going to just do a bit of rearranging of the order."},{"Start":"04:12.455 ","End":"04:15.965","Text":"Sort of rearranging this with an eye on this."},{"Start":"04:15.965 ","End":"04:20.960","Text":"We see that we need here c plus c. We take this first and then this,"},{"Start":"04:20.960 ","End":"04:22.970","Text":"and then the 2 minuses,"},{"Start":"04:22.970 ","End":"04:25.640","Text":"but I can take the minus outside the brackets."},{"Start":"04:25.640 ","End":"04:28.130","Text":"Similarly, on the right-hand side,"},{"Start":"04:28.130 ","End":"04:32.570","Text":"I can take it as this plus this and then the 2 minuses."},{"Start":"04:32.570 ","End":"04:35.470","Text":"That\u0027s exactly what\u0027s written here."},{"Start":"04:35.470 ","End":"04:37.650","Text":"This means, in other words,"},{"Start":"04:37.650 ","End":"04:43.275","Text":"that u plus v is in W. That\u0027s the third check mark."},{"Start":"04:43.275 ","End":"04:51.310","Text":"You would then write yes W is a subspace of R^3 and we\u0027re done."}],"ID":10100},{"Watched":false,"Name":"Exercise 7","Duration":"5m 55s","ChapterTopicVideoID":9893,"CourseChapterTopicPlaylistID":31673,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9893.jpeg","UploadDate":"2017-08-07T11:25:55.7170000","DurationForVideoObject":"PT5M55S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:06.540","Text":"In this exercise, we have to check if W is a subspace of R^3,"},{"Start":"00:06.540 ","End":"00:08.970","Text":"and W is defined as follows."},{"Start":"00:08.970 ","End":"00:12.960","Text":"It\u0027s the set of all 3D vectors with 3 components; a, b,"},{"Start":"00:12.960 ","End":"00:21.050","Text":"c. With the condition in this case that b is equal to a times q for some q,"},{"Start":"00:21.050 ","End":"00:26.780","Text":"and c is equal to a times q squared for that q."},{"Start":"00:26.780 ","End":"00:31.715","Text":"There is another way of saying this in simpler terms."},{"Start":"00:31.715 ","End":"00:35.240","Text":"This is in fact a way of saying that the 3 numbers a,"},{"Start":"00:35.240 ","End":"00:37.820","Text":"b, c form a geometric sequence."},{"Start":"00:37.820 ","End":"00:40.940","Text":"You see b is a times q."},{"Start":"00:40.940 ","End":"00:43.810","Text":"From here to here, I multiply by q."},{"Start":"00:43.810 ","End":"00:48.855","Text":"Then if I multiply by q again here q here q."},{"Start":"00:48.855 ","End":"00:50.285","Text":"From here to here,"},{"Start":"00:50.285 ","End":"00:53.420","Text":"it\u0027s like multiplying by q squared."},{"Start":"00:53.420 ","End":"00:55.640","Text":"When we have a constant ratio,"},{"Start":"00:55.640 ","End":"00:57.570","Text":"then it\u0027s a geometric sequence."},{"Start":"00:57.570 ","End":"01:02.705","Text":"Another question is, does the geometric sequences of R^3 form a subspace?"},{"Start":"01:02.705 ","End":"01:05.300","Text":"We have to check 3 things."},{"Start":"01:05.300 ","End":"01:10.835","Text":"As you know, the first thing is to check that the 0 vector belongs to the subspace."},{"Start":"01:10.835 ","End":"01:12.240","Text":"Here I\u0027m putting a question mark,"},{"Start":"01:12.240 ","End":"01:13.880","Text":"this is the 0 vector."},{"Start":"01:13.880 ","End":"01:16.625","Text":"Does it satisfy this condition?"},{"Start":"01:16.625 ","End":"01:18.200","Text":"Well certainly, yes,"},{"Start":"01:18.200 ","End":"01:19.610","Text":"in fact, any q will do."},{"Start":"01:19.610 ","End":"01:21.470","Text":"I\u0027m just taking, for instance,"},{"Start":"01:21.470 ","End":"01:25.950","Text":"let\u0027s take q is 7 just randomly."},{"Start":"01:26.570 ","End":"01:30.535","Text":"This b, this 0,"},{"Start":"01:30.535 ","End":"01:32.590","Text":"is equal to q times this 0,"},{"Start":"01:32.590 ","End":"01:34.805","Text":"the 0 times 7 is 0."},{"Start":"01:34.805 ","End":"01:40.265","Text":"This is also equal to this times q squared because 0 times 7 squared is 0."},{"Start":"01:40.265 ","End":"01:43.010","Text":"The 0 vector is n, that\u0027s the first 1."},{"Start":"01:43.010 ","End":"01:48.495","Text":"Now let\u0027s check the second condition which relates to scalar multiplication."},{"Start":"01:48.495 ","End":"01:53.360","Text":"We have to show that if we have a vector in the subspace and we multiply it by"},{"Start":"01:53.360 ","End":"01:58.120","Text":"a scalar that remains in the subspace. Well, let\u0027s see."},{"Start":"01:58.120 ","End":"02:00.545","Text":"Let\u0027s write it in component form."},{"Start":"02:00.545 ","End":"02:02.810","Text":"U, let it be a, b,"},{"Start":"02:02.810 ","End":"02:06.990","Text":"C then ku is just ka, kb, kc."},{"Start":"02:07.730 ","End":"02:10.880","Text":"Another way of expressing that something is in"},{"Start":"02:10.880 ","End":"02:14.060","Text":"the subspace without using the q is just to"},{"Start":"02:14.060 ","End":"02:20.215","Text":"say like with the geometric sequences that this divided by this is this divided by this."},{"Start":"02:20.215 ","End":"02:22.250","Text":"We can write it this way."},{"Start":"02:22.250 ","End":"02:28.520","Text":"It\u0027s not perfect because this doesn\u0027t quite work when we have a/b as 0,"},{"Start":"02:28.520 ","End":"02:30.785","Text":"but if we have 1 of them is 0,"},{"Start":"02:30.785 ","End":"02:35.540","Text":"then it automatically is a geometric sequence anyway."},{"Start":"02:35.540 ","End":"02:38.410","Text":"I don\u0027t want to get into these technical details,"},{"Start":"02:38.410 ","End":"02:40.510","Text":"let\u0027s just say that they\u0027re not 0."},{"Start":"02:40.510 ","End":"02:44.435","Text":"Believe me, it can be straightened out if 1 of them is."},{"Start":"02:44.435 ","End":"02:48.010","Text":"We have this over this equals this over this."},{"Start":"02:48.010 ","End":"02:53.450","Text":"To say that ku is a geometric sequence or that it\u0027s in"},{"Start":"02:53.450 ","End":"02:56.090","Text":"the subspace is to say that this over"},{"Start":"02:56.090 ","End":"02:59.500","Text":"this equals this over this and this is what we have."},{"Start":"02:59.500 ","End":"03:06.080","Text":"But that\u0027s not that difficult to show because you remember how the council fraction,"},{"Start":"03:06.080 ","End":"03:07.760","Text":"this is numerator and denominator."},{"Start":"03:07.760 ","End":"03:11.060","Text":"We can cancel the k in the top and bottom here,"},{"Start":"03:11.060 ","End":"03:14.650","Text":"and we can cancel the k in the top and bottom here."},{"Start":"03:14.650 ","End":"03:20.100","Text":"This will tell us exactly that c/b is equal to"},{"Start":"03:20.100 ","End":"03:26.989","Text":"b/a which is this and so ku is in the subspace."},{"Start":"03:26.989 ","End":"03:29.660","Text":"I don\u0027t know why I left too much gap here."},{"Start":"03:29.660 ","End":"03:31.310","Text":"Anyway, we\u0027ve checked Part 2,"},{"Start":"03:31.310 ","End":"03:32.885","Text":"we\u0027ve checked Part 1,"},{"Start":"03:32.885 ","End":"03:37.070","Text":"and now we have to move on to Part 3 which"},{"Start":"03:37.070 ","End":"03:43.805","Text":"is what we call closure under addition that if u and v are in w,"},{"Start":"03:43.805 ","End":"03:46.880","Text":"then the sum must be also,"},{"Start":"03:46.880 ","End":"03:49.880","Text":"what do you think is the sum of 2 geometric sequences,"},{"Start":"03:49.880 ","End":"03:52.330","Text":"also a geometric sequence?"},{"Start":"03:52.330 ","End":"03:56.030","Text":"Let\u0027s see, this is a bit more delicate this time."},{"Start":"03:56.030 ","End":"03:58.190","Text":"Let\u0027s say that u is little a, b,"},{"Start":"03:58.190 ","End":"04:00.320","Text":"c and v is capital a, b,"},{"Start":"04:00.320 ","End":"04:08.505","Text":"c. Then we want to also compute u plus v which is this component-wise addition."},{"Start":"04:08.505 ","End":"04:11.970","Text":"I want to interpret what this means."},{"Start":"04:11.970 ","End":"04:17.130","Text":"To say that u is in w means c/b is b/a."},{"Start":"04:17.130 ","End":"04:20.270","Text":"V is in w is the same thing with capital letters."},{"Start":"04:20.270 ","End":"04:21.680","Text":"Like I said, we\u0027re not going to get into"},{"Start":"04:21.680 ","End":"04:25.040","Text":"the technical obscure cases where the 0 and the denominator,"},{"Start":"04:25.040 ","End":"04:26.180","Text":"it can be worked out."},{"Start":"04:26.180 ","End":"04:28.895","Text":"That\u0027s just not complicated."},{"Start":"04:28.895 ","End":"04:31.700","Text":"What about the right-hand side?"},{"Start":"04:31.700 ","End":"04:37.070","Text":"That would give us that this condition would have to be true."},{"Start":"04:37.070 ","End":"04:38.585","Text":"That\u0027s equivalent to this."},{"Start":"04:38.585 ","End":"04:44.195","Text":"It\u0027s almost like we\u0027re adding fractions by adding numerators and adding denominators."},{"Start":"04:44.195 ","End":"04:49.150","Text":"It\u0027s not immediately clear but this is false."},{"Start":"04:49.150 ","End":"04:55.400","Text":"The best thing is to provide an example, a counterexample."},{"Start":"04:55.400 ","End":"04:59.210","Text":"Here\u0027s an example of a geometric sequence; 2, 4, 8."},{"Start":"04:59.210 ","End":"05:02.730","Text":"Clearly, 4/2 is 8/4,"},{"Start":"05:02.730 ","End":"05:06.225","Text":"I\u0027m multiplying by 2 each time, that\u0027s my q."},{"Start":"05:06.225 ","End":"05:12.185","Text":"If I take v to be, let\u0027s say, 2, 6, 18,"},{"Start":"05:12.185 ","End":"05:14.945","Text":"then I\u0027m multiplying by 3 each time."},{"Start":"05:14.945 ","End":"05:16.100","Text":"2 times 3 is 6,"},{"Start":"05:16.100 ","End":"05:17.260","Text":"6 times 3 is 18,"},{"Start":"05:17.260 ","End":"05:19.550","Text":"that\u0027s also a geometric sequence."},{"Start":"05:19.550 ","End":"05:21.170","Text":"But if you add them,"},{"Start":"05:21.170 ","End":"05:22.410","Text":"you get 4, 10,"},{"Start":"05:22.410 ","End":"05:26.870","Text":"26 and then it\u0027s no longer an equal ratio."},{"Start":"05:26.870 ","End":"05:30.940","Text":"From 4-10, I need to multiply by what?"},{"Start":"05:30.940 ","End":"05:35.240","Text":"2.5. From here to here by 2.6,"},{"Start":"05:35.240 ","End":"05:37.615","Text":"though it\u0027s not the same."},{"Start":"05:37.615 ","End":"05:41.300","Text":"We came close, we had 2 check marks,"},{"Start":"05:41.300 ","End":"05:43.625","Text":"but on the third 1 we failed."},{"Start":"05:43.625 ","End":"05:47.900","Text":"W is not a subspace because the sum of"},{"Start":"05:47.900 ","End":"05:55.050","Text":"geometric sequences basically is not necessarily a geometric sequence. We\u0027re done."}],"ID":10101}],"Thumbnail":null,"ID":31673},{"Name":"Linear Combination, Dependence and Span","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Linear Combinations","Duration":"19m 7s","ChapterTopicVideoID":9901,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9901.jpeg","UploadDate":"2017-08-07T11:31:05.5330000","DurationForVideoObject":"PT19M7S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.350 ","End":"00:07.410","Text":"We\u0027re still in vector spaces and the nexttopic will be linear combinations and I\u0027ll"},{"Start":"00:07.410 ","End":"00:11.910","Text":"try to illustrate it by way of example,and then go to a formal definition."},{"Start":"00:13.440 ","End":"00:17.040","Text":"So we\u0027ll take our familiar vectorspace, our three, and that\u0027s"},{"Start":"00:17.040 ","End":"00:21.450","Text":"considered the following threevectors, UV, and w as written."},{"Start":"00:22.110 ","End":"00:27.540","Text":"And if you stare at them a while,you might notice a certain pattern."},{"Start":"00:27.870 ","End":"00:32.760","Text":"I, for example, notice that the middleone is the average between the left"},{"Start":"00:32.760 ","End":"00:38.900","Text":"and the right one, but in any eventa one way or another, you might reach"},{"Start":"00:38.930 ","End":"00:42.150","Text":"the following or anyway, just note."},{"Start":"00:42.290 ","End":"00:48.170","Text":"And we\u0027re going to verify that thefirst vector U is twice the vector V."},{"Start":"00:49.380 ","End":"00:54.420","Text":"Plus minus one times the vectorw I actually cooked it too."},{"Start":"00:54.420 ","End":"00:55.470","Text":"It works that way."},{"Start":"00:55.500 ","End":"00:58.980","Text":"But turn also follows from the factthat if this is the average of these"},{"Start":"00:58.980 ","End":"01:01.080","Text":"two, then twice, this is this plus this."},{"Start":"01:01.500 ","End":"01:05.120","Text":"And so twice this minus,this gives me this anyway."},{"Start":"01:05.900 ","End":"01:10.280","Text":"So notice that we have constants hereto set some constant times, one of"},{"Start":"01:10.280 ","End":"01:11.810","Text":"them, plus another constant times."},{"Start":"01:11.810 ","End":"01:13.940","Text":"The other gives me this one."},{"Start":"01:14.690 ","End":"01:19.790","Text":"And in this case, we would say that thevector you, the one on the left hand"},{"Start":"01:19.790 ","End":"01:26.120","Text":"side is the linear combination of V andw just to give you the idea, but they"},{"Start":"01:26.130 ","End":"01:28.070","Text":"will go with another couple of examples."},{"Start":"01:29.150 ","End":"01:34.190","Text":"So once again, we\u0027re staying with ourthree and we take three different vectors."},{"Start":"01:34.730 ","End":"01:39.740","Text":"And this is WV and youand stare at them a while."},{"Start":"01:39.830 ","End":"01:46.670","Text":"You might notice that V is exactlydouble, of w and if that\u0027s the"},{"Start":"01:46.670 ","End":"01:51.170","Text":"case, I can just write that V astwice w but I\u0027d prefer to look at it"},{"Start":"01:51.170 ","End":"01:53.990","Text":"differently to say that w is half of V."},{"Start":"01:55.070 ","End":"01:57.260","Text":"So here w is half a V."},{"Start":"01:57.530 ","End":"02:00.920","Text":"And just so you doesn\u0027t feelleft out, I can include you,"},{"Start":"02:00.920 ","End":"02:02.750","Text":"but I\u0027ll put a zero in front."},{"Start":"02:03.185 ","End":"02:07.505","Text":"So once again, we have that onevector is related to the other"},{"Start":"02:07.505 ","End":"02:11.345","Text":"two vectors by having a constantin front of each and then adding."},{"Start":"02:12.245 ","End":"02:16.985","Text":"And so once again, we see thatone is a linear combination of"},{"Start":"02:16.985 ","End":"02:21.665","Text":"the other two in this case, wis a linear combination of VNU."},{"Start":"02:22.055 ","End":"02:25.385","Text":"Now what I started to say beforethat, I first noticed that V"},{"Start":"02:25.385 ","End":"02:28.595","Text":"is twice w that\u0027s also good."},{"Start":"02:28.595 ","End":"02:32.015","Text":"I mean, you could also saythat V is a combination of."},{"Start":"02:32.329 ","End":"02:33.409","Text":"WMU."},{"Start":"02:33.649 ","End":"02:42.860","Text":"I mean, if I wrote that V isequal to twice, w plus zero times"},{"Start":"02:42.860 ","End":"02:49.810","Text":"you, then it would show me thatV linear combination of w and U."},{"Start":"02:50.079 ","End":"02:50.320","Text":"Okay."},{"Start":"02:50.320 ","End":"02:53.500","Text":"I didn\u0027t put these in blue, likehere, but I highlighted them."},{"Start":"02:53.860 ","End":"02:57.939","Text":"So, It doesn\u0027t always have tobe that one specifically is"},{"Start":"02:57.939 ","End":"02:59.559","Text":"the combination of the others."},{"Start":"03:00.129 ","End":"03:01.179","Text":"Sometimes it works."},{"Start":"03:01.179 ","End":"03:03.019","Text":"there\u0027s more than one possibility."},{"Start":"03:03.709 ","End":"03:04.129","Text":"Okay."},{"Start":"03:04.159 ","End":"03:07.609","Text":"Let\u0027s take a, a third example."},{"Start":"03:09.739 ","End":"03:16.429","Text":"And this time we\u0027ll take also threevectors, but this time in our four, it"},{"Start":"03:16.429 ","End":"03:21.529","Text":"means that each vector has four componentshere again, U V and w as written."},{"Start":"03:22.219 ","End":"03:29.439","Text":"and this time, even if you stare atit, awhile, pattern may not emerge, but"},{"Start":"03:29.439 ","End":"03:36.579","Text":"since I cooked it up, I\u0027ll share with youthat the following holds that I made it."},{"Start":"03:36.579 ","End":"03:41.409","Text":"So that w is twice you plus four times V."},{"Start":"03:42.044 ","End":"03:44.484","Text":"let\u0027s just check, or atleast start checking."},{"Start":"03:44.904 ","End":"03:46.664","Text":"Eh, we do this in our heads."},{"Start":"03:46.664 ","End":"03:48.644","Text":"Looks look at the firstcomponent on the right."},{"Start":"03:48.644 ","End":"03:54.434","Text":"We have twice minus one plus fourtimes one minus two, plus four is two"},{"Start":"03:54.764 ","End":"04:00.704","Text":"as to another one, twice 2.5 is fiveand then four times minus four is"},{"Start":"04:00.704 ","End":"04:03.524","Text":"minus 16, five minus 16 is minus 11."},{"Start":"04:03.854 ","End":"04:05.654","Text":"I\u0027ll leave you to check the other two."},{"Start":"04:06.134 ","End":"04:10.304","Text":"So here also we have a, anexample of a linear combination."},{"Start":"04:11.844 ","End":"04:17.364","Text":"Specifically w is a linear combinationof the vectors UNV of course you"},{"Start":"04:17.364 ","End":"04:22.524","Text":"could rearrange it maybe to makeyou a combination of w and V or"},{"Start":"04:23.034 ","End":"04:28.435","Text":"maybe V the combination of WMU it\u0027spossible, generalize still further"},{"Start":"04:29.635 ","End":"04:32.575","Text":"let\u0027s suppose that we have UV and w."},{"Start":"04:33.130 ","End":"04:39.580","Text":"Not necessarily in our three or ourfour and any spec vector space V though,"},{"Start":"04:39.590 ","End":"04:42.820","Text":"perhaps the only ones, you know, areRN, but whatever\u0027s back to spaces,"},{"Start":"04:42.820 ","End":"04:45.880","Text":"you know, UV in w or the, from there."},{"Start":"04:46.810 ","End":"04:54.340","Text":"And then we would say as follows, ifthere are scalers a and B, now these"},{"Start":"04:54.340 ","End":"04:58.330","Text":"a and B are going to be the oneslike in blue here, such that you."},{"Start":"04:58.750 ","End":"05:05.650","Text":"Is a times V plus B times wand some of them could be zero."},{"Start":"05:05.650 ","End":"05:07.450","Text":"Like we had an example would be a zero."},{"Start":"05:07.780 ","End":"05:14.080","Text":"Then we say that U is a linearcombination of V and w and we\u0027re going"},{"Start":"05:14.080 ","End":"05:18.580","Text":"to generalize this still further, butat least we\u0027ve defined what it means in"},{"Start":"05:18.580 ","End":"05:21.070","Text":"the case of three vectors, UV, and w."},{"Start":"05:22.719 ","End":"05:26.650","Text":"Now in the previous example, if youremember, I wrote you, what was it?"},{"Start":"05:26.650 ","End":"05:31.870","Text":"Twice V plus four times w and I justlike pulled the two and the four out"},{"Start":"05:31.870 ","End":"05:37.580","Text":"of a hat, but, when I cooked it up, butthat raises the question of how would you"},{"Start":"05:37.580 ","End":"05:42.590","Text":"go about finding them if I didn\u0027t tellyou two and four, how would you write a"},{"Start":"05:42.590 ","End":"05:45.979","Text":"certain vector as a linear combination of."},{"Start":"05:46.775 ","End":"05:50.764","Text":"Other veterans, I didn\u0027t say to othervectors because actually it could be"},{"Start":"05:50.764 ","End":"05:54.934","Text":"extended to more, but we\u0027ve only seencombination of two other vectors."},{"Start":"05:55.504 ","End":"05:57.484","Text":"So let me just show you."},{"Start":"05:58.414 ","End":"06:01.474","Text":"How you might go about itin that very same example."},{"Start":"06:01.474 ","End":"06:02.674","Text":"I\u0027ve just repeated that example."},{"Start":"06:02.784 ","End":"06:04.914","Text":"Well, here, this is just copied, right?"},{"Start":"06:04.974 ","End":"06:09.714","Text":"We had the UV in w and I gave youthe, a result that this was twice"},{"Start":"06:09.714 ","End":"06:14.364","Text":"system four times this, but thistime I don\u0027t, you don\u0027t know that"},{"Start":"06:14.364 ","End":"06:18.864","Text":"that\u0027s a two and a four here, andyou have to guess, not guess SRE."},{"Start":"06:19.194 ","End":"06:22.374","Text":"I mean, how to systematicallyfind the two and the four."},{"Start":"06:23.484 ","End":"06:25.974","Text":"So this might be givento you as an exercise."},{"Start":"06:26.405 ","End":"06:30.784","Text":"I didn\u0027t write out the wording of theexercise, but it would say something like"},{"Start":"06:31.085 ","End":"06:36.455","Text":"show that a w is a linear combination."},{"Start":"06:36.455 ","End":"06:46.344","Text":"You might as well learn the abbreviationElsie linear combination, eh, you and V."},{"Start":"06:46.344 ","End":"06:51.054","Text":"So that\u0027s the exercise and I\u0027m actuallygoing to show you two ways of solving it."},{"Start":"06:51.874 ","End":"06:53.854","Text":"The first method is pretty intuitive."},{"Start":"06:53.854 ","End":"06:57.334","Text":"It just, we just say, okay, wedon\u0027t know that it\u0027s two and full"},{"Start":"06:57.344 ","End":"07:03.034","Text":"here, but there are two scales here,so let\u0027s just call them X and Y."},{"Start":"07:04.354 ","End":"07:08.554","Text":"So I just copied this here and Iput X and Y and we get an equation"},{"Start":"07:08.554 ","End":"07:10.564","Text":"and we have to find X and Y."},{"Start":"07:11.074 ","End":"07:15.224","Text":"So the way we do this is firstof all we expect and the, right"},{"Start":"07:15.224 ","End":"07:17.544","Text":"on side here, do it in two steps."},{"Start":"07:17.544 ","End":"07:21.624","Text":"First of all, I multiply the scalersand then I\u0027ll do the addition."},{"Start":"07:21.655 ","End":"07:26.334","Text":"So X times this, remember wemultiply X by each one of them."},{"Start":"07:26.364 ","End":"07:30.294","Text":"So we get minus X, 2.5 X and so on."},{"Start":"07:30.294 ","End":"07:35.064","Text":"And here, everything is times Y andnow we add these component wise."},{"Start":"07:36.229 ","End":"07:40.489","Text":"I mean, this is an excellent practicefor scalar multiplication and addition"},{"Start":"07:40.489 ","End":"07:46.309","Text":"of vectors, so that see first componentand first component minus X plus Y"},{"Start":"07:46.669 ","End":"07:52.519","Text":"second component 2.5 X minus four Y andsimilarly for components three and four."},{"Start":"07:54.199 ","End":"07:57.739","Text":"Now we have equality oftwo vectors in our full."},{"Start":"07:58.114 ","End":"08:02.794","Text":"I went, two vectors are equal and eachcorresponding pair of components is equal."},{"Start":"08:03.005 ","End":"08:05.135","Text":"So we get actually four equations."},{"Start":"08:05.765 ","End":"08:09.094","Text":"But to help you see this, a bitof coloring will do the trick."},{"Start":"08:09.304 ","End":"08:11.434","Text":"See we\u0027re going to comparethe ones with the same color."},{"Start":"08:11.614 ","End":"08:17.375","Text":"First component here is the two in thiscolor, and then we\u0027ve got minus X plus Y."},{"Start":"08:17.614 ","End":"08:19.354","Text":"So color comparing."},{"Start":"08:20.194 ","End":"08:22.354","Text":"And the lost the color."},{"Start":"08:22.354 ","End":"08:23.824","Text":"Now we don\u0027t need it."},{"Start":"08:24.094 ","End":"08:24.874","Text":"We\u0027ll get that."},{"Start":"08:24.924 ","End":"08:28.284","Text":"Oh, and I\u0027m also reversing the directionsthat are saying two equals minus X,"},{"Start":"08:28.284 ","End":"08:32.634","Text":"plus Y I\u0027m saying that the right equalsthe left, it\u0027s just a more convenient."},{"Start":"08:32.904 ","End":"08:36.564","Text":"And then this 2.5 X minusfour, Y equals minus 11."},{"Start":"08:36.564 ","End":"08:39.564","Text":"That\u0027s here similarly with thethird and the fourth components."},{"Start":"08:39.894 ","End":"08:42.144","Text":"So that\u0027s straightforward now."},{"Start":"08:42.144 ","End":"08:44.684","Text":"It\u0027s, a system of linear equations."},{"Start":"08:46.124 ","End":"08:50.494","Text":"And easiest probably is to do itwith VEC, not vectors matrices."},{"Start":"08:51.064 ","End":"08:52.724","Text":"So here\u0027s the matrix for this."},{"Start":"08:52.724 ","End":"08:56.384","Text":"We get an augmented matrix, thevertical bar that separates the left"},{"Start":"08:56.384 ","End":"08:57.704","Text":"hand side from the right hand side."},{"Start":"08:58.214 ","End":"09:01.604","Text":"And we don\u0027t write the X and Yjust the coefficients, actually"},{"Start":"09:01.604 ","End":"09:03.644","Text":"four equations and two unknowns."},{"Start":"09:03.844 ","End":"09:05.764","Text":"so I usually wouldn\u0027thave a solution, but."},{"Start":"09:05.994 ","End":"09:07.224","Text":"I know that here it will."},{"Start":"09:07.824 ","End":"09:13.224","Text":"so we get minus one, one, two fromhere, then 2.5 minus four, minus 11."},{"Start":"09:13.224 ","End":"09:18.644","Text":"And so on, we want to do is bring this toechelon formats, the way we, solve things."},{"Start":"09:19.304 ","End":"09:24.734","Text":"so what I\u0027m going to do is okay,minus one, just as good as a one."},{"Start":"09:25.114 ","End":"09:28.925","Text":"I\u0027m going to add or subtractmultiples of the first row from"},{"Start":"09:28.925 ","End":"09:31.084","Text":"the second, third and fourth rows."},{"Start":"09:32.155 ","End":"09:32.905","Text":"changing my mind."},{"Start":"09:33.115 ","End":"09:35.755","Text":"I see this fractions here, liketwo and a half and the half."},{"Start":"09:35.785 ","End":"09:37.075","Text":"I\u0027d rather not have fractions."},{"Start":"09:37.255 ","End":"09:37.615","Text":"You know what?"},{"Start":"09:37.615 ","End":"09:40.555","Text":"Let\u0027s multiply second andthird row by two first."},{"Start":"09:41.575 ","End":"09:44.635","Text":"And here it is in rownotation and the result."},{"Start":"09:45.850 ","End":"09:47.050","Text":"Is here."},{"Start":"09:47.690 ","End":"09:49.700","Text":"we\u0027re going to do this a bitquicker, cause we\u0027re not here to"},{"Start":"09:49.700 ","End":"09:51.710","Text":"learn, systems and linear equations."},{"Start":"09:52.010 ","End":"09:53.580","Text":"I\u0027m going to go to the next page."},{"Start":"09:54.750 ","End":"09:55.920","Text":"Ah, here we are."},{"Start":"09:56.280 ","End":"10:01.080","Text":"And now we\u0027ll add and subtract multiplesof the first row from the second, third"},{"Start":"10:01.080 ","End":"10:07.890","Text":"and fourth, to be precise, I\u0027m goingto add five times this row to this"},{"Start":"10:07.890 ","End":"10:09.990","Text":"row and you\u0027ll check all the rest."},{"Start":"10:09.990 ","End":"10:11.490","Text":"I would just want to rush through this."},{"Start":"10:12.435 ","End":"10:16.214","Text":"We get this matrix andthat was all zeros here."},{"Start":"10:16.454 ","End":"10:21.885","Text":"And now we want to, get rid of, these,but, ah, again, I can see that we can,"},{"Start":"10:21.935 ","End":"10:24.515","Text":"multiply or divide this second row."},{"Start":"10:24.545 ","End":"10:28.115","Text":"Everything that is all by three,ah, here, everything\u0027s divisible"},{"Start":"10:28.115 ","End":"10:30.155","Text":"by seven and here by five."},{"Start":"10:30.305 ","End":"10:33.814","Text":"So let\u0027s simplify and you can check."},{"Start":"10:33.814 ","End":"10:37.295","Text":"This is, these are the raw operationswe\u0027re going to do and what we get."},{"Start":"10:38.675 ","End":"10:42.064","Text":"Is this now look at thesecond, third and fourth rows."},{"Start":"10:42.875 ","End":"10:43.865","Text":"They\u0027re the same."},{"Start":"10:43.865 ","End":"10:48.605","Text":"So the most obvious thing todo now, we didn\u0027t even bother"},{"Start":"10:48.605 ","End":"10:50.045","Text":"writing the row operations."},{"Start":"10:50.045 ","End":"10:52.985","Text":"I mean, we subtractrow two from row three."},{"Start":"10:52.985 ","End":"10:56.645","Text":"We subtract row two from rowfour that makes these two zero."},{"Start":"10:56.795 ","End":"11:00.084","Text":"So we can really throwthese, two rows out."},{"Start":"11:00.385 ","End":"11:01.765","Text":"And if we go back."},{"Start":"11:02.800 ","End":"11:05.320","Text":"To a system of linear equations."},{"Start":"11:05.470 ","End":"11:06.280","Text":"This is what it is."},{"Start":"11:06.280 ","End":"11:08.320","Text":"Minus X plus Y is two."},{"Start":"11:08.530 ","End":"11:09.640","Text":"Why is full?"},{"Start":"11:10.120 ","End":"11:15.550","Text":"This is one of those back substitutions,but it\u0027s so already, almost there."},{"Start":"11:15.760 ","End":"11:20.860","Text":"I mean, we see that Y is four alreadyand we put Y equals four in here."},{"Start":"11:21.220 ","End":"11:23.080","Text":"Then we get minus X is minus two."},{"Start":"11:23.080 ","End":"11:24.130","Text":"So X is two."},{"Start":"11:25.165 ","End":"11:26.634","Text":"So this is the solution."},{"Start":"11:26.634 ","End":"11:29.864","Text":"And if you actually look back, you\u0027llsee, see, this is what we had before."},{"Start":"11:30.045 ","End":"11:32.545","Text":"Anyway, that\u0027s just,straighten things out."},{"Start":"11:33.565 ","End":"11:37.555","Text":"This was the equation that weoriginally had to solve for X and Y."},{"Start":"11:37.825 ","End":"11:41.275","Text":"And now that we found what theyare, let\u0027s plug those values in."},{"Start":"11:41.275 ","End":"11:43.435","Text":"We\u0027ll write two here andwe\u0027ll write four here."},{"Start":"11:44.255 ","End":"11:46.055","Text":"And so this is what we get."},{"Start":"11:46.055 ","End":"11:50.615","Text":"And this shows that w is alinear combination of U and V."},{"Start":"11:50.825 ","End":"11:55.465","Text":"And specifically that the scalers,the constant, two and four,"},{"Start":"11:56.605 ","End":"11:59.785","Text":"just want to say something, it,could have been the equation."},{"Start":"11:59.785 ","End":"12:01.965","Text":"Didn\u0027t have any, solution."},{"Start":"12:02.475 ","End":"12:09.605","Text":"And maybe w isn\u0027t a, I mean, in this case,it is, but it might not have been possible"},{"Start":"12:09.605 ","End":"12:11.735","Text":"to find it as a combination of U and V."},{"Start":"12:11.975 ","End":"12:13.505","Text":"Just saying that could happen."},{"Start":"12:14.115 ","End":"12:15.875","Text":"It could also be that there are."},{"Start":"12:16.575 ","End":"12:18.765","Text":"Multiple solutions to assist them."},{"Start":"12:19.185 ","End":"12:22.905","Text":"In that case, you can just take yourpick of any, a particular solution"},{"Start":"12:23.145 ","End":"12:28.965","Text":"to show that a, this is somethingtimes you plus something times V okay."},{"Start":"12:28.965 ","End":"12:30.045","Text":"That\u0027s the first method."},{"Start":"12:30.045 ","End":"12:30.135","Text":"Now."},{"Start":"12:30.145 ","End":"12:31.365","Text":"I said, there\u0027d be two methods."},{"Start":"12:31.365 ","End":"12:33.045","Text":"So let\u0027s jump to method two."},{"Start":"12:33.225 ","End":"12:38.485","Text":"And as a refresher, here\u0027s the exerciseagain that we had, you was this V was this"},{"Start":"12:38.485 ","End":"12:44.005","Text":"and w was this, and we wanted to writew as a linear combination of U and V."},{"Start":"12:44.380 ","End":"12:48.660","Text":"Now we saw earlier that it\u0027spossible and you know, even got"},{"Start":"12:48.660 ","End":"12:51.060","Text":"the coefficients, the scalers."},{"Start":"12:51.900 ","End":"12:53.370","Text":"Now here\u0027s the second method."},{"Start":"12:54.535 ","End":"13:01.915","Text":"The technique is to write the data, theUV and w here like U is equal to this."},{"Start":"13:01.915 ","End":"13:08.665","Text":"So we put you here and the fourcomponents go here in a matrix."},{"Start":"13:09.295 ","End":"13:14.065","Text":"Similarly, the next one is V let\u0027scomponents are one minus four,"},{"Start":"13:14.245 ","End":"13:17.275","Text":"0.5 and one, and the same for a."},{"Start":"13:17.565 ","End":"13:19.665","Text":"W it\u0027s components."},{"Start":"13:19.815 ","End":"13:25.365","Text":"So we get an augmented matrix, but of adifferent kind than when we solve an SLE."},{"Start":"13:27.440 ","End":"13:29.360","Text":"the important thing is the order."},{"Start":"13:29.600 ","End":"13:32.570","Text":"The one that we want to make alinear combination of the other"},{"Start":"13:32.570 ","End":"13:34.100","Text":"two should be at the bottom."},{"Start":"13:34.370 ","End":"13:37.670","Text":"In this case, we want w asa combination of U and V."},{"Start":"13:37.670 ","End":"13:39.260","Text":"So w goes less."},{"Start":"13:39.500 ","End":"13:41.780","Text":"Other than that, the orderdoesn\u0027t really matter."},{"Start":"13:42.230 ","End":"13:48.380","Text":"And the strategy now is to do rowoperations in such a way that the."},{"Start":"13:48.780 ","End":"13:53.830","Text":"last row is all zeros, in therestricted part, to the left of the"},{"Start":"13:54.040 ","End":"13:57.290","Text":"separator, UV and w stay as letters."},{"Start":"13:57.310 ","End":"13:59.240","Text":"Well, when we\u0027ve gonethrough it, once you\u0027ll see."},{"Start":"14:00.125 ","End":"14:04.705","Text":"And all this of course is assumingthat, it is possible to write w as a"},{"Start":"14:04.705 ","End":"14:10.435","Text":"combination of U and V, but we know itis, and in practice, you won\u0027t be able to"},{"Start":"14:10.615 ","End":"14:14.125","Text":"get all zeros in the bottom if it isn\u0027t."},{"Start":"14:14.605 ","End":"14:14.995","Text":"Okay."},{"Start":"14:15.025 ","End":"14:15.745","Text":"Let\u0027s start."},{"Start":"14:16.375 ","End":"14:21.145","Text":"So going to add the firstrow to the second row."},{"Start":"14:21.490 ","End":"14:29.890","Text":"And also twice the first row to thelast row as written here and notice that"},{"Start":"14:29.890 ","End":"14:34.270","Text":"we just leave the letters as is whenwe say add the first row to the second"},{"Start":"14:34.270 ","End":"14:40.120","Text":"row I add you to V just stays V plusyou, or if I had twice you to the last"},{"Start":"14:40.750 ","End":"14:44.760","Text":"one, w we get w plus two, you, okay."},{"Start":"14:44.760 ","End":"14:48.000","Text":"So let\u0027s see, how am Igoing to get old zeros here?"},{"Start":"14:49.140 ","End":"14:53.560","Text":"I was deliberating whether I shouldmultiply, but here by two to get rid of"},{"Start":"14:53.560 ","End":"14:58.930","Text":"fractions and it decided not worth it,that\u0027s just subtract four times this"},{"Start":"14:59.020 ","End":"15:01.720","Text":"row from this row and make a zero here."},{"Start":"15:02.530 ","End":"15:04.030","Text":"This is the row operation."},{"Start":"15:04.030 ","End":"15:10.705","Text":"And if we do that, Then we getreally lucky because it turns"},{"Start":"15:10.705 ","End":"15:12.835","Text":"out that all these become zero."},{"Start":"15:12.835 ","End":"15:18.415","Text":"You see 20 minus four times five iszero negative 14, minus four times."},{"Start":"15:18.415 ","End":"15:24.775","Text":"This zeros here, what we havehere is this w plus two, you"},{"Start":"15:24.865 ","End":"15:26.935","Text":"minus four times, what was this?"},{"Start":"15:27.595 ","End":"15:28.105","Text":"So."},{"Start":"15:29.350 ","End":"15:30.310","Text":"This is very good."},{"Start":"15:30.310 ","End":"15:33.790","Text":"Now, when we get to the point wherethey\u0027re all zeros, then we just"},{"Start":"15:33.790 ","End":"15:41.500","Text":"say that this expression is zero,like, so, and w is on its own."},{"Start":"15:41.870 ","End":"15:45.440","Text":"I mean, we can separate it,put the rest on the right."},{"Start":"15:46.190 ","End":"15:47.180","Text":"And here we are."},{"Start":"15:47.390 ","End":"15:51.230","Text":"It\u0027s the same as before we hadthe two and the four earlier."},{"Start":"15:51.230 ","End":"15:53.090","Text":"So it\u0027s exactly the same answer."},{"Start":"15:53.585 ","End":"15:54.875","Text":"So, this is how it works."},{"Start":"15:55.145 ","End":"16:00.575","Text":"And once again of caution the cautionthat it may not always be possible"},{"Start":"16:00.665 ","End":"16:05.525","Text":"if w wasn\u0027t a linear combination of Uand V you wouldn\u0027t be able to get all"},{"Start":"16:05.525 ","End":"16:08.585","Text":"zeros here, but we knew that we could."},{"Start":"16:10.205 ","End":"16:10.445","Text":"Okay."},{"Start":"16:10.445 ","End":"16:12.125","Text":"That\u0027s enough with that exercise."},{"Start":"16:12.345 ","End":"16:16.355","Text":"Just want to conclude thisclip, with a definition."},{"Start":"16:17.840 ","End":"16:22.820","Text":"So I\u0027m going to give a proper definitionof linear combination up til now."},{"Start":"16:23.240 ","End":"16:26.540","Text":"We haven\u0027t really done a general case."},{"Start":"16:26.660 ","End":"16:29.690","Text":"We\u0027ve always had something like that."},{"Start":"16:29.690 ","End":"16:34.880","Text":"You was a combination like."},{"Start":"16:36.140 ","End":"16:46.160","Text":"A times V plus B times w it was always acombination of two vectors now in general."},{"Start":"16:46.545 ","End":"16:50.235","Text":"we can have a linear combinationof many vectors, so I\u0027m"},{"Start":"16:50.235 ","End":"16:51.555","Text":"going to run out of letters."},{"Start":"16:51.795 ","End":"16:55.845","Text":"So what we usually do is we\u0027regoing to define when you is a"},{"Start":"16:55.845 ","End":"17:00.045","Text":"linear combination of V one V two."},{"Start":"17:00.495 ","End":"17:05.505","Text":"And so on up to V N effect, vectorcan be a linear combination of"},{"Start":"17:05.505 ","End":"17:07.815","Text":"three or four or any number."},{"Start":"17:08.310 ","End":"17:13.050","Text":"of other vectors, actually, it could evenbe a linear combination of one vector"},{"Start":"17:13.050 ","End":"17:20.450","Text":"if you was exactly twice, that wouldalso, so any, any natural and of vectors."},{"Start":"17:20.930 ","End":"17:23.840","Text":"So here goes a vector."},{"Start":"17:23.840 ","End":"17:30.680","Text":"You is called a linear combinationof the other vectors V1 through VN."},{"Start":"17:31.160 ","End":"17:34.340","Text":"I could be just two Vone V two or V and w."},{"Start":"17:34.830 ","End":"17:39.270","Text":"But could be more if there exists."},{"Start":"17:39.270 ","End":"17:44.010","Text":"Scale is a one through AAN and this isgoing to be like our a and B in the case"},{"Start":"17:44.010 ","End":"17:50.910","Text":"with two vectors, such that, so the analogof this, when we have V1 through VPN is"},{"Start":"17:50.910 ","End":"17:56.730","Text":"that you is some constant, some scalartimes V one, another scale of times, V2."},{"Start":"17:56.970 ","End":"18:05.565","Text":"And so on up to scalar Ntimes the vector, An Vn."},{"Start":"18:06.285 ","End":"18:11.555","Text":"of course, some of these could be a zeroand we had a case, where some of them,"},{"Start":"18:11.595 ","End":"18:18.005","Text":"some of the coefficients, the scalers, Imean a zero and we just would omit them."},{"Start":"18:18.425 ","End":"18:24.635","Text":"Like if you was exactly twice V then Bwould be zero here and we\u0027d just put zero"},{"Start":"18:24.635 ","End":"18:26.525","Text":"w or you just wouldn\u0027t write it at all."},{"Start":"18:28.110 ","End":"18:30.870","Text":"We\u0027ll take a break now we\u0027llcontinue in the next clip."}],"ID":10012},{"Watched":false,"Name":"Span","Duration":"5m 52s","ChapterTopicVideoID":9902,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9902.jpeg","UploadDate":"2017-08-07T11:31:37.0000000","DurationForVideoObject":"PT5M52S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.330","Text":"In this clip, we\u0027ll learn a new concept,"},{"Start":"00:03.330 ","End":"00:07.845","Text":"the linear span of a set of vectors."},{"Start":"00:07.845 ","End":"00:10.559","Text":"I\u0027ll introduce it by way of example."},{"Start":"00:10.559 ","End":"00:13.725","Text":"Let\u0027s take the vector space R^3,"},{"Start":"00:13.725 ","End":"00:16.950","Text":"and let\u0027s take a set of 3 vectors,"},{"Start":"00:16.950 ","End":"00:18.090","Text":"doesn\u0027t have to be 3,"},{"Start":"00:18.090 ","End":"00:19.500","Text":"but in this case, u,"},{"Start":"00:19.500 ","End":"00:22.650","Text":"v, and w as follows."},{"Start":"00:22.650 ","End":"00:26.625","Text":"This set of 3 vectors will give it a name a."},{"Start":"00:26.625 ","End":"00:29.040","Text":"Using these 3 vectors,"},{"Start":"00:29.040 ","End":"00:38.770","Text":"I can build a whole lot more vectors in R^3 just by taking linear combinations of these."},{"Start":"00:38.770 ","End":"00:40.700","Text":"I\u0027ll give a few examples."},{"Start":"00:40.700 ","End":"00:42.950","Text":"Here\u0027s 1 linear combination,"},{"Start":"00:42.950 ","End":"00:49.280","Text":"1 times u plus 1 times v plus 1 times w. Computed comes out to be this."},{"Start":"00:49.280 ","End":"00:53.060","Text":"Another example, minus u plus v minus 4w,"},{"Start":"00:53.060 ","End":"00:55.805","Text":"that\u0027s a linear combination of the 3,"},{"Start":"00:55.805 ","End":"00:58.060","Text":"turns out to be this."},{"Start":"00:58.060 ","End":"01:03.410","Text":"Yet another example, I could just take twice w and if you double this,"},{"Start":"01:03.410 ","End":"01:05.105","Text":"it comes out to be this."},{"Start":"01:05.105 ","End":"01:09.590","Text":"That\u0027s also a linear combination and I could write it with the 0,"},{"Start":"01:09.590 ","End":"01:14.810","Text":"u and 0 v. You can see that u and v are also there somehow."},{"Start":"01:14.810 ","End":"01:21.165","Text":"They\u0027re endlessly infinitely many linear combinations."},{"Start":"01:21.165 ","End":"01:25.060","Text":"Now if I take the set of all possible linear combinations,"},{"Start":"01:25.060 ","End":"01:28.675","Text":"and it\u0027s an infinite set, it has a name."},{"Start":"01:28.675 ","End":"01:33.550","Text":"Well, this set is called the linear span of a."},{"Start":"01:33.550 ","End":"01:38.120","Text":"Usually the word linear is omitted when the context is clear,"},{"Start":"01:38.120 ","End":"01:41.490","Text":"and we just say the span of a."},{"Start":"01:41.490 ","End":"01:45.010","Text":"We write it sometimes like this,"},{"Start":"01:45.010 ","End":"01:49.860","Text":"but more commonly, we just use the SP to abbreviate the span."},{"Start":"01:49.860 ","End":"01:51.835","Text":"We write SP of a,"},{"Start":"01:51.835 ","End":"01:54.040","Text":"to mean the span of a."},{"Start":"01:54.040 ","End":"01:57.430","Text":"If you haven\u0027t given a name to the set of this 3 vectors,"},{"Start":"01:57.430 ","End":"02:00.325","Text":"you could just say the span of the vectors u, v,"},{"Start":"02:00.325 ","End":"02:05.435","Text":"and w. Although this is a set,"},{"Start":"02:05.435 ","End":"02:09.560","Text":"it actually turns out to be a vector subspace,"},{"Start":"02:09.560 ","End":"02:10.835","Text":"so we\u0027ll see this later."},{"Start":"02:10.835 ","End":"02:18.560","Text":"Sometimes we call it the subspace spanned by a or the subspace spanned by u,"},{"Start":"02:18.560 ","End":"02:19.910","Text":"v, and w,"},{"Start":"02:19.910 ","End":"02:22.800","Text":"many different ways to say the same thing."},{"Start":"02:22.900 ","End":"02:25.490","Text":"The span of a, it\u0027s an infinite set,"},{"Start":"02:25.490 ","End":"02:27.800","Text":"and account of course write all the members,"},{"Start":"02:27.800 ","End":"02:31.235","Text":"but I can write a whole bunch more examples."},{"Start":"02:31.235 ","End":"02:35.300","Text":"Also notice that u and v and w themselves are in the span."},{"Start":"02:35.300 ","End":"02:37.670","Text":"For example, u is 1, u plus 0,"},{"Start":"02:37.670 ","End":"02:38.930","Text":"v plus 0 w,"},{"Start":"02:38.930 ","End":"02:40.820","Text":"and here\u0027s a lot more."},{"Start":"02:40.820 ","End":"02:47.750","Text":"I also wanted you to note that the 0 vector is also in the span because it\u0027s 0 u plus 0,"},{"Start":"02:47.750 ","End":"02:48.950","Text":"v plus 0,"},{"Start":"02:48.950 ","End":"02:53.630","Text":"w. Now, I just want to write it more formally."},{"Start":"02:53.630 ","End":"02:58.430","Text":"We would say that the span of a for our particular a is the set of"},{"Start":"02:58.430 ","End":"03:03.305","Text":"all au plus bv plus cw,"},{"Start":"03:03.305 ","End":"03:05.780","Text":"where the scalars a, b,"},{"Start":"03:05.780 ","End":"03:13.405","Text":"and c belong to the set of real numbers."},{"Start":"03:13.405 ","End":"03:20.330","Text":"This set, the span of a is a subset of our vector space,"},{"Start":"03:20.330 ","End":"03:23.420","Text":"which happens to be R^3 in this case."},{"Start":"03:23.420 ","End":"03:28.415","Text":"As I said later on, we\u0027ll show that it\u0027s actually a subspace."},{"Start":"03:28.415 ","End":"03:35.570","Text":"I gave this definition for the specific set a and a specific space 3,"},{"Start":"03:35.570 ","End":"03:38.195","Text":"and it can be broadened and I\u0027ll do that in a moment."},{"Start":"03:38.195 ","End":"03:41.240","Text":"But I just want to observe something."},{"Start":"03:41.240 ","End":"03:43.360","Text":"I can make a corollary,"},{"Start":"03:43.360 ","End":"03:50.960","Text":"is that we can rephrase the concept of linear combination in terms of the span."},{"Start":"03:50.960 ","End":"03:56.360","Text":"To ask if u is a linear combination of v and w,"},{"Start":"03:56.360 ","End":"04:03.560","Text":"is the same as asking if u is in the span of v and w. The span of 2 vectors,"},{"Start":"04:03.560 ","End":"04:06.920","Text":"v and w is the set of all linear combinations of v and"},{"Start":"04:06.920 ","End":"04:10.760","Text":"w. To ask if something is in the set of"},{"Start":"04:10.760 ","End":"04:15.575","Text":"linear combinations is the same thing as saying that it is a linear combination."},{"Start":"04:15.575 ","End":"04:20.285","Text":"Now, yes, about that broadening of the definition."},{"Start":"04:20.285 ","End":"04:22.205","Text":"Let\u0027s take a set a,"},{"Start":"04:22.205 ","End":"04:24.140","Text":"not just of u v w,"},{"Start":"04:24.140 ","End":"04:30.350","Text":"but any n vectors and say U_1 through U_n."},{"Start":"04:30.350 ","End":"04:32.240","Text":"It doesn\u0027t have to be R^3,"},{"Start":"04:32.240 ","End":"04:35.490","Text":"could be another vector space."},{"Start":"04:35.740 ","End":"04:41.240","Text":"Then the span of a is still just the set of linear combinations of these."},{"Start":"04:41.240 ","End":"04:43.355","Text":"But written out more specifically,"},{"Start":"04:43.355 ","End":"04:49.490","Text":"it\u0027s the set of all a_1u_1 plus a_2u_2, and so on."},{"Start":"04:49.490 ","End":"04:58.070","Text":"Where the a\u0027s are scalars and they are in R. That\u0027s really all I wanted to say just to"},{"Start":"04:58.070 ","End":"05:01.670","Text":"introduce you to the concept of a span and we\u0027ll be"},{"Start":"05:01.670 ","End":"05:06.949","Text":"discussing this later on in future clips."},{"Start":"05:06.949 ","End":"05:10.385","Text":"I want to say a few more words to the advanced students,"},{"Start":"05:10.385 ","End":"05:14.040","Text":"other than that we are done here."},{"Start":"05:15.440 ","End":"05:19.105","Text":"When I say advanced student,"},{"Start":"05:19.105 ","End":"05:24.095","Text":"I just mean that you\u0027ve studied other fields besides the real numbers."},{"Start":"05:24.095 ","End":"05:29.285","Text":"For example, the complex numbers, rational numbers,"},{"Start":"05:29.285 ","End":"05:39.020","Text":"integers modulo p. All we do is we replace the reals by any field F,"},{"Start":"05:39.020 ","End":"05:44.260","Text":"where we take a vector space over a field F not necessarily"},{"Start":"05:44.260 ","End":"05:52.920","Text":"R. This is really almost the same as this. I\u0027m done."}],"ID":10013},{"Watched":false,"Name":"Linear dependence - pair of vectors","Duration":"3m 32s","ChapterTopicVideoID":9900,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9900.jpeg","UploadDate":"2017-08-07T11:29:29.7570000","DurationForVideoObject":"PT3M32S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.750","Text":"We\u0027re still in the chapter on vector spaces."},{"Start":"00:03.750 ","End":"00:06.345","Text":"And I\u0027m going to introduce another concept,"},{"Start":"00:06.345 ","End":"00:10.020","Text":"which is not really a new concept, just new terminology."},{"Start":"00:10.020 ","End":"00:12.240","Text":"The 1 of linear dependence."},{"Start":"00:12.240 ","End":"00:17.470","Text":"When we say that 1 vector depends on other vectors."},{"Start":"00:17.470 ","End":"00:20.780","Text":"I\u0027ll start with an example that we\u0027ve seen before."},{"Start":"00:20.780 ","End":"00:22.790","Text":"My favorite example from R3,"},{"Start":"00:22.790 ","End":"00:24.020","Text":"we take 3 vectors, u, v,"},{"Start":"00:24.020 ","End":"00:26.180","Text":"w and 1, 2, 3,"},{"Start":"00:26.180 ","End":"00:29.080","Text":"4, 5, 6, 7, 8, 9, whatever."},{"Start":"00:29.080 ","End":"00:33.950","Text":"We noted previously that the following equality holds."},{"Start":"00:33.950 ","End":"00:42.050","Text":"If we take twice vector v and minus 1 vector w and add them together,"},{"Start":"00:42.050 ","End":"00:44.330","Text":"then we get vector u."},{"Start":"00:44.330 ","End":"00:46.805","Text":"We\u0027ve seen this in a previous clip."},{"Start":"00:46.805 ","End":"00:50.510","Text":"And then we defined it that u is"},{"Start":"00:50.510 ","End":"00:56.240","Text":"a linear combination of v and w. That was the term we used, linear combination."},{"Start":"00:56.240 ","End":"01:00.690","Text":"There\u0027s another term for the same thing pretty much,"},{"Start":"01:00.690 ","End":"01:05.049","Text":"and we sometimes say that u is linearly"},{"Start":"01:05.049 ","End":"01:09.910","Text":"dependent on v and w. I\u0027m not even sure what the difference is."},{"Start":"01:09.910 ","End":"01:13.329","Text":"For historical reasons, it evolved."},{"Start":"01:13.329 ","End":"01:15.870","Text":"There were 2 terms with the same thing."},{"Start":"01:15.870 ","End":"01:22.780","Text":"Anyway, you should know that the say that u is a linear combination of v and w. It\u0027s"},{"Start":"01:22.780 ","End":"01:30.095","Text":"the same thing that u is linearly dependent on v and w. Previously,"},{"Start":"01:30.095 ","End":"01:36.625","Text":"we already tied in the concept of linear combination with the concept of a span,"},{"Start":"01:36.625 ","End":"01:43.030","Text":"of a linear span of 2 or more vectors."},{"Start":"01:43.490 ","End":"01:48.790","Text":"The 1 I\u0027m leading up to is 3 ways of saying the same thing."},{"Start":"01:48.790 ","End":"01:51.850","Text":"Let\u0027s say in general that we have 3 vectors, u, v,"},{"Start":"01:51.850 ","End":"01:53.530","Text":"w in some vector space,"},{"Start":"01:53.530 ","End":"01:58.235","Text":"doesn\u0027t have to be the ones we defined in R3 in general."},{"Start":"01:58.235 ","End":"02:01.390","Text":"Suppose there are scalars, a and b,"},{"Start":"02:01.390 ","End":"02:07.000","Text":"such that u is a times v plus v times w. Then we say"},{"Start":"02:07.000 ","End":"02:12.670","Text":"that u is linearly dependent on v and w. We had this exactly the same thing before,"},{"Start":"02:12.670 ","End":"02:16.300","Text":"except there we said that u was a linear combination of v and w,"},{"Start":"02:16.300 ","End":"02:18.085","Text":"and I\u0027m repeating myself."},{"Start":"02:18.085 ","End":"02:23.365","Text":"But these occur frequently and you\u0027ll see it in both forms."},{"Start":"02:23.365 ","End":"02:26.800","Text":"Now let\u0027s tie it into the concept of span."},{"Start":"02:26.800 ","End":"02:28.660","Text":"I\u0027m going to give you not 2,"},{"Start":"02:28.660 ","End":"02:31.440","Text":"but 3 ways of saying the same thing."},{"Start":"02:31.440 ","End":"02:39.035","Text":"U is a linear combination of v and w. I\u0027ll go to 3 next,"},{"Start":"02:39.035 ","End":"02:41.705","Text":"you is linearly dependent on v and w. Well,"},{"Start":"02:41.705 ","End":"02:43.370","Text":"this is what we just said,"},{"Start":"02:43.370 ","End":"02:44.930","Text":"1 and 3 are the same."},{"Start":"02:44.930 ","End":"02:46.595","Text":"But in a previous clip,"},{"Start":"02:46.595 ","End":"02:50.450","Text":"we also said that for you to be a linear combination of"},{"Start":"02:50.450 ","End":"02:54.920","Text":"v and w means that u belongs to the span of v and w,"},{"Start":"02:54.920 ","End":"03:00.535","Text":"because the span is the set of all linear combinations of these 2."},{"Start":"03:00.535 ","End":"03:03.800","Text":"Now we have 3 ways of saying the same thing."},{"Start":"03:03.800 ","End":"03:08.870","Text":"Of course, we could extend it to more than just u being"},{"Start":"03:08.870 ","End":"03:14.615","Text":"dependent on v and w. We could have v_1 through v_n,"},{"Start":"03:14.615 ","End":"03:17.495","Text":"but we\u0027re just keeping it simple."},{"Start":"03:17.495 ","End":"03:24.335","Text":"I\u0027m taking just dependent on 2 vectors or being a linear combination of 2 vectors."},{"Start":"03:24.335 ","End":"03:27.500","Text":"I don\u0027t know, I\u0027m repeating myself."},{"Start":"03:27.500 ","End":"03:32.550","Text":"There we are, 3 ways of saying the same thing and that\u0027s all there is to this clip."}],"ID":10014},{"Watched":false,"Name":"Exercise 1","Duration":"3m 38s","ChapterTopicVideoID":9903,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9903.jpeg","UploadDate":"2017-08-07T11:31:48.4200000","DurationForVideoObject":"PT3M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:06.000","Text":"This exercise is the 1st in a series of about 7 exercises,"},{"Start":"00:06.000 ","End":"00:10.590","Text":"and they all start out with the same given u_1,"},{"Start":"00:10.590 ","End":"00:16.305","Text":"u_2, u_3, u_4 vectors in 4-dimensional space."},{"Start":"00:16.305 ","End":"00:19.950","Text":"In this particular 1, we only use u_1 and u_4,"},{"Start":"00:19.950 ","End":"00:23.460","Text":"but u_2 and u_3 stay along for the ride."},{"Start":"00:23.460 ","End":"00:25.665","Text":"There are 3 questions."},{"Start":"00:25.665 ","End":"00:27.750","Text":"Let\u0027s take them 1 at a time."},{"Start":"00:27.750 ","End":"00:31.845","Text":"Is u_1 a linear combination of u_4?"},{"Start":"00:31.845 ","End":"00:35.355","Text":"Is this 1 a linear combination of this 1?"},{"Start":"00:35.355 ","End":"00:37.860","Text":"There\u0027s a shortcut,"},{"Start":"00:37.860 ","End":"00:38.940","Text":"and a long answer,"},{"Start":"00:38.940 ","End":"00:41.540","Text":"and I want to go for the long answer because it\u0027s more general,"},{"Start":"00:41.540 ","End":"00:43.440","Text":"and I\u0027ll mention the shortcut."},{"Start":"00:43.440 ","End":"00:47.920","Text":"The linear combination of 1 vector just means a multiple of it."},{"Start":"00:47.920 ","End":"00:51.270","Text":"Is u_1 a multiple of u_4,"},{"Start":"00:51.270 ","End":"00:53.055","Text":"in other words, a constant times this?"},{"Start":"00:53.055 ","End":"00:54.630","Text":"The answer has to be no."},{"Start":"00:54.630 ","End":"00:56.655","Text":"If this is something times this,"},{"Start":"00:56.655 ","End":"01:01.340","Text":"the 1st has to be 4 times this to get the 1st component right."},{"Start":"01:01.340 ","End":"01:04.255","Text":"But 4 times this doesn\u0027t give this."},{"Start":"01:04.255 ","End":"01:07.090","Text":"That\u0027s the shortcut answer."},{"Start":"01:07.090 ","End":"01:09.335","Text":"But to be more general,"},{"Start":"01:09.335 ","End":"01:15.485","Text":"let me check if u_1 and u_4 are linearly dependent."},{"Start":"01:15.485 ","End":"01:18.460","Text":"It\u0027s almost the same thing."},{"Start":"01:18.460 ","End":"01:21.535","Text":"To check linear dependence,"},{"Start":"01:21.535 ","End":"01:23.730","Text":"we put them in a matrix,"},{"Start":"01:23.730 ","End":"01:25.740","Text":"and do row operations,"},{"Start":"01:25.740 ","End":"01:28.460","Text":"and see if we get a row of zeros."},{"Start":"01:28.460 ","End":"01:31.820","Text":"We could use the restricted matrix,"},{"Start":"01:31.820 ","End":"01:35.500","Text":"but we\u0027re using the augmented because this will also tell"},{"Start":"01:35.500 ","End":"01:39.340","Text":"us if they are linearly dependent."},{"Start":"01:39.340 ","End":"01:45.220","Text":"It actually gives us a formula for how to get 0 as a linear combination of u_1 and u_4."},{"Start":"01:45.220 ","End":"01:48.150","Text":"But we\u0027re going to do everything the long way."},{"Start":"01:48.150 ","End":"01:52.900","Text":"What we can do here now is to take"},{"Start":"01:52.900 ","End":"01:59.230","Text":"4 times the last row,"},{"Start":"01:59.230 ","End":"02:01.865","Text":"and subtract from it the 1st row,"},{"Start":"02:01.865 ","End":"02:03.995","Text":"but to put it in the last row."},{"Start":"02:03.995 ","End":"02:05.990","Text":"Let\u0027s do the right side 1st,"},{"Start":"02:05.990 ","End":"02:07.340","Text":"so you can see what I mean."},{"Start":"02:07.340 ","End":"02:09.640","Text":"I\u0027m taking 4 times this,"},{"Start":"02:09.640 ","End":"02:12.165","Text":"minus this, and putting it here."},{"Start":"02:12.165 ","End":"02:14.795","Text":"4 times 1 minus 4 is 0,"},{"Start":"02:14.795 ","End":"02:18.260","Text":"4 times 3 minus 1 is 11,"},{"Start":"02:18.260 ","End":"02:21.810","Text":"and so on, so that\u0027s our second step."},{"Start":"02:22.810 ","End":"02:27.605","Text":"This already is in row echelon form."},{"Start":"02:27.605 ","End":"02:30.320","Text":"We didn\u0027t get any rows of 0 \u0027s,"},{"Start":"02:30.320 ","End":"02:33.330","Text":"so it means that u_1,"},{"Start":"02:33.330 ","End":"02:37.785","Text":"u_4 are linearly independent."},{"Start":"02:37.785 ","End":"02:40.725","Text":"We\u0027ve answered 3 really."},{"Start":"02:40.725 ","End":"02:47.115","Text":"But it also answers number 1,"},{"Start":"02:47.115 ","End":"02:50.450","Text":"because if u_1 was a linear combination,"},{"Start":"02:50.450 ","End":"02:52.685","Text":"then they would be linearly dependent,"},{"Start":"02:52.685 ","End":"02:54.110","Text":"and they aren\u0027t,"},{"Start":"02:54.110 ","End":"03:00.240","Text":"so now u_1 is not a multiple of u_4."},{"Start":"03:00.240 ","End":"03:06.785","Text":"Number 2 is really just a rephrasing of the same thing with the notation span."},{"Start":"03:06.785 ","End":"03:14.780","Text":"The span is the space of all linear combinations,"},{"Start":"03:14.780 ","End":"03:18.635","Text":"and the linear combination of 1 vector is just a multiple of it."},{"Start":"03:18.635 ","End":"03:20.995","Text":"2 is the same as 1,"},{"Start":"03:20.995 ","End":"03:23.190","Text":"so that\u0027s also a no."},{"Start":"03:23.190 ","End":"03:24.980","Text":"The last question, as we mentioned,"},{"Start":"03:24.980 ","End":"03:27.110","Text":"if they were linearly dependent,"},{"Start":"03:27.110 ","End":"03:29.330","Text":"when you brought to row echelon form,"},{"Start":"03:29.330 ","End":"03:32.050","Text":"we\u0027d get a row of 0 \u0027s, which we didn\u0027t."},{"Start":"03:32.050 ","End":"03:34.875","Text":"Here, again, the answer is no,"},{"Start":"03:34.875 ","End":"03:39.700","Text":"and that concludes this 1st in the series."}],"ID":10015},{"Watched":false,"Name":"Exercise 2","Duration":"4m 17s","ChapterTopicVideoID":9904,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9904.jpeg","UploadDate":"2017-08-07T11:32:05.0030000","DurationForVideoObject":"PT4M17S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"This is a continuation,"},{"Start":"00:02.280 ","End":"00:05.040","Text":"it\u0027s part of a series of exercises with the same u_1,"},{"Start":"00:05.040 ","End":"00:06.645","Text":"u_2, u_3, u_4."},{"Start":"00:06.645 ","End":"00:08.325","Text":"This time we ask,"},{"Start":"00:08.325 ","End":"00:14.080","Text":"is u_3 a linear combination of u_1 and u_2?"},{"Start":"00:14.210 ","End":"00:17.129","Text":"Actually, peeking ahead,"},{"Start":"00:17.129 ","End":"00:19.910","Text":"I see that the first part of Question 3,"},{"Start":"00:19.910 ","End":"00:21.660","Text":"is to ask if u_1, u_2,"},{"Start":"00:21.660 ","End":"00:23.840","Text":"u_3 are linearly dependent?"},{"Start":"00:23.840 ","End":"00:27.725","Text":"I\u0027ll actually like to do that first because this"},{"Start":"00:27.725 ","End":"00:32.345","Text":"could have an influence on this and you will see that in a moment."},{"Start":"00:32.345 ","End":"00:34.200","Text":"If they\u0027re linearly dependent,"},{"Start":"00:34.200 ","End":"00:36.500","Text":"1 of them is a linear combination of the other 2,"},{"Start":"00:36.500 ","End":"00:38.020","Text":"possibly more than 1,"},{"Start":"00:38.020 ","End":"00:40.630","Text":"and we might just be lucky and get this."},{"Start":"00:40.630 ","End":"00:47.490","Text":"Let\u0027s start with the matrix form of u_1, u_2,"},{"Start":"00:47.490 ","End":"00:50.355","Text":"u_3, I\u0027ll write them as an augmented matrix,"},{"Start":"00:50.355 ","End":"00:55.520","Text":"and then we\u0027re going to start doing row operations to bring to echelon form."},{"Start":"00:55.520 ","End":"01:02.075","Text":"What I\u0027m going do is double this row and then subtract the top row from it."},{"Start":"01:02.075 ","End":"01:06.035","Text":"Twice u_3 minus u_1 goes here,"},{"Start":"01:06.035 ","End":"01:09.350","Text":"twice 2 minus 4 gives me this 0,"},{"Start":"01:09.350 ","End":"01:11.555","Text":"and so 1 for the rest of them,"},{"Start":"01:11.555 ","End":"01:13.535","Text":"now we\u0027re up to this point."},{"Start":"01:13.535 ","End":"01:20.815","Text":"Now, let\u0027s add the second row to the third row."},{"Start":"01:20.815 ","End":"01:26.060","Text":"Notice that after that we have a row of zeros,"},{"Start":"01:26.060 ","End":"01:35.310","Text":"and what that means is that this bit is equal to 0, let\u0027s write that."},{"Start":"01:35.310 ","End":"01:38.300","Text":"For 1 thing, it means that you\u0027ve answered"},{"Start":"01:38.300 ","End":"01:40.730","Text":"that first part of Question 3, is that, yes,"},{"Start":"01:40.730 ","End":"01:47.705","Text":"they are linearly dependent and this actually gives us the dependency."},{"Start":"01:47.705 ","End":"01:52.430","Text":"Now, notice that all 3 of the vectors appear u_1,"},{"Start":"01:52.430 ","End":"01:54.005","Text":"u_2, and u_3,"},{"Start":"01:54.005 ","End":"01:56.175","Text":"if 1 of them wasn\u0027t present,"},{"Start":"01:56.175 ","End":"01:58.370","Text":"meaning it has a coefficient of 0,"},{"Start":"01:58.370 ","End":"02:01.670","Text":"we wouldn\u0027t be able to get it as a combination of the others."},{"Start":"02:01.670 ","End":"02:03.815","Text":"But in this case,"},{"Start":"02:03.815 ","End":"02:07.640","Text":"any one of them will be a linear combination of the other 2."},{"Start":"02:07.640 ","End":"02:09.410","Text":"What were we asked for?"},{"Start":"02:09.410 ","End":"02:15.250","Text":"We were asked, is u_3 a linear combination of u_1 and u_2?"},{"Start":"02:15.250 ","End":"02:22.080","Text":"The idea is to extract u_3,"},{"Start":"02:22.080 ","End":"02:27.820","Text":"which we can do because u_3 appears as a coefficient which is not 0."},{"Start":"02:27.820 ","End":"02:32.135","Text":"We bring these 2 to the other side and divide by 2"},{"Start":"02:32.135 ","End":"02:37.020","Text":"and we get that u_3 is equal to a half of u_1 minus u_2,"},{"Start":"02:37.020 ","End":"02:42.330","Text":"so it is a combination of the other 2 and the answer to 1 is yes."},{"Start":"02:42.500 ","End":"02:50.850","Text":"Part 2, I remember just asks if u_3 is in the span of u_1 and"},{"Start":"02:50.850 ","End":"02:53.780","Text":"u_2 and that\u0027s essentially"},{"Start":"02:53.780 ","End":"02:59.150","Text":"the same thing because the span is the set of all linear combinations."},{"Start":"02:59.150 ","End":"03:04.560","Text":"1 and 2 are just different ways of phrasing the same question,"},{"Start":"03:04.560 ","End":"03:06.690","Text":"so it also, yes."},{"Start":"03:06.690 ","End":"03:11.870","Text":"Like we mentioned, getting this formula shows that we did have"},{"Start":"03:11.870 ","End":"03:16.130","Text":"linear dependence and we also were able to"},{"Start":"03:16.130 ","End":"03:21.875","Text":"extract u_3 from this equation as a linear combination of the other 2."},{"Start":"03:21.875 ","End":"03:23.480","Text":"Now if you go back, in fact,"},{"Start":"03:23.480 ","End":"03:26.450","Text":"I\u0027ll just scroll back and see that we were asked,"},{"Start":"03:26.450 ","End":"03:28.280","Text":"that if they are linearly dependent,"},{"Start":"03:28.280 ","End":"03:35.255","Text":"we\u0027re going to try and write each vector of these as a combination of the other 2."},{"Start":"03:35.255 ","End":"03:43.335","Text":"Back here, this time I\u0027ll try and extract u_1,"},{"Start":"03:43.335 ","End":"03:45.945","Text":"I\u0027ll just bring it to the other side,"},{"Start":"03:45.945 ","End":"03:49.905","Text":"and then we get u_1 in terms of the other 2."},{"Start":"03:49.905 ","End":"03:54.360","Text":"Finally, we can extract u_2,"},{"Start":"03:54.360 ","End":"03:57.240","Text":"just throw the other stuff to the right,"},{"Start":"03:57.240 ","End":"04:01.350","Text":"and we have that u_2 is this."},{"Start":"04:01.350 ","End":"04:03.620","Text":"We\u0027ve done all 3 bits,"},{"Start":"04:03.620 ","End":"04:05.570","Text":"each 1 in terms of the others,"},{"Start":"04:05.570 ","End":"04:08.180","Text":"and like I said, it\u0027s not always possible."},{"Start":"04:08.180 ","End":"04:11.420","Text":"Only if all are present in this formula,"},{"Start":"04:11.420 ","End":"04:14.360","Text":"if 1 of them doesn\u0027t appear or appears with coefficient 0,"},{"Start":"04:14.360 ","End":"04:18.690","Text":"we couldn\u0027t. We\u0027re done."}],"ID":10016},{"Watched":false,"Name":"Exercise 3","Duration":"2m 34s","ChapterTopicVideoID":9905,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9905.jpeg","UploadDate":"2017-08-07T11:32:14.9270000","DurationForVideoObject":"PT2M34S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.140","Text":"We\u0027re continuing in the series of"},{"Start":"00:03.140 ","End":"00:06.120","Text":"exercises and this is very similar to the previous 1,"},{"Start":"00:06.120 ","End":"00:07.875","Text":"so I\u0027ll go a bit quicker."},{"Start":"00:07.875 ","End":"00:12.165","Text":"It\u0027s practically the same thing except that we have u_4 instead of u_3."},{"Start":"00:12.165 ","End":"00:15.090","Text":"Just like in the previous time,"},{"Start":"00:15.090 ","End":"00:20.730","Text":"we found it more convenient to actually start with this question."},{"Start":"00:20.730 ","End":"00:26.010","Text":"If the sets of these 3 vectors is linearly dependent and we do that"},{"Start":"00:26.010 ","End":"00:33.000","Text":"using matrices and row echelon form."},{"Start":"00:33.000 ","End":"00:36.810","Text":"Here are the 3 vectors, u_1,"},{"Start":"00:36.810 ","End":"00:41.880","Text":"u_2, u_4. Let\u0027s see."},{"Start":"00:41.880 ","End":"00:47.965","Text":"I can get 0 here if I take 4 times this and subtract the first row,"},{"Start":"00:47.965 ","End":"00:51.040","Text":"4 times u_4 minus u_1 here."},{"Start":"00:51.040 ","End":"00:56.785","Text":"4 times 1 minus 4 gives me the 0 here, check the others."},{"Start":"00:56.785 ","End":"01:02.790","Text":"The next thing to do is to subtract this row from"},{"Start":"01:02.790 ","End":"01:09.115","Text":"this row and that already gives us 0s here."},{"Start":"01:09.115 ","End":"01:12.255","Text":"The row echelon form we have 0s."},{"Start":"01:12.255 ","End":"01:17.795","Text":"When that happens, we know that this expression here is going to equal 0,"},{"Start":"01:17.795 ","End":"01:22.495","Text":"let\u0027s write that, like so."},{"Start":"01:22.495 ","End":"01:26.539","Text":"Now because of this,"},{"Start":"01:26.539 ","End":"01:30.750","Text":"I can now answer the first question."},{"Start":"01:30.820 ","End":"01:38.725","Text":"As you can see, I just moved u_1 and u_2 to the other side and divided by 4."},{"Start":"01:38.725 ","End":"01:43.370","Text":"That gives me u_4 as a linear combination of u_1 and u_2,"},{"Start":"01:43.370 ","End":"01:46.585","Text":"so the answer to the first part is yes."},{"Start":"01:46.585 ","End":"01:53.130","Text":"As before 2 is a rephrasing of 1 to say that u_4 is in the span of"},{"Start":"01:53.130 ","End":"01:59.770","Text":"u_1 and u_2 is the same thing as to say it\u0027s a linear combination of them, so yes."},{"Start":"01:59.770 ","End":"02:03.440","Text":"Now that this linear combination is 0 means that yes,"},{"Start":"02:03.440 ","End":"02:06.515","Text":"they are linearly dependent."},{"Start":"02:06.515 ","End":"02:12.095","Text":"Previously, we already extracted u_4 in terms of the other 2,"},{"Start":"02:12.095 ","End":"02:19.590","Text":"so all we have to do is also to extract u_1 in terms of u_4 and u_2,"},{"Start":"02:19.590 ","End":"02:23.175","Text":"then u_2 in terms of the other 2."},{"Start":"02:23.175 ","End":"02:29.880","Text":"Here\u0027s u_1 just simple algebra and here\u0027s u_2."},{"Start":"02:29.880 ","End":"02:31.040","Text":"We got all the 3 of them,"},{"Start":"02:31.040 ","End":"02:34.770","Text":"each in terms of the others and we are done."}],"ID":10017},{"Watched":false,"Name":"Exercise 4","Duration":"4m 26s","ChapterTopicVideoID":9906,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9906.jpeg","UploadDate":"2017-08-07T11:32:37.5300000","DurationForVideoObject":"PT4M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"Continuing in this series of exercises,"},{"Start":"00:03.420 ","End":"00:06.800","Text":"they all have the same u_1, u_2, u_3,"},{"Start":"00:06.800 ","End":"00:10.945","Text":"u_4 in 4-dimensional space."},{"Start":"00:10.945 ","End":"00:16.380","Text":"The difference here is that we have a parameter thrown in,"},{"Start":"00:16.380 ","End":"00:19.335","Text":"we get an extra vector v, which is this,"},{"Start":"00:19.335 ","End":"00:24.180","Text":"with a parameter k. That makes things a bit more interesting."},{"Start":"00:24.180 ","End":"00:28.200","Text":"Part 1, we want to know what value of"},{"Start":"00:28.200 ","End":"00:35.265","Text":"k will make v a linear combination of u_1 and u_2."},{"Start":"00:35.265 ","End":"00:39.499","Text":"These 2 don\u0027t play a part in this particular exercise."},{"Start":"00:39.499 ","End":"00:42.590","Text":"Part 2, as we already know,"},{"Start":"00:42.590 ","End":"00:46.700","Text":"is going to be assigned as part 1 just in different words."},{"Start":"00:46.700 ","End":"00:55.480","Text":"To say that v is in the span or the vector space span by these 2 subspaces,"},{"Start":"00:55.480 ","End":"01:00.630","Text":"it\u0027s the same thing because this span means the set of all linear combinations,"},{"Start":"01:00.630 ","End":"01:02.670","Text":"so 1 and 2 are the same."},{"Start":"01:02.670 ","End":"01:06.000","Text":"3 is a little bit different;"},{"Start":"01:06.000 ","End":"01:07.630","Text":"it\u0027s actually more symmetrical."},{"Start":"01:07.630 ","End":"01:09.080","Text":"We are asking about u_1,"},{"Start":"01:09.080 ","End":"01:10.565","Text":"u_2, and u_3,"},{"Start":"01:10.565 ","End":"01:14.530","Text":"we want to know the conditions on k for these to be linearly dependent."},{"Start":"01:14.530 ","End":"01:20.660","Text":"Linear dependent means that at least 1 of them is a combination of the other 2,"},{"Start":"01:20.660 ","End":"01:23.795","Text":"not necessarily v in terms of u_1 and u_2."},{"Start":"01:23.795 ","End":"01:30.305","Text":"But because this is a more balanced definition problem,"},{"Start":"01:30.305 ","End":"01:32.795","Text":"I prefer to start with 3,"},{"Start":"01:32.795 ","End":"01:37.280","Text":"and sometimes it comes out to be the same as 1 and 2."},{"Start":"01:37.280 ","End":"01:42.170","Text":"Let\u0027s start with the linearly dependence or independence,"},{"Start":"01:42.170 ","End":"01:45.545","Text":"and we do that, of course, with matrices."},{"Start":"01:45.545 ","End":"01:48.545","Text":"I\u0027m going to put this 1,"},{"Start":"01:48.545 ","End":"01:51.545","Text":"this 1, and this 1 in a matrix."},{"Start":"01:51.545 ","End":"01:55.390","Text":"I\u0027m going to lose them if I scroll, but it\u0027s okay."},{"Start":"01:55.390 ","End":"01:57.870","Text":"Anyway, they\u0027re off screen but I copied them."},{"Start":"01:57.870 ","End":"02:00.715","Text":"This is u_1, this is u_2 and this is v,"},{"Start":"02:00.715 ","End":"02:07.240","Text":"and we want to bring this to echelon form with row operations."},{"Start":"02:07.240 ","End":"02:13.820","Text":"The first thing that\u0027s obvious to do is to subtract the first row from the last row."},{"Start":"02:13.820 ","End":"02:16.070","Text":"That will give us a 0 here,"},{"Start":"02:16.070 ","End":"02:19.580","Text":"and notice that here we have v minus u_1."},{"Start":"02:19.580 ","End":"02:22.760","Text":"The next thing to do for echelon form,"},{"Start":"02:22.760 ","End":"02:24.965","Text":"we want a 0 here."},{"Start":"02:24.965 ","End":"02:31.230","Text":"Let\u0027s subtract the second row from the third row,"},{"Start":"02:31.570 ","End":"02:34.595","Text":"and then of course we get the 0 here,"},{"Start":"02:34.595 ","End":"02:38.760","Text":"that was the whole purpose of the echelon form,"},{"Start":"02:38.760 ","End":"02:42.415","Text":"and here we get this minus this which is this."},{"Start":"02:42.415 ","End":"02:45.350","Text":"Because we\u0027re in echelon form,"},{"Start":"02:45.350 ","End":"02:47.555","Text":"when we didn\u0027t get a row of zeros,"},{"Start":"02:47.555 ","End":"02:52.910","Text":"that means that the 3 are independent,"},{"Start":"02:52.910 ","End":"02:55.430","Text":"unless somehow this is a row of zeros,"},{"Start":"02:55.430 ","End":"02:57.875","Text":"it just doesn\u0027t look like all zeros."},{"Start":"02:57.875 ","End":"03:03.020","Text":"Because supposing that k plus 4 was 0 and minus 2k minus 8 was 0,"},{"Start":"03:03.020 ","End":"03:04.520","Text":"then it would be a row of zeros,"},{"Start":"03:04.520 ","End":"03:06.650","Text":"even though it doesn\u0027t look like it."},{"Start":"03:06.650 ","End":"03:11.060","Text":"I just took out k plus 4 from here,"},{"Start":"03:11.060 ","End":"03:13.340","Text":"so we can see that really there\u0027s"},{"Start":"03:13.340 ","End":"03:18.740","Text":"only 1 condition that k plus 4b is 0 and then this will be 0 and otherwise not."},{"Start":"03:18.740 ","End":"03:29.430","Text":"The only way this is going to be linearly dependent is k plus 4 to be 0."},{"Start":"03:29.430 ","End":"03:33.125","Text":"That if this is 0, then that means that k is minus 4."},{"Start":"03:33.125 ","End":"03:38.990","Text":"We already said the part 2 is the same as part 1 in other words,"},{"Start":"03:38.990 ","End":"03:42.580","Text":"and also the linearly dependence,"},{"Start":"03:42.580 ","End":"03:45.920","Text":"because actually what we just did here is the same thing."},{"Start":"03:45.920 ","End":"03:48.710","Text":"Now I realize I haven\u0027t been precise enough,"},{"Start":"03:48.710 ","End":"03:52.520","Text":"because 1 and 2 we\u0027re not just that these 3 are dependent,"},{"Start":"03:52.520 ","End":"03:57.155","Text":"it was specifically that v is a combination of u_1 and u_2."},{"Start":"03:57.155 ","End":"03:59.630","Text":"Because the coefficient of v is not 0,"},{"Start":"03:59.630 ","End":"04:01.085","Text":"that v is present here,"},{"Start":"04:01.085 ","End":"04:05.730","Text":"this 1 we can write as v equals u_1 plus"},{"Start":"04:05.730 ","End":"04:12.505","Text":"u_2 which is really the more precise thing for parts 1 and 2."},{"Start":"04:12.505 ","End":"04:16.010","Text":"If we didn\u0027t get a v appearing explicitly here,"},{"Start":"04:16.010 ","End":"04:17.030","Text":"that would be a problem;"},{"Start":"04:17.030 ","End":"04:18.770","Text":"we\u0027d have to analyze it differently."},{"Start":"04:18.770 ","End":"04:26.370","Text":"But as it is, all 3 come out the same and k is minus 4. We\u0027re done."}],"ID":10018},{"Watched":false,"Name":"Exercise 5","Duration":"8m 33s","ChapterTopicVideoID":9907,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9907.jpeg","UploadDate":"2017-08-07T11:33:23.1870000","DurationForVideoObject":"PT8M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"We\u0027re continuing with the series of exercises."},{"Start":"00:03.030 ","End":"00:09.190","Text":"I think this is about the 5th in the series or something like that."},{"Start":"00:09.860 ","End":"00:14.340","Text":"This time we have an extra vector v thrown in,"},{"Start":"00:14.340 ","End":"00:17.400","Text":"which contains 4 parameters, a, b, c, and d."},{"Start":"00:17.400 ","End":"00:23.745","Text":"We want to know the conditions on the parameters such that"},{"Start":"00:23.745 ","End":"00:31.035","Text":"part 1 we want v to be a linear combination of u_1 and u_2."},{"Start":"00:31.035 ","End":"00:33.870","Text":"We\u0027re not using u_3 and u_4 here."},{"Start":"00:33.870 ","End":"00:37.860","Text":"Then we want something which is actually equivalent to"},{"Start":"00:37.860 ","End":"00:42.890","Text":"1 for v to be in the span of u_1 and u_2."},{"Start":"00:42.890 ","End":"00:46.205","Text":"It\u0027s actually the same thing to be in the spanning subspace"},{"Start":"00:46.205 ","End":"00:49.790","Text":"means to be in the set of linear combinations,"},{"Start":"00:49.790 ","End":"00:51.290","Text":"which is the same thing."},{"Start":"00:51.290 ","End":"00:56.300","Text":"In 3, which is similar to 1, we want u_1,"},{"Start":"00:56.300 ","End":"01:00.880","Text":"u_2 and v to be linearly dependent,"},{"Start":"01:00.880 ","End":"01:05.810","Text":"which means that 1 of them at least is a combination of the other 2,"},{"Start":"01:05.810 ","End":"01:11.360","Text":"not necessarily v is a combination of u_1 and u_2, although probably is."},{"Start":"01:11.360 ","End":"01:15.770","Text":"Let\u0027s see. As usual,"},{"Start":"01:15.770 ","End":"01:18.220","Text":"the ones we\u0027re interested in is u_1, u_2,"},{"Start":"01:18.220 ","End":"01:26.279","Text":"and v, and we put them in a matrix and bring to row echelon form."},{"Start":"01:26.650 ","End":"01:29.090","Text":"Before I get into matrices,"},{"Start":"01:29.090 ","End":"01:31.850","Text":"I just want to emphasize a subtle point,"},{"Start":"01:31.850 ","End":"01:37.340","Text":"but it\u0027s important to say that these 3 are linearly dependent is"},{"Start":"01:37.340 ","End":"01:43.430","Text":"not quite the same as to say that v is a combination of u_1 and u_2."},{"Start":"01:43.430 ","End":"01:49.820","Text":"This is a bit more broad because it could be that u_2 is a combination of u_1 and v,"},{"Start":"01:49.820 ","End":"01:55.370","Text":"or that u_1 is a combination of u_2 and v. It\u0027s a subtle point,"},{"Start":"01:55.370 ","End":"02:00.400","Text":"I\u0027ll return to that after we\u0027ve done the matrix computations."},{"Start":"02:00.400 ","End":"02:03.060","Text":"Here\u0027s the matrix,"},{"Start":"02:03.060 ","End":"02:04.865","Text":"I started a new page."},{"Start":"02:04.865 ","End":"02:07.640","Text":"There\u0027s another subtle point."},{"Start":"02:07.640 ","End":"02:11.000","Text":"Normally or usually,"},{"Start":"02:11.000 ","End":"02:15.590","Text":"we would also include the vector name."},{"Start":"02:15.590 ","End":"02:17.180","Text":"Here I would put u_1,"},{"Start":"02:17.180 ","End":"02:18.320","Text":"u_2, v,"},{"Start":"02:18.320 ","End":"02:20.540","Text":"and use an augmented matrix."},{"Start":"02:20.540 ","End":"02:24.320","Text":"This just complicates the computations and it\u0027s not really"},{"Start":"02:24.320 ","End":"02:29.165","Text":"necessary because we weren\u0027t asked to exactly find the coefficients."},{"Start":"02:29.165 ","End":"02:30.920","Text":"If they are linearly dependent,"},{"Start":"02:30.920 ","End":"02:37.685","Text":"we don\u0027t actually need to know how many u plus how many u_2 plus how many v equals 0."},{"Start":"02:37.685 ","End":"02:40.780","Text":"We\u0027re just throwing that out."},{"Start":"02:40.780 ","End":"02:44.675","Text":"Let\u0027s get to it. You want to get to row echelon form."},{"Start":"02:44.675 ","End":"02:48.110","Text":"We start off by doing 2 things at once,"},{"Start":"02:48.110 ","End":"02:52.780","Text":"multiplying this by 4 and then subtracting a times this,"},{"Start":"02:52.780 ","End":"02:54.420","Text":"and this is what we get,"},{"Start":"02:54.420 ","End":"02:55.995","Text":"and a 0 here, of course."},{"Start":"02:55.995 ","End":"02:59.060","Text":"Feel free to pause the clip at any point"},{"Start":"02:59.060 ","End":"03:02.330","Text":"to check the computations because I\u0027m going to be going quickly."},{"Start":"03:02.330 ","End":"03:07.450","Text":"After this, we want to get 0\u0027s here and here."},{"Start":"03:07.450 ","End":"03:14.845","Text":"We\u0027ll multiply the last row by 11 and subtract 4b minus a times this row."},{"Start":"03:14.845 ","End":"03:18.710","Text":"If we do that, then this is what we get,"},{"Start":"03:18.710 ","End":"03:22.600","Text":"which looks a mess and we better tidy it up a bit."},{"Start":"03:22.600 ","End":"03:24.840","Text":"Here it is a bit better, and of course,"},{"Start":"03:24.840 ","End":"03:29.230","Text":"it is already in row echelon form."},{"Start":"03:29.230 ","End":"03:32.470","Text":"The question is, do we have a row of 0\u0027s?"},{"Start":"03:32.470 ","End":"03:34.030","Text":"Well, certainly not here and here."},{"Start":"03:34.030 ","End":"03:36.040","Text":"But the question is,"},{"Start":"03:36.040 ","End":"03:38.530","Text":"could the last row be all 0\u0027s?"},{"Start":"03:38.530 ","End":"03:41.799","Text":"In other words, what if this is 0 and this is 0?"},{"Start":"03:41.799 ","End":"03:50.835","Text":"That\u0027s going to be exactly the condition we need for the linear dependency."},{"Start":"03:50.835 ","End":"03:56.430","Text":"That gives us a pair of 2 equations and 4 unknowns, a, b, c, d."},{"Start":"03:56.430 ","End":"03:58.985","Text":"Straight away,"},{"Start":"03:58.985 ","End":"04:02.255","Text":"I\u0027ll simplify by dividing by minus 4,"},{"Start":"04:02.255 ","End":"04:05.115","Text":"and here we are."},{"Start":"04:05.115 ","End":"04:08.880","Text":"We have 2 equations and 4 unknowns."},{"Start":"04:08.880 ","End":"04:11.865","Text":"I don\u0027t think I\u0027ll use matrices, you could,"},{"Start":"04:11.865 ","End":"04:15.815","Text":"let\u0027s just do row operations as is."},{"Start":"04:15.815 ","End":"04:18.915","Text":"I just need to get a 0 here."},{"Start":"04:18.915 ","End":"04:25.790","Text":"We\u0027ll take 4 times the second row and subtract 13 times the first row from it."},{"Start":"04:25.790 ","End":"04:31.700","Text":"This is what we get and indeed there\u0027s a 0 or missing a here."},{"Start":"04:31.700 ","End":"04:34.070","Text":"I\u0027ll leave you to check the computations."},{"Start":"04:34.070 ","End":"04:36.869","Text":"Like I said, you can always pause."},{"Start":"04:37.320 ","End":"04:44.110","Text":"At this point, this second row is divisible by 11,"},{"Start":"04:44.110 ","End":"04:48.500","Text":"so let\u0027s divide the second row by 11."},{"Start":"04:48.500 ","End":"04:53.210","Text":"For some reason, I forget why did I make this row negative."},{"Start":"04:53.210 ","End":"04:55.550","Text":"You could leave it as plus,"},{"Start":"04:55.550 ","End":"04:59.510","Text":"minus, minus, no harm done anyway."},{"Start":"04:59.510 ","End":"05:03.770","Text":"The free variables, c and d,"},{"Start":"05:03.770 ","End":"05:11.335","Text":"and the pivot entries or the dependent constraint variables are a and b."},{"Start":"05:11.335 ","End":"05:16.520","Text":"As usual, we assign the free variables 2 letters"},{"Start":"05:16.520 ","End":"05:21.530","Text":"like s and t. But when I tried it first time,"},{"Start":"05:21.530 ","End":"05:25.950","Text":"it came out messy because of this 7 here."},{"Start":"05:25.950 ","End":"05:31.395","Text":"From experience, I substituted instead of t and s,"},{"Start":"05:31.395 ","End":"05:32.925","Text":"7t and 7s,"},{"Start":"05:32.925 ","End":"05:36.780","Text":"which is just as much a general number,"},{"Start":"05:36.780 ","End":"05:40.170","Text":"and then the computations come out easier."},{"Start":"05:40.170 ","End":"05:42.840","Text":"The 7 is just for convenience."},{"Start":"05:42.840 ","End":"05:47.600","Text":"First we get the b from here after the substitution,"},{"Start":"05:47.600 ","End":"05:50.120","Text":"and this is what it comes out to be."},{"Start":"05:50.120 ","End":"05:51.620","Text":"Then we\u0027ve got b, c,"},{"Start":"05:51.620 ","End":"05:54.310","Text":"and d to plug into here."},{"Start":"05:54.310 ","End":"05:56.805","Text":"For this, we can extract a."},{"Start":"05:56.805 ","End":"05:59.060","Text":"Again, I\u0027ll leave you to check the computations,"},{"Start":"05:59.060 ","End":"06:00.875","Text":"you can pause the clip."},{"Start":"06:00.875 ","End":"06:03.440","Text":"We have a, b, c, and d."},{"Start":"06:03.440 ","End":"06:08.320","Text":"Now let\u0027s return to the previous page,"},{"Start":"06:08.320 ","End":"06:10.300","Text":"and here we are."},{"Start":"06:10.300 ","End":"06:12.060","Text":"Let\u0027s look at the questions."},{"Start":"06:12.060 ","End":"06:15.675","Text":"We actually answered number 3,"},{"Start":"06:15.675 ","End":"06:20.370","Text":"which is that these 3 are linearly dependent if a, b, c,"},{"Start":"06:20.370 ","End":"06:22.530","Text":"and d are equal to,"},{"Start":"06:22.530 ","End":"06:28.125","Text":"what\u0027s written here for sum t and s. If for sum t and s we can get this,"},{"Start":"06:28.125 ","End":"06:30.454","Text":"then they\u0027re linearly dependent."},{"Start":"06:30.454 ","End":"06:32.060","Text":"Now what about 1 and 2,"},{"Start":"06:32.060 ","End":"06:34.615","Text":"which are the same as we know?"},{"Start":"06:34.615 ","End":"06:39.510","Text":"Here, under what conditions is v a linear combination of u_1 and u_2?"},{"Start":"06:39.510 ","End":"06:42.090","Text":"I claim it\u0027s the same as this,"},{"Start":"06:42.090 ","End":"06:44.505","Text":"and I\u0027ll show you why."},{"Start":"06:44.505 ","End":"06:46.980","Text":"It\u0027s a bit delicate."},{"Start":"06:46.980 ","End":"06:49.970","Text":"Sorry, before that I forgot I was going to show you"},{"Start":"06:49.970 ","End":"06:54.660","Text":"an example of a solution, a, b, c, d."},{"Start":"06:54.660 ","End":"06:59.550","Text":"For example, if we let s equal 1 and t equals 1 here,"},{"Start":"06:59.550 ","End":"07:03.390","Text":"and a will be 5 plus 3 is 8, and so on."},{"Start":"07:03.390 ","End":"07:06.360","Text":"This is just an example of a, b, c, d."},{"Start":"07:06.360 ","End":"07:10.175","Text":"Now back to the delicate point I was going to make."},{"Start":"07:10.175 ","End":"07:13.925","Text":"Like I said, this answers part 3."},{"Start":"07:13.925 ","End":"07:16.915","Text":"We said that 2 is the same as 1,"},{"Start":"07:16.915 ","End":"07:19.890","Text":"so we just have to show part 1,"},{"Start":"07:19.890 ","End":"07:25.350","Text":"that was when v is a combination of u_1 and u_2."},{"Start":"07:25.690 ","End":"07:30.580","Text":"1 of the definitions of linearly dependent means you can find constants,"},{"Start":"07:30.580 ","End":"07:35.350","Text":"not all of them 0, such that when you multiply these,"},{"Start":"07:35.350 ","End":"07:38.125","Text":"you get 0, just like I would have written here."},{"Start":"07:38.125 ","End":"07:41.300","Text":"Now I claim that c is not 0,"},{"Start":"07:41.300 ","End":"07:43.530","Text":"because if c was 0,"},{"Start":"07:43.530 ","End":"07:45.525","Text":"this whole thing would be missing."},{"Start":"07:45.525 ","End":"07:52.450","Text":"Then you\u0027d get that u_1 and u_2 are linearly independent, but they\u0027re not."},{"Start":"07:52.450 ","End":"07:55.390","Text":"This is similar to the first in the series where we showed"},{"Start":"07:55.390 ","End":"07:58.995","Text":"that u_1 and u_4 and not linearly dependent."},{"Start":"07:58.995 ","End":"08:00.360","Text":"If they were linearly dependent,"},{"Start":"08:00.360 ","End":"08:02.300","Text":"1 would be a multiple of the other,"},{"Start":"08:02.300 ","End":"08:04.410","Text":"and if you look back, you\u0027ll see that it\u0027s not."},{"Start":"08:04.410 ","End":"08:06.130","Text":"I don\u0027t want to get too deeply into this,"},{"Start":"08:06.130 ","End":"08:08.205","Text":"we\u0027re already on to the fine points."},{"Start":"08:08.205 ","End":"08:13.080","Text":"But u_1 and u_2 are not linearly dependent,"},{"Start":"08:13.080 ","End":"08:14.865","Text":"so this can\u0027t be 0."},{"Start":"08:14.865 ","End":"08:16.530","Text":"If it\u0027s not 0,"},{"Start":"08:16.530 ","End":"08:20.795","Text":"and what we get since we can divide by c if it\u0027s not 0,"},{"Start":"08:20.795 ","End":"08:26.880","Text":"is we\u0027ve got v as something times u_1 plus something times u_2 from here,"},{"Start":"08:26.880 ","End":"08:30.890","Text":"so v is a linear combination of u_1 and u_2."},{"Start":"08:30.890 ","End":"08:33.990","Text":"That\u0027s it, we\u0027re done."}],"ID":10019},{"Watched":false,"Name":"Linear dependence - set of vectors","Duration":"5m 2s","ChapterTopicVideoID":25066,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25066.jpeg","UploadDate":"2021-06-27T07:04:37.7570000","DurationForVideoObject":"PT5M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.440","Text":"In this clip, I will be talking about what it means"},{"Start":"00:04.440 ","End":"00:09.780","Text":"for a set of vectors to be linearly dependent."},{"Start":"00:09.780 ","End":"00:15.825","Text":"We\u0027ve already encountered the term linearly dependent but in a different context."},{"Start":"00:15.825 ","End":"00:22.570","Text":"We had 1 vector was linearly dependent on another bunch of vectors."},{"Start":"00:22.570 ","End":"00:26.300","Text":"This time we\u0027re talking about the set of vectors as a whole,"},{"Start":"00:26.300 ","End":"00:28.609","Text":"so it\u0027s slightly different."},{"Start":"00:28.609 ","End":"00:33.800","Text":"A set of vectors is called linearly dependent if"},{"Start":"00:33.800 ","End":"00:40.675","Text":"at least 1 vector in the set is a linear combination of other vectors in the set."},{"Start":"00:40.675 ","End":"00:44.855","Text":"These 2 variants of the concept, linearly dependent."},{"Start":"00:44.855 ","End":"00:49.415","Text":"You can have an individual vector being linearly dependent on some others,"},{"Start":"00:49.415 ","End":"00:54.640","Text":"or you could have the set as a whole being linearly dependent or not."},{"Start":"00:54.640 ","End":"00:57.030","Text":"When it\u0027s not, there\u0027s a name for that also,"},{"Start":"00:57.030 ","End":"01:00.550","Text":"then we say it\u0027s linearly independent."},{"Start":"01:01.910 ","End":"01:04.655","Text":"Let\u0027s go for an example."},{"Start":"01:04.655 ","End":"01:08.315","Text":"We\u0027re talking about the space R^3."},{"Start":"01:08.315 ","End":"01:10.640","Text":"I should have mentioned that,"},{"Start":"01:10.640 ","End":"01:15.340","Text":"but it\u0027s clear that we have 3D vectors here."},{"Start":"01:15.340 ","End":"01:17.780","Text":"We\u0027ve seen these vectors before."},{"Start":"01:17.780 ","End":"01:19.490","Text":"They are linearly dependent."},{"Start":"01:19.490 ","End":"01:26.550","Text":"I\u0027ll remind you that u is equal to twice v minus w. We"},{"Start":"01:26.550 ","End":"01:34.590","Text":"have that 1 of the vectors in the set is a linear combination of 2 others,"},{"Start":"01:34.590 ","End":"01:37.545","Text":"and so that fits our definition, so,"},{"Start":"01:37.545 ","End":"01:42.075","Text":"this set is linearly dependent."},{"Start":"01:42.075 ","End":"01:48.950","Text":"This could be done in any number of ways to show that 1 is a combination of others."},{"Start":"01:48.950 ","End":"01:54.210","Text":"For example, I could have said that w was 2v minus u"},{"Start":"01:54.210 ","End":"02:00.215","Text":"or even the v is 1/2u plus a 1/2w, doesn\u0027t matter."},{"Start":"02:00.215 ","End":"02:07.640","Text":"All you have to do is give 1 example and then that makes it linearly dependent."},{"Start":"02:07.640 ","End":"02:13.085","Text":"Just a notation, we don\u0027t always have them arranged as a set with the name of the set."},{"Start":"02:13.085 ","End":"02:15.980","Text":"Sometimes you just say the vectors themselves u, v,"},{"Start":"02:15.980 ","End":"02:19.165","Text":"w, are linearly dependent."},{"Start":"02:19.165 ","End":"02:27.560","Text":"I just want to emphasize the business of at least 1 and all you need is 1."},{"Start":"02:27.560 ","End":"02:30.470","Text":"For example, it\u0027s sufficient just to find"},{"Start":"02:30.470 ","End":"02:34.240","Text":"one vector in the set which is a linear combination."},{"Start":"02:34.240 ","End":"02:40.245","Text":"Did I say that lc is often used for linear combination?"},{"Start":"02:40.245 ","End":"02:45.170","Text":"Yes. We just have to find 1 which is the linear combination of some of"},{"Start":"02:45.170 ","End":"02:51.725","Text":"the others and then we can say that the set is linearly dependent."},{"Start":"02:51.725 ","End":"02:54.590","Text":"An example, if I have 10 vectors,"},{"Start":"02:54.590 ","End":"02:56.645","Text":"v_1 through v_10,"},{"Start":"02:56.645 ","End":"03:05.910","Text":"and suppose that we know that v_4 is equal to v_8 plus v_10,"},{"Start":"03:05.910 ","End":"03:09.885","Text":"you happen to discover this or whatever."},{"Start":"03:09.885 ","End":"03:15.140","Text":"This is all we need in order to be able to say that A is a linearly independent set,"},{"Start":"03:15.140 ","End":"03:17.140","Text":"you don\u0027t need to have that every one"},{"Start":"03:17.140 ","End":"03:19.520","Text":"is the combination of the others or anything like that,"},{"Start":"03:19.520 ","End":"03:21.775","Text":"just 1 single example."},{"Start":"03:21.775 ","End":"03:23.580","Text":"Another question arises,"},{"Start":"03:23.580 ","End":"03:27.380","Text":"of how can we check in the cases that we had,"},{"Start":"03:27.380 ","End":"03:32.345","Text":"I just knew the answers or I pulled things out of the hat like this."},{"Start":"03:32.345 ","End":"03:36.095","Text":"Suppose I gave you a set of vectors."},{"Start":"03:36.095 ","End":"03:38.815","Text":"We\u0027re talking now vectors in R^n."},{"Start":"03:38.815 ","End":"03:43.250","Text":"How do you check if they\u0027re linearly dependent or independent?"},{"Start":"03:43.250 ","End":"03:46.430","Text":"I\u0027ll give you the picture and I\u0027ll give you an example."},{"Start":"03:46.430 ","End":"03:53.270","Text":"The basic recipe is you create the matrix with rows from the vectors in the set."},{"Start":"03:53.270 ","End":"03:58.160","Text":"Next, we bring the matrix to row echelon form in"},{"Start":"03:58.160 ","End":"04:04.385","Text":"the usual way just mean by doing a combination of the elementary row operations,"},{"Start":"04:04.385 ","End":"04:09.685","Text":"like adding 3 times 1 row to another row and so on."},{"Start":"04:09.685 ","End":"04:15.515","Text":"If at some point during this process you get a row of all 0s,"},{"Start":"04:15.515 ","End":"04:19.220","Text":"then you can stop there and say yes,"},{"Start":"04:19.220 ","End":"04:22.365","Text":"the set is linearly dependent."},{"Start":"04:22.365 ","End":"04:26.660","Text":"If you get to the process the whole way and you get it to row"},{"Start":"04:26.660 ","End":"04:33.080","Text":"echelon form but there\u0027s no row of 0s in the end,"},{"Start":"04:33.080 ","End":"04:36.995","Text":"then the set is not linearly dependent,"},{"Start":"04:36.995 ","End":"04:39.905","Text":"meaning it\u0027s linearly independent."},{"Start":"04:39.905 ","End":"04:42.020","Text":"Now I\u0027m going to give you an example,"},{"Start":"04:42.020 ","End":"04:49.895","Text":"but the example does more than what I said here because if it happens to be dependent,"},{"Start":"04:49.895 ","End":"04:53.350","Text":"this technique actually gives you the combination."},{"Start":"04:53.350 ","End":"04:56.675","Text":"Like here we had that this plus this, plus this."},{"Start":"04:56.675 ","End":"05:03.030","Text":"I\u0027m going to expand on what I wrote here."}],"ID":25834},{"Watched":false,"Name":"Formal Definition - Linear Independence","Duration":"11m 25s","ChapterTopicVideoID":25065,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25065.jpeg","UploadDate":"2021-06-27T07:04:04.1000000","DurationForVideoObject":"PT11M25S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.740","Text":"We\u0027re heading towards a more formal definition of what it means for a set of"},{"Start":"00:04.740 ","End":"00:09.945","Text":"vectors to be linearly independent or linearly dependent."},{"Start":"00:09.945 ","End":"00:13.320","Text":"But I\u0027m going to start off light so to speak,"},{"Start":"00:13.320 ","End":"00:16.335","Text":"with a preface introduction,"},{"Start":"00:16.335 ","End":"00:20.445","Text":"and let\u0027s return to our familiar set of vectors,"},{"Start":"00:20.445 ","End":"00:21.780","Text":"call the set A."},{"Start":"00:21.780 ","End":"00:26.250","Text":"The vectors u, v, and w as described here."},{"Start":"00:26.250 ","End":"00:29.940","Text":"Let\u0027s write an equation where a,"},{"Start":"00:29.940 ","End":"00:31.950","Text":"b, and c are the unknowns,"},{"Start":"00:31.950 ","End":"00:39.360","Text":"and we want to write a times u plus b times v plus c times w equals 0,"},{"Start":"00:39.360 ","End":"00:41.040","Text":"and look at it as an equation in a,"},{"Start":"00:41.040 ","End":"00:43.320","Text":"b and c. U,"},{"Start":"00:43.320 ","End":"00:44.790","Text":"v and w are not unknown,"},{"Start":"00:44.790 ","End":"00:47.325","Text":"they\u0027re the ones here so let\u0027s replace them,"},{"Start":"00:47.325 ","End":"00:50.375","Text":"called it expanding substituting,"},{"Start":"00:50.375 ","End":"00:54.970","Text":"a times this plus b times this plus c times this is vector 0,"},{"Start":"00:54.970 ","End":"00:58.620","Text":"which is 0, 0, 0."},{"Start":"00:58.620 ","End":"01:01.075","Text":"There\u0027s 1 solution to this equation,"},{"Start":"01:01.075 ","End":"01:05.200","Text":"which is obvious and it\u0027s called the trivial solution."},{"Start":"01:05.200 ","End":"01:09.870","Text":"The file at a equals 0 and b equals 0 and c equals 0,"},{"Start":"01:09.870 ","End":"01:13.850","Text":"certainly, this will be true without any computations."},{"Start":"01:13.850 ","End":"01:17.910","Text":"This solution, and that\u0027s why we call it the trivial solution,"},{"Start":"01:17.910 ","End":"01:20.250","Text":"it\u0027s obvious and it\u0027s always there,"},{"Start":"01:20.250 ","End":"01:23.954","Text":"and whenever we have a linear combination of vectors,"},{"Start":"01:23.954 ","End":"01:27.940","Text":"z is 0, then we can always choose coefficients"},{"Start":"01:27.940 ","End":"01:32.135","Text":"0 and be guaranteed that at least it has the trivial solution."},{"Start":"01:32.135 ","End":"01:33.995","Text":"But the interesting question is,"},{"Start":"01:33.995 ","End":"01:38.005","Text":"does it have any other solutions besides the trivial solution?"},{"Start":"01:38.005 ","End":"01:39.890","Text":"The question I asked first is,"},{"Start":"01:39.890 ","End":"01:43.040","Text":"does it have only the trivial solution or maybe some others?"},{"Start":"01:43.040 ","End":"01:47.120","Text":"If it has gotten on non-trivial solution means not 0,"},{"Start":"01:47.120 ","End":"01:49.105","Text":"0, 0, not all three 0."},{"Start":"01:49.105 ","End":"01:52.579","Text":"What can we conclude if it has a non-trivial solution?"},{"Start":"01:52.579 ","End":"01:54.350","Text":"Just say for instance,"},{"Start":"01:54.350 ","End":"01:57.470","Text":"that we solve and we get a, b,"},{"Start":"01:57.470 ","End":"01:59.240","Text":"and c equal to 1, 2,"},{"Start":"01:59.240 ","End":"02:02.360","Text":"and 3 respectively, which is actually not a solution,"},{"Start":"02:02.360 ","End":"02:04.580","Text":"but I\u0027m just saying, for instance,"},{"Start":"02:04.580 ","End":"02:06.985","Text":"suppose that we had that,"},{"Start":"02:06.985 ","End":"02:13.640","Text":"then what would we conclude about the vectors u, v, and w?"},{"Start":"02:13.640 ","End":"02:18.380","Text":"Turns out that the answer is that they are linearly dependent."},{"Start":"02:18.380 ","End":"02:21.650","Text":"Whenever you find a non-trivial solution,"},{"Start":"02:21.650 ","End":"02:24.980","Text":"then the 3 vectors u, v,"},{"Start":"02:24.980 ","End":"02:28.625","Text":"and w are linearly dependent."},{"Start":"02:28.625 ","End":"02:31.790","Text":"I\u0027ll give you a brief explanation of why this is so that"},{"Start":"02:31.790 ","End":"02:35.435","Text":"they are linearly dependent in such a case."},{"Start":"02:35.435 ","End":"02:37.400","Text":"We chose 1, 2, 3,"},{"Start":"02:37.400 ","End":"02:40.440","Text":"but I mean any nontrivial case,"},{"Start":"02:40.440 ","End":"02:42.210","Text":"so 1 of them is not 0,"},{"Start":"02:42.210 ","End":"02:44.450","Text":"so let\u0027s say b is not 0."},{"Start":"02:44.450 ","End":"02:46.175","Text":"In this case, all 3 are not 0,"},{"Start":"02:46.175 ","End":"02:48.820","Text":"but if b was not 0,"},{"Start":"02:48.820 ","End":"02:51.210","Text":"I circled it is not a 0,"},{"Start":"02:51.210 ","End":"02:53.700","Text":"then we can get 1 of the vectors,"},{"Start":"02:53.700 ","End":"02:55.380","Text":"this 1 in terms of the others,"},{"Start":"02:55.380 ","End":"03:00.215","Text":"we just bring this part and this part to the other side of the equation,"},{"Start":"03:00.215 ","End":"03:01.670","Text":"and since b is not 0,"},{"Start":"03:01.670 ","End":"03:02.970","Text":"we divide by it,"},{"Start":"03:02.970 ","End":"03:08.855","Text":"so we can isolate and get this 1 as a linear combination of the other 2."},{"Start":"03:08.855 ","End":"03:11.525","Text":"This will always be the case with a non-trivial solution,"},{"Start":"03:11.525 ","End":"03:14.150","Text":"you pick 1 of the coefficients that isn\u0027t 0,"},{"Start":"03:14.150 ","End":"03:19.280","Text":"and then you can isolate that vector in terms of the others,"},{"Start":"03:19.280 ","End":"03:21.925","Text":"so that\u0027s the reason."},{"Start":"03:21.925 ","End":"03:25.520","Text":"This next paragraph, if I just summarize it,"},{"Start":"03:25.520 ","End":"03:27.995","Text":"just says that the opposite is also true."},{"Start":"03:27.995 ","End":"03:33.590","Text":"If the only solution to this equation is the trivial solution,"},{"Start":"03:33.590 ","End":"03:35.120","Text":"the trivial solution is always there,"},{"Start":"03:35.120 ","End":"03:36.680","Text":"but if it\u0027s the only 1 there,"},{"Start":"03:36.680 ","End":"03:38.465","Text":"0, 0, 0 solution,"},{"Start":"03:38.465 ","End":"03:42.295","Text":"then they\u0027re going to be independent linearly,"},{"Start":"03:42.295 ","End":"03:44.880","Text":"and I won\u0027t even read out what this says."},{"Start":"03:44.880 ","End":"03:48.800","Text":"Basically, if we find a solution other than the trivial 1,"},{"Start":"03:48.800 ","End":"03:50.630","Text":"then they are dependent,"},{"Start":"03:50.630 ","End":"03:53.870","Text":"and if we can show that that\u0027s the only solution,"},{"Start":"03:53.870 ","End":"03:56.845","Text":"the trivial 1 in they\u0027re independent."},{"Start":"03:56.845 ","End":"04:04.070","Text":"Let\u0027s actually check if this equation has a non-trivial solution where not all of a,"},{"Start":"04:04.070 ","End":"04:06.040","Text":"b, and c is 0 or not."},{"Start":"04:06.040 ","End":"04:10.445","Text":"This was just a made-up hypothetical instance,"},{"Start":"04:10.445 ","End":"04:14.160","Text":"so let\u0027s solve this equation."},{"Start":"04:14.380 ","End":"04:18.485","Text":"Here it is, we\u0027re going to actually try and solve it."},{"Start":"04:18.485 ","End":"04:24.380","Text":"First, we do the scalar multiplications and end with this,"},{"Start":"04:24.380 ","End":"04:26.165","Text":"and now we\u0027re going to add"},{"Start":"04:26.165 ","End":"04:31.760","Text":"component-wise like the first component will be a plus 4b plus 7c,"},{"Start":"04:31.760 ","End":"04:34.484","Text":"and we\u0027ll compare that to 0."},{"Start":"04:34.484 ","End":"04:36.535","Text":"I\u0027ve skip the step."},{"Start":"04:36.535 ","End":"04:39.920","Text":"But you can see how we end up with these 3 equations,"},{"Start":"04:39.920 ","End":"04:42.040","Text":"I showed you how we got the first 1,"},{"Start":"04:42.040 ","End":"04:48.215","Text":"and we want to see if this has non-trivial solutions."},{"Start":"04:48.215 ","End":"04:52.325","Text":"I don\u0027t want to break the flow with tedious computations,"},{"Start":"04:52.325 ","End":"04:54.785","Text":"so let me just tell you."},{"Start":"04:54.785 ","End":"04:58.595","Text":"I\u0027ll show you that there is a non-trivial solution,"},{"Start":"04:58.595 ","End":"05:01.460","Text":"and I\u0027ll leave the computations to the end,"},{"Start":"05:01.460 ","End":"05:06.585","Text":"so I owe you to show you how I got this."},{"Start":"05:06.585 ","End":"05:09.260","Text":"In our particular case,"},{"Start":"05:09.260 ","End":"05:13.395","Text":"we do have a non-trivial solution there."},{"Start":"05:13.395 ","End":"05:16.755","Text":"Here, I just copied the equation and,"},{"Start":"05:16.755 ","End":"05:20.910","Text":"again, remember that this was u, this was v,"},{"Start":"05:20.910 ","End":"05:24.480","Text":"this was w. We do have actual numbers here,"},{"Start":"05:24.480 ","End":"05:26.274","Text":"I put a is 1,"},{"Start":"05:26.274 ","End":"05:30.350","Text":"b is minus 2, and c is 1,"},{"Start":"05:30.350 ","End":"05:38.155","Text":"so we actually have found a non-trivial solution for the equation."},{"Start":"05:38.155 ","End":"05:46.580","Text":"But we wanted it in the form like au plus bv plus cw equals 0,"},{"Start":"05:46.580 ","End":"05:48.940","Text":"so here it is."},{"Start":"05:48.940 ","End":"05:54.625","Text":"Let\u0027s get back to our formal definition of linear dependence,"},{"Start":"05:54.625 ","End":"05:59.330","Text":"and there I want to start off with a general situation with n vectors,"},{"Start":"05:59.330 ","End":"06:01.730","Text":"so let\u0027s start off with 3 vectors,"},{"Start":"06:01.730 ","End":"06:06.120","Text":"and then I\u0027ll generalize to any number of vectors."},{"Start":"06:06.280 ","End":"06:08.945","Text":"For 3 vectors,"},{"Start":"06:08.945 ","End":"06:14.060","Text":"I\u0027m pretty much restating what we had before that given 3 vectors, u, v, w,"},{"Start":"06:14.060 ","End":"06:20.110","Text":"if this equation has only the trivial solution,"},{"Start":"06:20.110 ","End":"06:23.865","Text":"then the vectors are linearly independent,"},{"Start":"06:23.865 ","End":"06:27.490","Text":"there we did dependent first and then independent this way here."},{"Start":"06:27.490 ","End":"06:30.275","Text":"This is the only solution,"},{"Start":"06:30.275 ","End":"06:32.410","Text":"trivial solution is always a solution."},{"Start":"06:32.410 ","End":"06:36.080","Text":"The only question is whether it\u0027s the only one or whether there were others."},{"Start":"06:36.210 ","End":"06:38.890","Text":"Here\u0027s the other side of the coin."},{"Start":"06:38.890 ","End":"06:43.839","Text":"If the equation has a solution which is not all 0,"},{"Start":"06:43.839 ","End":"06:48.460","Text":"it could be that just 1 of them is non-zero or 2 of them are non-zero,"},{"Start":"06:48.460 ","End":"06:50.170","Text":"or even 3 of them non-zero,"},{"Start":"06:50.170 ","End":"06:52.180","Text":"but not all 3 of them are 0,"},{"Start":"06:52.180 ","End":"06:53.770","Text":"at least 1 is non-zero."},{"Start":"06:53.770 ","End":"06:55.705","Text":"Then it\u0027s not trivial,"},{"Start":"06:55.705 ","End":"06:59.920","Text":"then the solution and the vectors turn out to be dependent."},{"Start":"06:59.920 ","End":"07:01.540","Text":"That\u0027s for 3 vectors,"},{"Start":"07:01.540 ","End":"07:05.420","Text":"and now I want to generalize a bit,"},{"Start":"07:05.420 ","End":"07:09.565","Text":"maybe not a bit let\u0027s generalize to any number of vectors."},{"Start":"07:09.565 ","End":"07:13.575","Text":"This time we take n vectors,"},{"Start":"07:13.575 ","End":"07:16.230","Text":"we need to use subscripts, u_1,"},{"Start":"07:16.230 ","End":"07:18.045","Text":"u_2 up to u_n,"},{"Start":"07:18.045 ","End":"07:20.975","Text":"and here\u0027s the equation that we\u0027re interested in."},{"Start":"07:20.975 ","End":"07:23.120","Text":"Put a coefficient in front of each 1."},{"Start":"07:23.120 ","End":"07:24.740","Text":"Take a linear combination."},{"Start":"07:24.740 ","End":"07:28.880","Text":"If this equation and the equation is in a_1 through a_n,"},{"Start":"07:28.880 ","End":"07:33.670","Text":"has only the trivial solution where all the a\u0027s are 0,"},{"Start":"07:33.670 ","End":"07:39.290","Text":"then the vectors u_1 through u_n are linearly independent,"},{"Start":"07:39.290 ","End":"07:43.055","Text":"or the set containing these vectors is linearly independent,"},{"Start":"07:43.055 ","End":"07:45.710","Text":"and on the other side,"},{"Start":"07:45.710 ","End":"07:50.795","Text":"if the equation has a solution where not all of them is 0,"},{"Start":"07:50.795 ","End":"07:52.670","Text":"could be some of them 0,"},{"Start":"07:52.670 ","End":"07:54.875","Text":"so long as not all of them are 0,"},{"Start":"07:54.875 ","End":"07:59.330","Text":"then this is a non-trivial solution and the vectors are linearly dependent"},{"Start":"07:59.330 ","End":"08:04.630","Text":"or the set of vectors is linearly dependent."},{"Start":"08:04.630 ","End":"08:07.590","Text":"That\u0027s the formal definition,"},{"Start":"08:07.590 ","End":"08:11.570","Text":"and it\u0027s equivalent to the 1 we had before where 1"},{"Start":"08:11.570 ","End":"08:15.380","Text":"is linear combination of some of the others,"},{"Start":"08:15.380 ","End":"08:16.910","Text":"because like I said before,"},{"Start":"08:16.910 ","End":"08:19.805","Text":"as soon as you have 1 of these which is non 0,"},{"Start":"08:19.805 ","End":"08:23.900","Text":"you can extract it as a combination of the others at putting everything to"},{"Start":"08:23.900 ","End":"08:28.750","Text":"the other side and dividing by the non 0 coefficient."},{"Start":"08:28.750 ","End":"08:32.045","Text":"We\u0027re not quite finished because I have this debt,"},{"Start":"08:32.045 ","End":"08:35.375","Text":"I didn\u0027t solve this for you."},{"Start":"08:35.375 ","End":"08:41.580","Text":"I wanted to find a non-trivial solution where not all of a, b, c is 0,"},{"Start":"08:41.580 ","End":"08:43.835","Text":"I remember I produced out of a hat,"},{"Start":"08:43.835 ","End":"08:49.295","Text":"rotated here faintly, a solution and non-trivial."},{"Start":"08:49.295 ","End":"08:54.435","Text":"Let\u0027s see how I could go about this."},{"Start":"08:54.435 ","End":"08:58.110","Text":"We convert the equation to matrix form."},{"Start":"08:58.110 ","End":"09:01.120","Text":"Now, I have to tell you that when it\u0027s homogeneous,"},{"Start":"09:01.120 ","End":"09:03.310","Text":"meaning the right-hand sides are all 0,"},{"Start":"09:03.310 ","End":"09:08.860","Text":"it\u0027s not customary to use the augmented matrix to separate on all 0 \u0027s"},{"Start":"09:08.860 ","End":"09:15.055","Text":"because these things are styled 0 and it\u0027s just the way it\u0027s is dragging these with you."},{"Start":"09:15.055 ","End":"09:16.270","Text":"But in this case,"},{"Start":"09:16.270 ","End":"09:19.405","Text":"we\u0027ll leave them in, it doesn\u0027t hurt."},{"Start":"09:19.405 ","End":"09:22.540","Text":"This is the matrix form,"},{"Start":"09:22.540 ","End":"09:27.065","Text":"the augmented, and now we do row operations."},{"Start":"09:27.065 ","End":"09:32.595","Text":"Subtract twice the first from the second and 3 times the first from the third."},{"Start":"09:32.595 ","End":"09:35.320","Text":"I\u0027m going to leave you to check if this is what we get,"},{"Start":"09:35.320 ","End":"09:42.095","Text":"and now we subtract twice this row from this row."},{"Start":"09:42.095 ","End":"09:46.845","Text":"We get this and all these are 0\u0027s so we just basically throw that out,"},{"Start":"09:46.845 ","End":"09:52.970","Text":"and now we want to go back from the matrix to the equation form,"},{"Start":"09:52.970 ","End":"09:55.539","Text":"and this is what we get."},{"Start":"09:55.539 ","End":"09:59.290","Text":"I put boxes around the a and the b"},{"Start":"09:59.290 ","End":"10:03.550","Text":"because the leading entries are the ones that become dependent."},{"Start":"10:03.550 ","End":"10:06.850","Text":"But the other variable c is free,"},{"Start":"10:06.850 ","End":"10:08.110","Text":"can be whatever you like."},{"Start":"10:08.110 ","End":"10:09.790","Text":"You can just let c be anything,"},{"Start":"10:09.790 ","End":"10:15.445","Text":"and then a and b are determined from c. When I say c equals anything,"},{"Start":"10:15.445 ","End":"10:20.535","Text":"not quite because the context c equals 0."},{"Start":"10:20.535 ","End":"10:24.020","Text":"It\u0027s possible to take c is 0,"},{"Start":"10:24.020 ","End":"10:25.295","Text":"will give me a solution,"},{"Start":"10:25.295 ","End":"10:29.930","Text":"but it\u0027s not the 1 I want because that will be the trivial solution, because if c is 0,"},{"Start":"10:29.930 ","End":"10:32.390","Text":"then for me a b is 0, and if c and b is 0,"},{"Start":"10:32.390 ","End":"10:35.960","Text":"then a is 0, so anything except c equals 0."},{"Start":"10:35.960 ","End":"10:42.395","Text":"In fact, I chose c equals 1 in the solution I showed you earlier,"},{"Start":"10:42.395 ","End":"10:45.740","Text":"and that if you check by back substitution,"},{"Start":"10:45.740 ","End":"10:49.030","Text":"gives us b is minus 2 from this equation,"},{"Start":"10:49.030 ","End":"10:51.895","Text":"and then from this 1 we get a is 1,"},{"Start":"10:51.895 ","End":"10:54.270","Text":"and recall that a, b, and c,"},{"Start":"10:54.270 ","End":"10:58.305","Text":"are what we wanted to plug in to this equation,"},{"Start":"10:58.305 ","End":"11:04.510","Text":"so this is what we get, which really is a non-trivial solution for this,"},{"Start":"11:04.510 ","End":"11:06.950","Text":"and like I said, if you choose any 1,"},{"Start":"11:06.950 ","End":"11:08.060","Text":"well they\u0027re all non 0,"},{"Start":"11:08.060 ","End":"11:10.340","Text":"say this 1, then I can isolate,"},{"Start":"11:10.340 ","End":"11:12.260","Text":"it\u0027s the easiest, 1, 2,"},{"Start":"11:12.260 ","End":"11:16.370","Text":"3 as a linear combination of the others which shows that"},{"Start":"11:16.370 ","End":"11:21.035","Text":"even in the previous definition that 1 is a linear combination of the others."},{"Start":"11:21.035 ","End":"11:26.340","Text":"Anyway, I\u0027ve cleared my debt and we are really done."}],"ID":25833},{"Watched":false,"Name":"Sufficient Condition for Linear Independence","Duration":"4m 40s","ChapterTopicVideoID":25067,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25067.jpeg","UploadDate":"2021-06-27T07:05:06.2370000","DurationForVideoObject":"PT4M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"There\u0027s a small proposition,"},{"Start":"00:03.045 ","End":"00:07.050","Text":"which is very useful at times."},{"Start":"00:07.050 ","End":"00:11.820","Text":"It\u0027s a sufficient condition for linear dependence that under"},{"Start":"00:11.820 ","End":"00:17.415","Text":"certain conditions we can conclude that set of vectors is dependent."},{"Start":"00:17.415 ","End":"00:22.230","Text":"There is a more general formulation for vector spaces in general,"},{"Start":"00:22.230 ","End":"00:23.760","Text":"but we won\u0027t bring that here."},{"Start":"00:23.760 ","End":"00:27.720","Text":"I\u0027ll just show you how it expresses itself in"},{"Start":"00:27.720 ","End":"00:32.670","Text":"the case of R^n and in 2 other examples that you\u0027ll see."},{"Start":"00:32.670 ","End":"00:39.020","Text":"In R^n, the proposition says that if our set contains more than n vectors,"},{"Start":"00:39.020 ","End":"00:42.675","Text":"where this n is the same as this n,"},{"Start":"00:42.675 ","End":"00:46.950","Text":"then the set is linearly independent."},{"Start":"00:46.950 ","End":"00:50.395","Text":"For example, in R^2,"},{"Start":"00:50.395 ","End":"00:52.160","Text":"which is 2D space,"},{"Start":"00:52.160 ","End":"00:54.140","Text":"so we have vectors with 2 components,"},{"Start":"00:54.140 ","End":"00:56.240","Text":"if we take more than 2, say,"},{"Start":"00:56.240 ","End":"00:58.370","Text":"3 of them as we have here,"},{"Start":"00:58.370 ","End":"01:02.390","Text":"then these I know are going to be linearly dependent."},{"Start":"01:02.390 ","End":"01:07.170","Text":"You don\u0027t even have to do all the computations because 3 is bigger than 2."},{"Start":"01:08.000 ","End":"01:13.680","Text":"In 3D space, if I have something bigger than 3,"},{"Start":"01:13.680 ","End":"01:22.520","Text":"say 4 vectors, then I know that these are going to be linearly dependent."},{"Start":"01:22.520 ","End":"01:24.500","Text":"These 4 are all different,"},{"Start":"01:24.500 ","End":"01:26.990","Text":"even though there\u0027s a parameter here, b,"},{"Start":"01:26.990 ","End":"01:28.370","Text":"and another parameter here,"},{"Start":"01:28.370 ","End":"01:30.500","Text":"c. Doesn\u0027t matter what b and c are,"},{"Start":"01:30.500 ","End":"01:33.780","Text":"this is still going to be different from all the rest."},{"Start":"01:34.520 ","End":"01:40.515","Text":"This one also has a 1 if I let b equals 1, but still,"},{"Start":"01:40.515 ","End":"01:45.270","Text":"it\u0027s going to be different all 4 them no matter what b and c are,"},{"Start":"01:45.270 ","End":"01:48.460","Text":"so they\u0027re going to be dependent."},{"Start":"01:49.790 ","End":"01:55.560","Text":"A third example of this is in our 4D space"},{"Start":"01:55.560 ","End":"02:00.790","Text":"and for that I need at least 5 vectors so here\u0027s an example with 5 vectors."},{"Start":"02:00.790 ","End":"02:06.245","Text":"I know without doing any computation that they are linearly dependent."},{"Start":"02:06.245 ","End":"02:12.295","Text":"Now, we can extend this proposition to other spaces."},{"Start":"02:12.295 ","End":"02:19.730","Text":"If our vector space is this the set of all m by n matrices over"},{"Start":"02:19.730 ","End":"02:28.865","Text":"the reels then the rule is that any set of more than m times n vectors is dependent."},{"Start":"02:28.865 ","End":"02:33.520","Text":"Another specific example, let\u0027s take 2 by 2 matrices,"},{"Start":"02:33.520 ","End":"02:38.055","Text":"and here mn is 2 times 2 is 4."},{"Start":"02:38.055 ","End":"02:43.760","Text":"If I have more than 4 matrices and here I have 1, 2, 3, 4,"},{"Start":"02:43.760 ","End":"02:51.960","Text":"5, then these are going to be linearly dependent because this 5 is bigger than 2 times 2."},{"Start":"02:51.960 ","End":"02:58.160","Text":"It\u0027s similar to the previous one with n equals 4 because we already showed how we can"},{"Start":"02:58.160 ","End":"03:05.540","Text":"convert from 2 by 2 matrices to 4D vectors."},{"Start":"03:05.540 ","End":"03:07.715","Text":"Remember we had a snake method."},{"Start":"03:07.715 ","End":"03:11.160","Text":"Anyway, if you don\u0027t remember that it doesn\u0027t matter."},{"Start":"03:11.260 ","End":"03:18.065","Text":"I want to bring 1 more example from another kind of vector space of polynomials."},{"Start":"03:18.065 ","End":"03:21.375","Text":"Remember that when we write P_n of R,"},{"Start":"03:21.375 ","End":"03:27.725","Text":"this n is the degree or maximal degree of the polynomial."},{"Start":"03:27.725 ","End":"03:33.630","Text":"It\u0027s the set of all polynomials up to degree n then could be"},{"Start":"03:33.630 ","End":"03:39.815","Text":"less than n. The rule here is that if we have a set of more than n plus 1,"},{"Start":"03:39.815 ","End":"03:42.320","Text":"not the same as n, here is n plus 1,"},{"Start":"03:42.320 ","End":"03:45.445","Text":"then we\u0027re going to be linearly dependent."},{"Start":"03:45.445 ","End":"03:52.940","Text":"If I write P_2 of R polynomials with real coefficients of degree at most 2,"},{"Start":"03:52.940 ","End":"03:59.540","Text":"basically it means the ax squared plus bx plus c,"},{"Start":"03:59.540 ","End":"04:03.620","Text":"the set of all those a and b could be 0,"},{"Start":"04:03.620 ","End":"04:06.325","Text":"could be degree less than 2."},{"Start":"04:06.325 ","End":"04:10.730","Text":"You can see where the n plus 1 comes in because of degree 2,"},{"Start":"04:10.730 ","End":"04:14.880","Text":"there are actually 3 real numbers here."},{"Start":"04:15.010 ","End":"04:17.450","Text":"Anyway, to be bigger than 3,"},{"Start":"04:17.450 ","End":"04:22.010","Text":"we have to be at least 4 so here indeed are 4,"},{"Start":"04:22.010 ","End":"04:23.690","Text":"1, 2, 3,"},{"Start":"04:23.690 ","End":"04:27.070","Text":"4 polynomials of degree up to 2."},{"Start":"04:27.070 ","End":"04:30.305","Text":"You know that these are going to be linearly dependent"},{"Start":"04:30.305 ","End":"04:34.220","Text":"without any computations because of this proposition."},{"Start":"04:34.220 ","End":"04:36.305","Text":"That\u0027s the last example,"},{"Start":"04:36.305 ","End":"04:40.530","Text":"I\u0027ll finish the clip here."}],"ID":25835},{"Watched":false,"Name":"Exercise 6","Duration":"3m ","ChapterTopicVideoID":25068,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25068.jpeg","UploadDate":"2021-06-27T07:05:22.0200000","DurationForVideoObject":"PT3M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.255","Text":"I will take the same 3 vectors we had before;"},{"Start":"00:03.255 ","End":"00:06.000","Text":"the ones that we keep using u,"},{"Start":"00:06.000 ","End":"00:07.380","Text":"v, and w as 1, 2,"},{"Start":"00:07.380 ","End":"00:09.300","Text":"3, 4, 5, 6, 7, 8, 9."},{"Start":"00:09.300 ","End":"00:13.380","Text":"You want to check if they\u0027re linearly dependent, and more than that,"},{"Start":"00:13.380 ","End":"00:15.270","Text":"if they are linearly dependent,"},{"Start":"00:15.270 ","End":"00:21.375","Text":"to actually give 1 of them as a linear combination of some of the others."},{"Start":"00:21.375 ","End":"00:23.975","Text":"When I put them in a matrix,"},{"Start":"00:23.975 ","End":"00:26.644","Text":"this is the variation I mentioned."},{"Start":"00:26.644 ","End":"00:31.910","Text":"This part, 1,2,3,4,5,6,7,8,9 are the vectors arranged,"},{"Start":"00:31.910 ","End":"00:35.285","Text":"but I actually take an augmented matrix,"},{"Start":"00:35.285 ","End":"00:36.920","Text":"which means I have an extra bit."},{"Start":"00:36.920 ","End":"00:39.470","Text":"Here I put u, v, and w,"},{"Start":"00:39.470 ","End":"00:45.850","Text":"just the names for the 3 vectors in this case,"},{"Start":"00:45.850 ","End":"00:51.795","Text":"and then we get started on the row operations to bring to echelon form on form."},{"Start":"00:51.795 ","End":"00:56.400","Text":"What we want to do is subtract 4 times this row from this row,"},{"Start":"00:56.400 ","End":"00:58.745","Text":"7 times this row from this row."},{"Start":"00:58.745 ","End":"01:04.520","Text":"This is the description of the operations and this is the result."},{"Start":"01:04.520 ","End":"01:06.980","Text":"Notice, I don\u0027t just work on the numbers,"},{"Start":"01:06.980 ","End":"01:10.430","Text":"but also on u, v, w. For example,"},{"Start":"01:10.430 ","End":"01:14.985","Text":"here when I subtracted 7 times first row from third row,"},{"Start":"01:14.985 ","End":"01:19.300","Text":"7 times this from this gives us w minus 7u."},{"Start":"01:19.300 ","End":"01:26.850","Text":"Next, I\u0027m going to get a 0 here by noticing that twice this makes it minus 6 something,"},{"Start":"01:26.850 ","End":"01:31.535","Text":"so I can subtract twice the second row from the third row."},{"Start":"01:31.535 ","End":"01:34.310","Text":"If we do that,"},{"Start":"01:34.310 ","End":"01:37.510","Text":"let\u0027s see what we get."},{"Start":"01:37.510 ","End":"01:43.340","Text":"This, which is good because not only do we get a 0 here,"},{"Start":"01:43.340 ","End":"01:46.510","Text":"but we\u0027re also lucky we got a 0 here."},{"Start":"01:46.510 ","End":"01:49.310","Text":"Here got a little bit complicated,"},{"Start":"01:49.310 ","End":"01:51.530","Text":"but we\u0027ll soon simplify that."},{"Start":"01:51.530 ","End":"01:53.780","Text":"Because of these 0s,"},{"Start":"01:53.780 ","End":"02:02.990","Text":"we can straight away say that the set of vectors is linearly dependent."},{"Start":"02:02.990 ","End":"02:09.145","Text":"This is going to show us how to actually write the dependency."},{"Start":"02:09.145 ","End":"02:11.460","Text":"We got here the 0 0 0,"},{"Start":"02:11.460 ","End":"02:13.755","Text":"which is the 0 vector."},{"Start":"02:13.755 ","End":"02:16.245","Text":"Sometimes I write it with an underline."},{"Start":"02:16.245 ","End":"02:20.594","Text":"Anyway, expand and collect like terms."},{"Start":"02:20.594 ","End":"02:22.140","Text":"We\u0027ve got to check it."},{"Start":"02:22.140 ","End":"02:25.410","Text":"W minus 2v plus u is 0,"},{"Start":"02:25.410 ","End":"02:30.660","Text":"and then I can isolate w as 2v minus u."},{"Start":"02:30.660 ","End":"02:31.685","Text":"It\u0027s not the only choice,"},{"Start":"02:31.685 ","End":"02:34.660","Text":"I could have actually isolated u or v,"},{"Start":"02:34.660 ","End":"02:37.420","Text":"but just have to choose 1 of them."},{"Start":"02:37.420 ","End":"02:44.490","Text":"Now we have that 1 of the vectors is a linear combination of other vectors in the set,"},{"Start":"02:44.490 ","End":"02:48.130","Text":"so it\u0027s linearly dependent as we said."},{"Start":"02:48.130 ","End":"02:54.980","Text":"We even showed the reason of the actual linear combination."},{"Start":"02:55.070 ","End":"02:57.900","Text":"I think it\u0027s time to take a break."},{"Start":"02:57.900 ","End":"03:00.580","Text":"We\u0027ll continue after the break."}],"ID":25836},{"Watched":false,"Name":"Exercise 7","Duration":"1m 19s","ChapterTopicVideoID":25064,"CourseChapterTopicPlaylistID":31674,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25064.jpeg","UploadDate":"2021-06-27T07:02:39.0000000","DurationForVideoObject":"PT1M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.880","Text":"The next exercise looks very much like the previous 1,"},{"Start":"00:02.880 ","End":"00:07.215","Text":"but notice that there\u0027s a 10 here and there was a 9 here before,"},{"Start":"00:07.215 ","End":"00:08.490","Text":"otherwise the same,"},{"Start":"00:08.490 ","End":"00:11.970","Text":"so we tackle it similarly."},{"Start":"00:11.970 ","End":"00:19.170","Text":"We put these vectors into a matrix and I\u0027m using the augmented version,"},{"Start":"00:19.170 ","End":"00:22.500","Text":"as I explained before why."},{"Start":"00:22.500 ","End":"00:25.725","Text":"Row operations, just like before."},{"Start":"00:25.725 ","End":"00:27.765","Text":"I won\u0027t repeat."},{"Start":"00:27.765 ","End":"00:32.414","Text":"What we get this time is this,"},{"Start":"00:32.414 ","End":"00:34.620","Text":"and this is similar to what we had before,"},{"Start":"00:34.620 ","End":"00:37.730","Text":"but then we had minus 12 here."},{"Start":"00:37.730 ","End":"00:42.260","Text":"That makes a big difference because now when we do"},{"Start":"00:42.260 ","End":"00:48.005","Text":"the last row operations of subtracting twice this from this,"},{"Start":"00:48.005 ","End":"00:51.380","Text":"we don\u0027t get a 0 here like we did before."},{"Start":"00:51.380 ","End":"00:52.790","Text":"We get a 1,"},{"Start":"00:52.790 ","End":"00:57.900","Text":"which means that we finished the row echelon form."},{"Start":"00:58.330 ","End":"01:02.015","Text":"But we didn\u0027t get a row with all zeros."},{"Start":"01:02.015 ","End":"01:03.469","Text":"And when that happens,"},{"Start":"01:03.469 ","End":"01:07.940","Text":"then we know that the vectors U, V, and W,"},{"Start":"01:07.940 ","End":"01:10.070","Text":"or the Set A,"},{"Start":"01:10.070 ","End":"01:15.275","Text":"is linearly independent because we didn\u0027t get zeros."},{"Start":"01:15.275 ","End":"01:17.030","Text":"Okay, that\u0027s this example."},{"Start":"01:17.030 ","End":"01:20.609","Text":"Now let\u0027s move on to something a bit more formal."}],"ID":25832}],"Thumbnail":null,"ID":31674},{"Name":"Linear Dependence in R^n","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"A Linearly Dependent Set of Vectors","Duration":"5m 2s","ChapterTopicVideoID":13521,"CourseChapterTopicPlaylistID":31675,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/13521.jpeg","UploadDate":"2018-09-05T04:58:51.0400000","DurationForVideoObject":"PT5M2S","Description":null,"VideoComments":[],"Subtitles":[],"ID":14161},{"Watched":false,"Name":"Exercise 1","Duration":"3m ","ChapterTopicVideoID":13522,"CourseChapterTopicPlaylistID":31675,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/13522.jpeg","UploadDate":"2018-09-05T04:59:06.3200000","DurationForVideoObject":"PT3M","Description":null,"VideoComments":[],"Subtitles":[],"ID":14162},{"Watched":false,"Name":"Exercise 2","Duration":"1m 19s","ChapterTopicVideoID":13523,"CourseChapterTopicPlaylistID":31675,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/13523.jpeg","UploadDate":"2018-09-05T04:59:12.9930000","DurationForVideoObject":"PT1M19S","Description":null,"VideoComments":[],"Subtitles":[],"ID":14163},{"Watched":false,"Name":"Linear Dependence and Independence of a Set of Vectors","Duration":"11m 25s","ChapterTopicVideoID":13524,"CourseChapterTopicPlaylistID":31675,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/13524.jpeg","UploadDate":"2018-09-05T05:00:20.8070000","DurationForVideoObject":"PT11M25S","Description":null,"VideoComments":[],"Subtitles":[],"ID":14164},{"Watched":false,"Name":"Proposition - A Sufficient Condition for Linear Independence","Duration":"4m 40s","ChapterTopicVideoID":13525,"CourseChapterTopicPlaylistID":31675,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/13525.jpeg","UploadDate":"2018-09-05T22:28:53.5170000","DurationForVideoObject":"PT4M40S","Description":null,"VideoComments":[],"Subtitles":[],"ID":14165}],"Thumbnail":null,"ID":31675},{"Name":"Basis for Rn","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Definition and Examples","Duration":"15m 35s","ChapterTopicVideoID":9911,"CourseChapterTopicPlaylistID":31676,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9911.jpeg","UploadDate":"2019-06-27T01:40:22.8470000","DurationForVideoObject":"PT15M35S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"In this clip, I\u0027m going to be introducing the concept of"},{"Start":"00:03.150 ","End":"00:06.765","Text":"a basis for the vector space R^n."},{"Start":"00:06.765 ","End":"00:13.895","Text":"By the way, the plural for bases is basis, not bases."},{"Start":"00:13.895 ","End":"00:16.290","Text":"Anyway, but before that,"},{"Start":"00:16.290 ","End":"00:18.960","Text":"I have to introduce another concept."},{"Start":"00:18.960 ","End":"00:22.485","Text":"That is the concept of a spanning set."},{"Start":"00:22.485 ","End":"00:28.230","Text":"It\u0027s related to the concept of span that we talked about in a previous clip."},{"Start":"00:28.230 ","End":"00:31.845","Text":"Let\u0027s start with a specific n. Lets take n equals 3."},{"Start":"00:31.845 ","End":"00:35.580","Text":"We\u0027re considering 3D vectors,"},{"Start":"00:35.580 ","End":"00:36.810","Text":"they are 3,"},{"Start":"00:36.810 ","End":"00:42.505","Text":"and I\u0027ve just listed a few examples of this infinite set,"},{"Start":"00:42.505 ","End":"00:46.535","Text":"each of them being triples of 3 numbers."},{"Start":"00:46.535 ","End":"00:50.960","Text":"Now I\u0027d like to consider a specific set, call it E,"},{"Start":"00:50.960 ","End":"00:53.450","Text":"containing these 3 vectors, 1, 0,"},{"Start":"00:53.450 ","End":"00:57.540","Text":"0, 0, 1, 0 and 0, 0, 1."},{"Start":"00:57.650 ","End":"00:59.910","Text":"I\u0027m going to ask a question,"},{"Start":"00:59.910 ","End":"01:04.070","Text":"would it be true to say that every vector in R^3 can be"},{"Start":"01:04.070 ","End":"01:09.155","Text":"written as a linear combination of these 3 vectors?"},{"Start":"01:09.155 ","End":"01:11.870","Text":"It turns out that the answer is yes."},{"Start":"01:11.870 ","End":"01:14.000","Text":"I\u0027ll take an example,"},{"Start":"01:14.000 ","End":"01:17.285","Text":"the vector 4, 5, 6."},{"Start":"01:17.285 ","End":"01:20.810","Text":"If you look, I can write it as 4 times 1,"},{"Start":"01:20.810 ","End":"01:22.940","Text":"0, 0 plus 5 times 0, 1,"},{"Start":"01:22.940 ","End":"01:24.920","Text":"0 plus 6 times 0, 0,"},{"Start":"01:24.920 ","End":"01:28.790","Text":"1 is to quick computation will show you that this is so."},{"Start":"01:28.790 ","End":"01:31.910","Text":"But it\u0027s also clear that this can be generalized."},{"Start":"01:31.910 ","End":"01:36.265","Text":"I mean, nothing special about 4, 5, 6."},{"Start":"01:36.265 ","End":"01:38.820","Text":"I can replace the 4, 5,"},{"Start":"01:38.820 ","End":"01:40.440","Text":"6 by an arbitrary a,"},{"Start":"01:40.440 ","End":"01:42.405","Text":"b, c, and do the same thing."},{"Start":"01:42.405 ","End":"01:47.515","Text":"You see that any vector is a linear combination of these 3."},{"Start":"01:47.515 ","End":"01:50.330","Text":"When this happens for a set E,"},{"Start":"01:50.330 ","End":"01:56.760","Text":"that every a vector is a linear combination of it,"},{"Start":"01:56.760 ","End":"02:00.230","Text":"then we say that the set E spans,"},{"Start":"02:00.230 ","End":"02:01.520","Text":"in this case R^3,"},{"Start":"02:01.520 ","End":"02:04.550","Text":"and in general it would be R^n."},{"Start":"02:04.550 ","End":"02:07.760","Text":"This relates to a previous concept we learned."},{"Start":"02:07.760 ","End":"02:09.350","Text":"We learned SP,"},{"Start":"02:09.350 ","End":"02:11.750","Text":"the span of the set."},{"Start":"02:11.750 ","End":"02:15.210","Text":"We can also write it this way."},{"Start":"02:15.520 ","End":"02:21.755","Text":"Now what I want to do is instead of E to take a different set of vectors,"},{"Start":"02:21.755 ","End":"02:24.310","Text":"and I\u0027ll call it A."},{"Start":"02:24.310 ","End":"02:28.140","Text":"A consists of 4 vectors."},{"Start":"02:28.140 ","End":"02:31.910","Text":"You probably will notice that the first 3 are exactly what we had"},{"Start":"02:31.910 ","End":"02:36.095","Text":"in E. But let me ask the same question as I asked before."},{"Start":"02:36.095 ","End":"02:38.060","Text":"The question I asked about E,"},{"Start":"02:38.060 ","End":"02:40.160","Text":"I\u0027m asking the same question about A."},{"Start":"02:40.160 ","End":"02:46.090","Text":"If I can write every vector in R^3 as a linear combination of the vectors in A."},{"Start":"02:46.090 ","End":"02:49.085","Text":"Once again, the answer is yes."},{"Start":"02:49.085 ","End":"02:57.485","Text":"Turns out that throwing in extra vectors into a spanning set is not going to hurt."},{"Start":"02:57.485 ","End":"02:59.195","Text":"Because I can always,"},{"Start":"02:59.195 ","End":"03:01.249","Text":"well, you\u0027ll see from the example,"},{"Start":"03:01.249 ","End":"03:04.070","Text":"if I take the same vector I took before,"},{"Start":"03:04.070 ","End":"03:05.870","Text":"in the example 4, 5, 6,"},{"Start":"03:05.870 ","End":"03:07.070","Text":"we had 4 times this,"},{"Start":"03:07.070 ","End":"03:08.690","Text":"5 times this, 6 times this."},{"Start":"03:08.690 ","End":"03:10.400","Text":"Now we have this new vector."},{"Start":"03:10.400 ","End":"03:12.425","Text":"I can just put 0 in front of it,"},{"Start":"03:12.425 ","End":"03:16.760","Text":"which shows that adding an extra vector, or more than 1 doesn\u0027t hurt"},{"Start":"03:16.760 ","End":"03:21.650","Text":"the property, because I can always put zeros in front of the extra ones."},{"Start":"03:21.650 ","End":"03:23.420","Text":"Just like before, instead of 4, 5,"},{"Start":"03:23.420 ","End":"03:25.310","Text":"6, I could generalize it to A, B,"},{"Start":"03:25.310 ","End":"03:35.430","Text":"C. You see that the set a also spans R^3, and we can write that the span of A is R^3."},{"Start":"03:35.780 ","End":"03:39.830","Text":"Back to the concept of a basis of R^n."},{"Start":"03:39.830 ","End":"03:42.665","Text":"We just talked about a spanning set."},{"Start":"03:42.665 ","End":"03:44.845","Text":"I\u0027m getting close to this."},{"Start":"03:44.845 ","End":"03:48.770","Text":"To remind you, we just took a look at 2 sets in R^3."},{"Start":"03:48.770 ","End":"03:51.470","Text":"Both of them span R^3."},{"Start":"03:51.470 ","End":"03:56.740","Text":"1 of them was E, which was this, and 1 of them was A, which was this."},{"Start":"03:56.740 ","End":"04:02.185","Text":"There is an essential difference between E, and A that I\u0027d like to stress."},{"Start":"04:02.185 ","End":"04:07.690","Text":"That difference is that the members of E,"},{"Start":"04:07.690 ","End":"04:08.885","Text":"the vectors in it,"},{"Start":"04:08.885 ","End":"04:17.785","Text":"are linearly independent, but the vectors in A are linearly dependent."},{"Start":"04:17.785 ","End":"04:20.594","Text":"I think that\u0027s fairly clear."},{"Start":"04:20.594 ","End":"04:22.640","Text":"1 of the definitions of"},{"Start":"04:22.640 ","End":"04:29.510","Text":"linearly independent is that no 1 is a linear combination of the others."},{"Start":"04:29.510 ","End":"04:32.090","Text":"I think in this case, that\u0027s fairly clear."},{"Start":"04:32.090 ","End":"04:34.120","Text":"For example, you can\u0027t get 0, 0,"},{"Start":"04:34.120 ","End":"04:38.075","Text":"1 as a combination of these 2."},{"Start":"04:38.075 ","End":"04:41.475","Text":"Similarly, or contrarily,"},{"Start":"04:41.475 ","End":"04:44.780","Text":"A is linearly dependent, because you can"},{"Start":"04:44.780 ","End":"04:47.930","Text":"get 1 as a linear combination of the other.s For example,"},{"Start":"04:47.930 ","End":"04:49.520","Text":"this 1, 1, 2,"},{"Start":"04:49.520 ","End":"04:53.540","Text":"3 is certainly a combination of the first 3,"},{"Start":"04:53.540 ","End":"04:56.465","Text":"because everything is a combination of the first 3."},{"Start":"04:56.465 ","End":"05:01.355","Text":"This turns out to be an essential difference, and leads us to a definition."},{"Start":"05:01.355 ","End":"05:05.105","Text":"Finally coming to the term basis."},{"Start":"05:05.105 ","End":"05:09.870","Text":"E spans R^3 and so does A,"},{"Start":"05:09.870 ","End":"05:12.410","Text":"but E is linearly independent."},{"Start":"05:12.410 ","End":"05:18.200","Text":"Now when you have these 2 properties that it spans, and it\u0027s linearly independent,"},{"Start":"05:18.200 ","End":"05:20.815","Text":"then that makes it a basis."},{"Start":"05:20.815 ","End":"05:24.185","Text":"Let me just give a more general definition."},{"Start":"05:24.185 ","End":"05:30.215","Text":"We\u0027ll jump from R^3 to R^n, because there\u0027s nothing special about the number 3,"},{"Start":"05:30.215 ","End":"05:33.980","Text":"is that if we have a set of vectors from R^n, and you can think"},{"Start":"05:33.980 ","End":"05:38.055","Text":"of R^3, and you can think of this example."},{"Start":"05:38.055 ","End":"05:44.220","Text":"It\u0027s a basis of R^n if it has the 2 properties that it spans R^n,"},{"Start":"05:44.220 ","End":"05:46.620","Text":"and it\u0027s linearly independent."},{"Start":"05:46.620 ","End":"05:49.375","Text":"In our example, E had these 2 properties,"},{"Start":"05:49.375 ","End":"05:52.145","Text":"that it spans R^3, linearly dependent."},{"Start":"05:52.145 ","End":"05:54.745","Text":"It\u0027s a basis for R^3."},{"Start":"05:54.745 ","End":"05:56.945","Text":"Now let\u0027s do an exercise,"},{"Start":"05:56.945 ","End":"06:00.125","Text":"typical exercise to see if we understood."},{"Start":"06:00.125 ","End":"06:02.180","Text":"We have a set B,"},{"Start":"06:02.180 ","End":"06:05.265","Text":"capital B, consisting of these 2 vectors."},{"Start":"06:05.265 ","End":"06:10.415","Text":"We have to check whether it\u0027s a basis of R^2 or not."},{"Start":"06:10.415 ","End":"06:14.314","Text":"Remember, for a basis you have to check 2 things."},{"Start":"06:14.314 ","End":"06:21.070","Text":"Number 1 is that B spans R^2, and later we\u0027ll show that B is linearly independent."},{"Start":"06:21.070 ","End":"06:24.450","Text":"The way we show that it spans R^2,"},{"Start":"06:24.450 ","End":"06:27.850","Text":"is that I\u0027m going to show you that if you give me any a,"},{"Start":"06:27.850 ","End":"06:30.040","Text":"b from R^2,"},{"Start":"06:30.040 ","End":"06:33.310","Text":"I can always find an x, and a y so that a,"},{"Start":"06:33.310 ","End":"06:40.240","Text":"b is a linear combination of these 2 vectors using x and y as scalars here."},{"Start":"06:40.240 ","End":"06:42.490","Text":"I have show you that for every a, b,"},{"Start":"06:42.490 ","End":"06:45.590","Text":"I can find the solution x, y."},{"Start":"06:46.280 ","End":"06:51.040","Text":"Let\u0027s multiply the scalars, and get this."},{"Start":"06:51.040 ","End":"06:55.535","Text":"Next, the addition of these 2 gives us this."},{"Start":"06:55.535 ","End":"06:57.600","Text":"Then from the equality of vectors,"},{"Start":"06:57.600 ","End":"07:00.190","Text":"it means that they\u0027re equal component wise,"},{"Start":"07:00.190 ","End":"07:03.415","Text":"and so we get this pair of equations."},{"Start":"07:03.415 ","End":"07:07.690","Text":"The unknowns are x and y. I\u0027m assuming that a,"},{"Start":"07:07.690 ","End":"07:09.110","Text":"b are given arbitrary,"},{"Start":"07:09.110 ","End":"07:12.830","Text":"but given. Here we are again."},{"Start":"07:12.830 ","End":"07:16.400","Text":"I\u0027m going to solve it using matrices although that\u0027s not"},{"Start":"07:16.400 ","End":"07:20.530","Text":"usually the way you would go for such a small system of linear equations."},{"Start":"07:20.530 ","End":"07:23.525","Text":"Here\u0027s the augmented matrix for this system,"},{"Start":"07:23.525 ","End":"07:26.195","Text":"and we want to bring it into row echelon form."},{"Start":"07:26.195 ","End":"07:32.060","Text":"What we\u0027ll do is subtract twice the first row from the second."},{"Start":"07:32.060 ","End":"07:34.070","Text":"If you want it in notation,"},{"Start":"07:34.070 ","End":"07:39.455","Text":"I would say R^2 minus twice R^1 goes into R^2."},{"Start":"07:39.455 ","End":"07:41.810","Text":"The event we get the 0 here."},{"Start":"07:41.810 ","End":"07:44.510","Text":"Also, don\u0027t forget to do the right hand side,"},{"Start":"07:44.510 ","End":"07:45.710","Text":"to the right of the separators."},{"Start":"07:45.710 ","End":"07:48.090","Text":"Here we get b minus 2a"},{"Start":"07:48.270 ","End":"07:52.870","Text":"Now back from the matrix to the system of equations,"},{"Start":"07:52.870 ","End":"07:54.625","Text":"this is what we get."},{"Start":"07:54.625 ","End":"08:02.125","Text":"From the last 1, we can easily get y just by dividing by minus 2."},{"Start":"08:02.125 ","End":"08:08.470","Text":"Once we\u0027ve got y, we can substitute in the first equation."},{"Start":"08:08.470 ","End":"08:10.630","Text":"I\u0027ll spare you the boring details,"},{"Start":"08:10.630 ","End":"08:12.865","Text":"x comes out to be this."},{"Start":"08:12.865 ","End":"08:15.700","Text":"If you look back to where x and y came from,"},{"Start":"08:15.700 ","End":"08:19.585","Text":"we want today b as x times this plus y times this."},{"Start":"08:19.585 ","End":"08:24.850","Text":"This is what we get, and this in essence shows us that for any a and b,"},{"Start":"08:24.850 ","End":"08:28.360","Text":"we can find an x and an y, because we can always compute this,"},{"Start":"08:28.360 ","End":"08:31.615","Text":"there\u0027s no 0s in the denominator, or anything."},{"Start":"08:31.615 ","End":"08:38.455","Text":"Just for instance, if you threw at me the vector 4,5,"},{"Start":"08:38.455 ","End":"08:41.185","Text":"I would put a equals 4,"},{"Start":"08:41.185 ","End":"08:42.640","Text":"b equals 5 here,"},{"Start":"08:42.640 ","End":"08:44.785","Text":"and if you do the computations,"},{"Start":"08:44.785 ","End":"08:50.455","Text":"you\u0027ll get 1 over minus 2 here, and 3 over 2 here."},{"Start":"08:50.455 ","End":"08:58.600","Text":"It shows you that this 4,5 is a combination of the 2 vectors in set B."},{"Start":"08:58.600 ","End":"09:04.330","Text":"Another example, minus 1,2 similarly,"},{"Start":"09:04.330 ","End":"09:10.330","Text":"that is a equals minus 1b equals 2 here gives us this."},{"Start":"09:10.330 ","End":"09:14.680","Text":"It can also even have a double-check by just multiplying this"},{"Start":"09:14.680 ","End":"09:18.640","Text":"out and seeing that this really gives us this."},{"Start":"09:18.640 ","End":"09:24.760","Text":"I think we\u0027ve demonstrated that these 2 vectors really do span 2."},{"Start":"09:24.760 ","End":"09:28.480","Text":"But remember there was a part 2 that we"},{"Start":"09:28.480 ","End":"09:32.560","Text":"also have to show that the set B is linearly independent."},{"Start":"09:32.560 ","End":"09:38.650","Text":"One way of showing that these 2 are linearly independent is to take the equation of"},{"Start":"09:38.650 ","End":"09:45.115","Text":"the linear combination if this is 0, and show that necessarily the coefficients,"},{"Start":"09:45.115 ","End":"09:46.630","Text":"the scalars are both 0,"},{"Start":"09:46.630 ","End":"09:50.155","Text":"that only the trivial linear combination can give us 0."},{"Start":"09:50.155 ","End":"09:54.295","Text":"Well, let\u0027s see. Multiply the scalars,"},{"Start":"09:54.295 ","End":"09:57.400","Text":"do the vector addition,"},{"Start":"09:57.400 ","End":"09:59.260","Text":"compare the components,"},{"Start":"09:59.260 ","End":"10:01.855","Text":"first component, second component."},{"Start":"10:01.855 ","End":"10:03.760","Text":"I won\u0027t use matrices here,"},{"Start":"10:03.760 ","End":"10:05.020","Text":"just go ahead, and solve it."},{"Start":"10:05.020 ","End":"10:10.240","Text":"Subtract twice the first row from the second row that will give us this."},{"Start":"10:10.240 ","End":"10:13.570","Text":"From that, we can easily see that y is 0."},{"Start":"10:13.570 ","End":"10:17.740","Text":"If y is 0, plugging that in here gives us that x is 0 also,"},{"Start":"10:17.740 ","End":"10:24.490","Text":"the only linear combination that gives the 0 vector is the trivial combination,"},{"Start":"10:24.490 ","End":"10:27.490","Text":"which means that these 2 vectors are linearly independent."},{"Start":"10:27.490 ","End":"10:30.265","Text":"We already showed that they span out 2,"},{"Start":"10:30.265 ","End":"10:39.295","Text":"and so they are a basis for R2 and that\u0027s the exercise."},{"Start":"10:39.295 ","End":"10:43.900","Text":"Now, that was an awful lot of work for a simple exercise like that."},{"Start":"10:43.900 ","End":"10:45.580","Text":"Here there\u0027s a proposition,"},{"Start":"10:45.580 ","End":"10:51.085","Text":"proposition is like a mini theorem that\u0027s going to help us to do such exercises."},{"Start":"10:51.085 ","End":"10:55.390","Text":"It\u0027s going to give us a condition for something to be a basis of Rn,"},{"Start":"10:55.390 ","End":"10:57.745","Text":"could be R2, R3, whatever."},{"Start":"10:57.745 ","End":"11:01.780","Text":"What this proposition says is if I have a set of"},{"Start":"11:01.780 ","End":"11:06.100","Text":"exactly n linearly independent members of Rn,"},{"Start":"11:06.100 ","End":"11:08.455","Text":"then it\u0027s automatically a basis,"},{"Start":"11:08.455 ","End":"11:13.810","Text":"there\u0027s no need to do the first part of showing that they span Rn."},{"Start":"11:13.810 ","End":"11:16.540","Text":"Just have to make sure that\u0027s the right number of them."},{"Start":"11:16.540 ","End":"11:19.060","Text":"In our previous example,"},{"Start":"11:19.060 ","End":"11:22.390","Text":"where we worked quite hard to show that this set,"},{"Start":"11:22.390 ","End":"11:24.880","Text":"well, it was called B over there."},{"Start":"11:24.880 ","End":"11:27.639","Text":"We showed that this was a basis of R2."},{"Start":"11:27.639 ","End":"11:32.965","Text":"We showed that they span and they are linearly independent,"},{"Start":"11:32.965 ","End":"11:36.430","Text":"but we can settle for just showing that these are"},{"Start":"11:36.430 ","End":"11:39.805","Text":"linearly independent because of exactly 2 of them."},{"Start":"11:39.805 ","End":"11:43.540","Text":"This is the 2 from the R2, like here,"},{"Start":"11:43.540 ","End":"11:46.150","Text":"we have to show that this is exactly n"},{"Start":"11:46.150 ","End":"11:51.055","Text":"in linearly independent members of Rn and it\u0027s a basis."},{"Start":"11:51.055 ","End":"11:58.045","Text":"Another example, say you had the exercise show that this set is a basis of R3."},{"Start":"11:58.045 ","End":"12:00.190","Text":"Then if you did it the long way it showed that they"},{"Start":"12:00.190 ","End":"12:02.920","Text":"span R3 and they\u0027re linearly independent."},{"Start":"12:02.920 ","End":"12:04.690","Text":"But with this proposition,"},{"Start":"12:04.690 ","End":"12:06.250","Text":"because it\u0027s the right number,"},{"Start":"12:06.250 ","End":"12:08.185","Text":"there\u0027s 1,2,3 of them,"},{"Start":"12:08.185 ","End":"12:13.060","Text":"then all we have to do is show that they are linearly independent."},{"Start":"12:13.060 ","End":"12:14.830","Text":"I\u0027ll skip that part."},{"Start":"12:14.830 ","End":"12:16.825","Text":"I think we\u0027ve worked hard enough already,"},{"Start":"12:16.825 ","End":"12:20.660","Text":"it\u0027s not difficult to show that they\u0027re linearly independent."},{"Start":"12:20.660 ","End":"12:24.360","Text":"There\u0027s also a corollary,"},{"Start":"12:24.360 ","End":"12:27.240","Text":"a conclusion from the proposition."},{"Start":"12:27.240 ","End":"12:29.805","Text":"That corollary is that,"},{"Start":"12:29.805 ","End":"12:34.750","Text":"if I have n linearly independent members of Rn,"},{"Start":"12:34.750 ","End":"12:36.400","Text":"I just say n-dimensional vectors,"},{"Start":"12:36.400 ","End":"12:37.525","Text":"meaning members of Rn,"},{"Start":"12:37.525 ","End":"12:42.640","Text":"then they span Rn."},{"Start":"12:42.640 ","End":"12:46.230","Text":"That\u0027s because they are a basis,"},{"Start":"12:46.230 ","End":"12:49.695","Text":"if I have n linearly independent,"},{"Start":"12:49.695 ","End":"12:53.520","Text":"I should have said members of Rn, same thing,"},{"Start":"12:53.520 ","End":"12:56.385","Text":"then there are basis and because there are basis,"},{"Start":"12:56.385 ","End":"12:59.585","Text":"then they also span Rn."},{"Start":"12:59.585 ","End":"13:02.710","Text":"There is a companion proposition to this,"},{"Start":"13:02.710 ","End":"13:06.445","Text":"which shows when something is not a basis of Rn,"},{"Start":"13:06.445 ","End":"13:09.140","Text":"and I want to show you this."},{"Start":"13:09.960 ","End":"13:12.640","Text":"I phrased it in bad English,"},{"Start":"13:12.640 ","End":"13:16.210","Text":"a sufficient condition to not be a basis of Rn,"},{"Start":"13:16.210 ","End":"13:18.550","Text":"I just called it for not a basis."},{"Start":"13:18.550 ","End":"13:23.470","Text":"That is if I have a set of less than n and I say"},{"Start":"13:23.470 ","End":"13:28.375","Text":"n-dimensional vectors I just mean members of Rn cannot be a basis."},{"Start":"13:28.375 ","End":"13:31.180","Text":"If I have less than n vectors,"},{"Start":"13:31.180 ","End":"13:33.880","Text":"doesn\u0027t matter linearly independent or otherwise,"},{"Start":"13:33.880 ","End":"13:37.030","Text":"less than n will never be a basis."},{"Start":"13:37.030 ","End":"13:42.175","Text":"For example, if I take the set A which has just the single vector 3,4,"},{"Start":"13:42.175 ","End":"13:49.120","Text":"it will not be a basis of R2 because it has less than 2 members."},{"Start":"13:49.120 ","End":"13:54.700","Text":"Another example, this set containing these 2,3 dimensional"},{"Start":"13:54.700 ","End":"14:01.525","Text":"vectors will not be a basis of R3 because it has less than 3 members."},{"Start":"14:01.525 ","End":"14:06.475","Text":"Next I want to introduce 1 more concept."},{"Start":"14:06.475 ","End":"14:13.285","Text":"There was a very special basis for Rn and it\u0027s called the standard basis of Rn."},{"Start":"14:13.285 ","End":"14:18.010","Text":"I\u0027ll just demonstrate it on a particular example of n equals 4."},{"Start":"14:18.010 ","End":"14:22.675","Text":"The standard basis of R4 is the following set."},{"Start":"14:22.675 ","End":"14:26.380","Text":"If you look at it, you see that it always has a 1 and 3"},{"Start":"14:26.380 ","End":"14:30.850","Text":"0s and the 1 just has different position each time."},{"Start":"14:30.850 ","End":"14:33.430","Text":"One with all 0s,"},{"Start":"14:33.430 ","End":"14:35.635","Text":"there\u0027s 4 of these."},{"Start":"14:35.635 ","End":"14:38.980","Text":"That\u0027s the same 4 as the 4 because there\u0027s"},{"Start":"14:38.980 ","End":"14:42.280","Text":"4 places that the 1 can go and we can generalize,"},{"Start":"14:42.280 ","End":"14:44.275","Text":"nothing special about 4,"},{"Start":"14:44.275 ","End":"14:46.180","Text":"and we can generalize this,"},{"Start":"14:46.180 ","End":"14:49.510","Text":"I just didn\u0027t want to write it with dot, dot, dot."},{"Start":"14:49.510 ","End":"14:52.300","Text":"You could get the idea if I asked you to write"},{"Start":"14:52.300 ","End":"14:55.885","Text":"the standard basis for R3 or R5 or any other,"},{"Start":"14:55.885 ","End":"14:57.430","Text":"you would be able to do it."},{"Start":"14:57.430 ","End":"15:01.540","Text":"It would have n members if it"},{"Start":"15:01.540 ","End":"15:06.400","Text":"was Rn like standard basis for R3 would have 3 members,1,0,0,"},{"Start":"15:06.400 ","End":"15:09.940","Text":"0,1,0 and 0,0,1 for example."},{"Start":"15:09.940 ","End":"15:15.250","Text":"The final thing for this clip is just want to give you a theorem that will be useful in"},{"Start":"15:15.250 ","End":"15:21.730","Text":"future is that there are many possible basis for Rn as the standard,"},{"Start":"15:21.730 ","End":"15:23.960","Text":"but there\u0027s many others."},{"Start":"15:24.210 ","End":"15:27.280","Text":"But they all have the same number of elements,"},{"Start":"15:27.280 ","End":"15:31.735","Text":"every basis of Rn has n members."},{"Start":"15:31.735 ","End":"15:35.240","Text":"That concludes this clip."}],"ID":9962},{"Watched":false,"Name":"Exercise 1","Duration":"3m 2s","ChapterTopicVideoID":9913,"CourseChapterTopicPlaylistID":31676,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9913.jpeg","UploadDate":"2017-08-07T11:34:36.2130000","DurationForVideoObject":"PT3M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.975","Text":"This exercise is 3 in 1,"},{"Start":"00:03.975 ","End":"00:06.390","Text":"3 separate exercises really."},{"Start":"00:06.390 ","End":"00:13.485","Text":"In each 1, we\u0027re given a set of vectors in R^3,"},{"Start":"00:13.485 ","End":"00:18.880","Text":"and we have to decide if it\u0027s a basis or not for R^3."},{"Start":"00:18.980 ","End":"00:21.465","Text":"In the first 1,"},{"Start":"00:21.465 ","End":"00:25.030","Text":"we noticed that there are 2 vectors."},{"Start":"00:25.040 ","End":"00:27.360","Text":"We don\u0027t have to work very hard,"},{"Start":"00:27.360 ","End":"00:32.450","Text":"we just have to remember a theorem which in general says"},{"Start":"00:32.450 ","End":"00:40.505","Text":"that a basis for R^n can\u0027t contain less than n vectors."},{"Start":"00:40.505 ","End":"00:44.180","Text":"More specifically, if you have less than n vectors,"},{"Start":"00:44.180 ","End":"00:46.160","Text":"then they can\u0027t span R^n,"},{"Start":"00:46.160 ","End":"00:47.540","Text":"if they can\u0027t span,"},{"Start":"00:47.540 ","End":"00:49.640","Text":"then they can\u0027t be a basis."},{"Start":"00:49.640 ","End":"00:53.140","Text":"Now here we\u0027ll take n equals 3."},{"Start":"00:53.140 ","End":"00:57.150","Text":"Like I said, there\u0027s 2 vectors here and 2 is certainly"},{"Start":"00:57.150 ","End":"01:02.000","Text":"less than 3 so together with the theorem,"},{"Start":"01:02.000 ","End":"01:04.220","Text":"then the answer is no,"},{"Start":"01:04.220 ","End":"01:06.685","Text":"this is not a basis."},{"Start":"01:06.685 ","End":"01:08.790","Text":"For the second part."},{"Start":"01:08.790 ","End":"01:09.840","Text":"If we count 1,"},{"Start":"01:09.840 ","End":"01:11.605","Text":"2, 3, 4,"},{"Start":"01:11.605 ","End":"01:15.785","Text":"then again we can solve it with a theorem"},{"Start":"01:15.785 ","End":"01:24.935","Text":"because another theorem says that basis for R^n can\u0027t contain more than n vectors."},{"Start":"01:24.935 ","End":"01:29.000","Text":"More specifically, if you have more than n vectors in R^n,"},{"Start":"01:29.000 ","End":"01:31.655","Text":"they can\u0027t be linearly independent,"},{"Start":"01:31.655 ","End":"01:34.820","Text":"and therefore they can\u0027t be a basis."},{"Start":"01:34.820 ","End":"01:41.400","Text":"In our case, we have the 4 is bigger than 3,"},{"Start":"01:41.400 ","End":"01:45.570","Text":"and so these 4 vectors can\u0027t be a basis for R^3,"},{"Start":"01:45.570 ","End":"01:48.575","Text":"they won\u0027t be linearly independent."},{"Start":"01:48.575 ","End":"01:51.274","Text":"For the third part,"},{"Start":"01:51.274 ","End":"01:53.885","Text":"could use a theorem,"},{"Start":"01:53.885 ","End":"01:57.140","Text":"but actually it could work from the definition."},{"Start":"01:57.140 ","End":"02:03.235","Text":"Now, the definition of a basis for R^n and actually for any vector space,"},{"Start":"02:03.235 ","End":"02:08.885","Text":"says 2 things, the set has to be linearly independent and it has to span."},{"Start":"02:08.885 ","End":"02:12.600","Text":"Again linearly independent and span."},{"Start":"02:13.850 ","End":"02:17.540","Text":"We\u0027ve seen this before, but anyway,"},{"Start":"02:17.540 ","End":"02:20.900","Text":"I happen to know that these are not linearly independent,"},{"Start":"02:20.900 ","End":"02:25.750","Text":"so if we just show that they\u0027re not linearly independent then they won\u0027t be a basis."},{"Start":"02:25.750 ","End":"02:31.025","Text":"Let\u0027s show this even though it was in the previous exercise,"},{"Start":"02:31.025 ","End":"02:32.420","Text":"we\u0027ll start from scratch."},{"Start":"02:32.420 ","End":"02:38.075","Text":"Here\u0027s the matrix and let\u0027s do row operations."},{"Start":"02:38.075 ","End":"02:40.910","Text":"Subtract 4 times this from here,"},{"Start":"02:40.910 ","End":"02:44.900","Text":"subtract 7 times from here and this is what we"},{"Start":"02:44.900 ","End":"02:51.330","Text":"get and next will subtract twice this from this and then we get a row of 0s."},{"Start":"02:51.330 ","End":"02:53.275","Text":"Once we get a row of 0s,"},{"Start":"02:53.275 ","End":"03:02.070","Text":"we know that they\u0027re linearly dependent and so are not a basis. We\u0027re done."}],"ID":9964},{"Watched":false,"Name":"Exercise 2","Duration":"3m 54s","ChapterTopicVideoID":9914,"CourseChapterTopicPlaylistID":31676,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9914.jpeg","UploadDate":"2017-08-07T11:35:09.6470000","DurationForVideoObject":"PT3M54S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:07.575","Text":"This exercise deals with the space M_2 by 2 over R,"},{"Start":"00:07.575 ","End":"00:12.180","Text":"which is the set of 2 by 2 matrices over R. When it\u0027s a square matrix,"},{"Start":"00:12.180 ","End":"00:15.600","Text":"we sometimes just write it as M_2."},{"Start":"00:15.600 ","End":"00:18.720","Text":"There are 3 parts,"},{"Start":"00:18.720 ","End":"00:22.110","Text":"and each part we are given a set of matrices and we have to"},{"Start":"00:22.110 ","End":"00:27.060","Text":"decide if it\u0027s a basis or not for this space."},{"Start":"00:27.060 ","End":"00:29.970","Text":"We\u0027ll start with Part 1."},{"Start":"00:29.970 ","End":"00:36.570","Text":"Notice there are 3 members of this set."},{"Start":"00:36.570 ","End":"00:45.710","Text":"There\u0027s a theorem that in general,"},{"Start":"00:45.710 ","End":"00:48.500","Text":"if we have n by n matrices over R,"},{"Start":"00:48.500 ","End":"00:52.610","Text":"it can\u0027t contain less than n squared matrices."},{"Start":"00:52.610 ","End":"00:55.880","Text":"Most specifically, if you have less than n squared,"},{"Start":"00:55.880 ","End":"00:57.200","Text":"then they can\u0027t span."},{"Start":"00:57.200 ","End":"00:59.810","Text":"If they can\u0027t span, they can\u0027t be a basis."},{"Start":"00:59.810 ","End":"01:04.790","Text":"In our case, n squared is 2 squared, which is 4."},{"Start":"01:04.790 ","End":"01:09.735","Text":"Here we have 3 elements which is less than 4,"},{"Start":"01:09.735 ","End":"01:13.245","Text":"and so they can\u0027t be a basis."},{"Start":"01:13.245 ","End":"01:15.840","Text":"Let\u0027s move on to the next 1."},{"Start":"01:15.840 ","End":"01:17.610","Text":"This time, if you count,"},{"Start":"01:17.610 ","End":"01:21.150","Text":"you see there\u0027s 1,2,3,4,5."},{"Start":"01:21.500 ","End":"01:26.360","Text":"A basis for this space, in general,"},{"Start":"01:26.360 ","End":"01:34.355","Text":"n by n matrices also can\u0027t contain more than n squared matrices because in that case,"},{"Start":"01:34.355 ","End":"01:36.865","Text":"they\u0027re not going to be linearly independent,"},{"Start":"01:36.865 ","End":"01:40.210","Text":"and so there won\u0027t be a basis."},{"Start":"01:40.400 ","End":"01:48.209","Text":"In our case, n is 2 and we have more than 2 squared,"},{"Start":"01:48.209 ","End":"01:52.950","Text":"that\u0027s 5, so we don\u0027t have a basis."},{"Start":"01:52.950 ","End":"01:56.450","Text":"In the 3rd part we actually have the right number."},{"Start":"01:56.450 ","End":"01:59.665","Text":"We have 2 squared which is 4."},{"Start":"01:59.665 ","End":"02:03.050","Text":"There\u0027s a theorem that if you have the right number,"},{"Start":"02:03.050 ","End":"02:07.565","Text":"in this case n squared."},{"Start":"02:07.565 ","End":"02:13.925","Text":"The theorem helps us by saying that we just have to prove the linearly independent path."},{"Start":"02:13.925 ","End":"02:19.830","Text":"Normally a basis has to span and be linearly independent."},{"Start":"02:19.830 ","End":"02:24.800","Text":"This theorem says, you can just show the linearly independent part."},{"Start":"02:25.270 ","End":"02:28.250","Text":"How do we show that?"},{"Start":"02:28.250 ","End":"02:31.505","Text":"You might recall we\u0027ve previously done this,"},{"Start":"02:31.505 ","End":"02:38.020","Text":"so we can flatten out the matrices by taking them in the snake order."},{"Start":"02:38.020 ","End":"02:41.880","Text":"This, so this 1 would be like 0,"},{"Start":"02:41.880 ","End":"02:45.104","Text":"1, 1, 1 and 0, 0, 1, 1."},{"Start":"02:45.104 ","End":"02:51.900","Text":"If we take them and I have to scroll."},{"Start":"02:53.350 ","End":"02:59.160","Text":"But if we take them as in the snake form,"},{"Start":"02:59.160 ","End":"03:01.575","Text":"let\u0027s take 1 of them, I go back."},{"Start":"03:01.575 ","End":"03:03.540","Text":"First 1 would be 1,"},{"Start":"03:03.540 ","End":"03:05.130","Text":"1, 1, 1,"},{"Start":"03:05.130 ","End":"03:09.960","Text":"and that\u0027s here, 1,"},{"Start":"03:09.960 ","End":"03:11.130","Text":"1, 1, 1."},{"Start":"03:11.130 ","End":"03:13.040","Text":"Similarly for all the rest,"},{"Start":"03:13.040 ","End":"03:17.015","Text":"and we have to see if we do a row operations,"},{"Start":"03:17.015 ","End":"03:20.990","Text":"whether we get a 0 or not."},{"Start":"03:20.990 ","End":"03:23.090","Text":"Now we happen to be in luck."},{"Start":"03:23.090 ","End":"03:28.865","Text":"We don\u0027t have to bring it to row echelon form because it already is in row echelon form."},{"Start":"03:28.865 ","End":"03:36.680","Text":"As you can see, all zeros below the diagonal will certainly make it echelon."},{"Start":"03:36.680 ","End":"03:40.530","Text":"It\u0027s even upper triangular."},{"Start":"03:40.870 ","End":"03:44.800","Text":"There are no rows of zeros."},{"Start":"03:44.800 ","End":"03:48.545","Text":"That means it\u0027s linearly independent."},{"Start":"03:48.545 ","End":"03:54.990","Text":"Yes, the 4 vectors form a basis. We\u0027re done."}],"ID":9965},{"Watched":false,"Name":"Exercise 3","Duration":"5m 13s","ChapterTopicVideoID":9915,"CourseChapterTopicPlaylistID":31676,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9915.jpeg","UploadDate":"2017-08-07T11:35:45.8070000","DurationForVideoObject":"PT5M13S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.565","Text":"This exercise deals with the space P_2 over R,"},{"Start":"00:05.565 ","End":"00:13.830","Text":"meaning the space of polynomials with real coefficients of degree up to 2."},{"Start":"00:13.830 ","End":"00:17.880","Text":"Here we have 3 sets of polynomials."},{"Start":"00:17.880 ","End":"00:23.140","Text":"We have to decide for each of them whether it\u0027s a basis or not."},{"Start":"00:23.630 ","End":"00:28.530","Text":"What\u0027s going to help us is some theorems that we know about"},{"Start":"00:28.530 ","End":"00:33.885","Text":"how many elements there should be in a basis and so on."},{"Start":"00:33.885 ","End":"00:38.985","Text":"Now, there\u0027s a theorem that a basis for P_n,"},{"Start":"00:38.985 ","End":"00:40.380","Text":"n in general,"},{"Start":"00:40.380 ","End":"00:42.380","Text":"though in our case n is 2,"},{"Start":"00:42.380 ","End":"00:46.385","Text":"can\u0027t contain less than n plus 1 polynomials."},{"Start":"00:46.385 ","End":"00:48.320","Text":"If there\u0027s too few,"},{"Start":"00:48.320 ","End":"00:52.120","Text":"they can\u0027t span and so they can\u0027t be a basis."},{"Start":"00:52.120 ","End":"00:58.300","Text":"In our case, 2 plus 1 is 3."},{"Start":"00:58.760 ","End":"01:05.700","Text":"We have only 2 here and 2 is less than 3,"},{"Start":"01:05.700 ","End":"01:08.670","Text":"and so these 2 don\u0027t span and so they\u0027re not"},{"Start":"01:08.670 ","End":"01:12.570","Text":"a basis and the answer to 1 is no, not a basis."},{"Start":"01:12.570 ","End":"01:14.550","Text":"In the second set,"},{"Start":"01:14.550 ","End":"01:15.810","Text":"we actually have too many,"},{"Start":"01:15.810 ","End":"01:16.995","Text":"there\u0027s 4 of them,"},{"Start":"01:16.995 ","End":"01:19.020","Text":"which is bigger than 3,"},{"Start":"01:19.020 ","End":"01:23.630","Text":"and there\u0027s another theorem that says that if we have more than n plus 1,"},{"Start":"01:23.630 ","End":"01:25.460","Text":"in our case, n is 2,"},{"Start":"01:25.460 ","End":"01:27.290","Text":"so n plus 1 is 3,"},{"Start":"01:27.290 ","End":"01:28.880","Text":"and we have 4 of them,"},{"Start":"01:28.880 ","End":"01:30.680","Text":"which is bigger than 3."},{"Start":"01:30.680 ","End":"01:32.360","Text":"Again, it can\u0027t be a basis,"},{"Start":"01:32.360 ","End":"01:34.235","Text":"but this time for a different reason,"},{"Start":"01:34.235 ","End":"01:36.725","Text":"they won\u0027t be linearly independent."},{"Start":"01:36.725 ","End":"01:39.350","Text":"If you have too few, they don\u0027t span if you have too many,"},{"Start":"01:39.350 ","End":"01:41.345","Text":"they are not linearly independent."},{"Start":"01:41.345 ","End":"01:46.100","Text":"Although in both cases we could have used a shortcut and say that a basis has to"},{"Start":"01:46.100 ","End":"01:52.075","Text":"contain exactly n plus 1 and ruled them out in 1 go."},{"Start":"01:52.075 ","End":"01:54.270","Text":"Let\u0027s see what happens in the third case,"},{"Start":"01:54.270 ","End":"01:56.505","Text":"it appears to have the correct number,"},{"Start":"01:56.505 ","End":"01:58.485","Text":"2 plus 1 is 3,"},{"Start":"01:58.485 ","End":"02:00.450","Text":"and we do have 1,"},{"Start":"02:00.450 ","End":"02:04.740","Text":"2, 3 members in this."},{"Start":"02:04.740 ","End":"02:09.995","Text":"I just forgot the last sentence in the previous part."},{"Start":"02:09.995 ","End":"02:12.844","Text":"Now, in part 3,"},{"Start":"02:12.844 ","End":"02:14.990","Text":"we\u0027re going to use the theorem to help us,"},{"Start":"02:14.990 ","End":"02:17.205","Text":"but let me just give you some background."},{"Start":"02:17.205 ","End":"02:22.940","Text":"Normally, if I have a set of polynomials here,"},{"Start":"02:22.940 ","End":"02:24.500","Text":"to show that it\u0027s a basis,"},{"Start":"02:24.500 ","End":"02:28.565","Text":"it have to show that these, in this case,"},{"Start":"02:28.565 ","End":"02:35.640","Text":"both are linearly independent and that they span the vector space P_n."},{"Start":"02:35.640 ","End":"02:39.530","Text":"Now, sorry, this is not a definition."},{"Start":"02:39.530 ","End":"02:44.145","Text":"Sorry, this is actually a theorem."},{"Start":"02:44.145 ","End":"02:49.600","Text":"What the theorem does is it makes life easier."},{"Start":"02:49.600 ","End":"02:57.890","Text":"Normally, we would have to show that these vectors are polynomials and span,"},{"Start":"02:57.890 ","End":"03:00.780","Text":"and are linearly independent, like I said."},{"Start":"03:00.780 ","End":"03:05.110","Text":"The theorem says that we just have to check the linearly independent."},{"Start":"03:05.110 ","End":"03:10.790","Text":"We can save the time of showing the part about spanning P_n."},{"Start":"03:10.790 ","End":"03:13.645","Text":"In our case, yeah, we do have,"},{"Start":"03:13.645 ","End":"03:17.260","Text":"if n is 2, we have 3 polynomials."},{"Start":"03:17.260 ","End":"03:19.940","Text":"Let\u0027s just see,"},{"Start":"03:19.940 ","End":"03:24.250","Text":"so how do we show that our polynomials are linearly independent?"},{"Start":"03:24.250 ","End":"03:27.070","Text":"I\u0027ve lost them, let\u0027s scroll back."},{"Start":"03:27.070 ","End":"03:33.850","Text":"Here they are. What we do is we give the vector representation of these."},{"Start":"03:33.850 ","End":"03:36.710","Text":"This 1, I\u0027ll just do 1 of them,"},{"Start":"03:36.710 ","End":"03:37.890","Text":"let\u0027s say the middle 1,"},{"Start":"03:37.890 ","End":"03:43.025","Text":"we can write it as 4, 5, 6."},{"Start":"03:43.025 ","End":"03:47.365","Text":"To do this, you have to make sure that they\u0027re all in the right order."},{"Start":"03:47.365 ","End":"03:49.240","Text":"We have to agree on an order,"},{"Start":"03:49.240 ","End":"03:51.655","Text":"let\u0027s say increasing order of powers,"},{"Start":"03:51.655 ","End":"03:55.480","Text":"and put them in order and make sure that there\u0027s non missing."},{"Start":"03:55.480 ","End":"03:57.789","Text":"If there\u0027s a missing term, you put a 0."},{"Start":"03:57.789 ","End":"04:02.340","Text":"That way we get 3 vectors, 1 for each."},{"Start":"04:02.340 ","End":"04:04.020","Text":"Here we have 1, 2, 3, then 4, 5,"},{"Start":"04:04.020 ","End":"04:06.045","Text":"6, then 7, 8, 10."},{"Start":"04:06.045 ","End":"04:13.805","Text":"I\u0027m going to take those numbers and put them in a matrix. Here we are."},{"Start":"04:13.805 ","End":"04:15.410","Text":"I remembered the numbers, it\u0027s 1, 2,"},{"Start":"04:15.410 ","End":"04:16.820","Text":"3, 4, 5, 6, 7, 8,"},{"Start":"04:16.820 ","End":"04:17.960","Text":"and then not 9,"},{"Start":"04:17.960 ","End":"04:21.050","Text":"but 10. Let\u0027s see."},{"Start":"04:21.050 ","End":"04:24.215","Text":"If we bring this to row echelon form,"},{"Start":"04:24.215 ","End":"04:27.930","Text":"will we get a row of 0s or not?"},{"Start":"04:28.250 ","End":"04:31.965","Text":"Let\u0027s just do the row operations."},{"Start":"04:31.965 ","End":"04:36.815","Text":"Subtract 4 times this from this to get a 0 here,"},{"Start":"04:36.815 ","End":"04:39.740","Text":"subtract 7 times this row from this row,"},{"Start":"04:39.740 ","End":"04:42.325","Text":"and that gives us this 0."},{"Start":"04:42.325 ","End":"04:44.955","Text":"Now, we need a 0 here."},{"Start":"04:44.955 ","End":"04:49.600","Text":"Let\u0027s subtract twice the second row from the last row,"},{"Start":"04:49.600 ","End":"04:51.770","Text":"that gives us this."},{"Start":"04:51.770 ","End":"04:56.225","Text":"This is in row echelon form."},{"Start":"04:56.225 ","End":"04:59.030","Text":"But there are no 0 rows,"},{"Start":"04:59.030 ","End":"05:06.050","Text":"which means that these 3 vectors or the 3 polynomials are linearly independent."},{"Start":"05:06.050 ","End":"05:08.480","Text":"By that theorem they are basis,"},{"Start":"05:08.480 ","End":"05:11.149","Text":"we don\u0027t have to show that they span."},{"Start":"05:11.149 ","End":"05:14.160","Text":"We are done."}],"ID":9966},{"Watched":false,"Name":"Exercise 4","Duration":"3m 56s","ChapterTopicVideoID":9912,"CourseChapterTopicPlaylistID":31676,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9912.jpeg","UploadDate":"2017-08-07T11:34:13.9670000","DurationForVideoObject":"PT3M56S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.660","Text":"In this exercise, we\u0027re dealing with the vector space R,"},{"Start":"00:03.660 ","End":"00:06.270","Text":"3, 3-dimensional real space."},{"Start":"00:06.270 ","End":"00:08.940","Text":"We have a set of vectors."},{"Start":"00:08.940 ","End":"00:11.190","Text":"Notice that each has 3 components."},{"Start":"00:11.190 ","End":"00:17.595","Text":"In fact, there\u0027s 4 of them and there are 3 questions and I just take them 1 at a time."},{"Start":"00:17.595 ","End":"00:22.105","Text":"First question, is T a basis for R^3."},{"Start":"00:22.105 ","End":"00:26.680","Text":"Now, the answer is immediately a no,"},{"Start":"00:26.680 ","End":"00:29.100","Text":"because we have 4 vectors,"},{"Start":"00:29.100 ","End":"00:37.370","Text":"which is more than 3 and 4 vectors can never be linearly independent in R^3,"},{"Start":"00:37.370 ","End":"00:41.725","Text":"and so if they\u0027re not linearly independent then it\u0027s not a basis."},{"Start":"00:41.725 ","End":"00:43.510","Text":"I\u0027ll make a note to that,"},{"Start":"00:43.510 ","End":"00:45.320","Text":"it contains more than 3 vectors,"},{"Start":"00:45.320 ","End":"00:47.780","Text":"are not linearly independent."},{"Start":"00:47.780 ","End":"00:50.405","Text":"In part b,"},{"Start":"00:50.405 ","End":"00:52.490","Text":"if I rephrase this,"},{"Start":"00:52.490 ","End":"00:56.640","Text":"we want to take maybe less than T,"},{"Start":"00:56.640 ","End":"01:01.415","Text":"part of T that will be linearly independent."},{"Start":"01:01.415 ","End":"01:03.200","Text":"Maybe throw out the last 1,"},{"Start":"01:03.200 ","End":"01:09.050","Text":"maybe throw out the last 2 until we just get linearly independent ones."},{"Start":"01:09.050 ","End":"01:14.075","Text":"Now, how do we do that systematically?"},{"Start":"01:14.075 ","End":"01:19.130","Text":"What we do is we take these full vectors and put them as rows in"},{"Start":"01:19.130 ","End":"01:24.455","Text":"a matrix and then we bring this to row echelon form."},{"Start":"01:24.455 ","End":"01:31.790","Text":"Let\u0027s see. Subtract the first row times 4 from here,"},{"Start":"01:31.790 ","End":"01:33.620","Text":"subtract it times 7 from here,"},{"Start":"01:33.620 ","End":"01:36.205","Text":"subtract twice of it from here."},{"Start":"01:36.205 ","End":"01:38.760","Text":"That gives us 0s here."},{"Start":"01:38.760 ","End":"01:40.850","Text":"Now we want to keep going."},{"Start":"01:40.850 ","End":"01:44.795","Text":"I want to get a 0 here."},{"Start":"01:44.795 ","End":"01:47.945","Text":"But look at this,"},{"Start":"01:47.945 ","End":"01:49.220","Text":"they\u0027re all divisible."},{"Start":"01:49.220 ","End":"01:51.570","Text":"I can cancel."},{"Start":"01:51.830 ","End":"01:55.360","Text":"In fact, yeah,"},{"Start":"01:55.360 ","End":"02:03.280","Text":"the second row can be divided by minus 3 and this 1 can be divided by minus 6."},{"Start":"02:03.280 ","End":"02:07.185","Text":"Then we have 3 rows all the same,"},{"Start":"02:07.185 ","End":"02:12.590","Text":"so of course we could subtract the second from the third and the fourth and make those 0."},{"Start":"02:12.590 ","End":"02:18.740","Text":"There we are. Now, I\u0027d like to just take the top 2."},{"Start":"02:18.740 ","End":"02:24.710","Text":"But the thing is that these are not from the original set T. What I can"},{"Start":"02:24.710 ","End":"02:33.215","Text":"do is to notice that the top 2 rows just came from the top 2 rows that we originally had."},{"Start":"02:33.215 ","End":"02:36.050","Text":"If I just did row operations on the top 2 rows,"},{"Start":"02:36.050 ","End":"02:38.770","Text":"I\u0027ll get from here to here, to here, to here."},{"Start":"02:38.770 ","End":"02:42.755","Text":"These are the 2 vectors that I want to take."},{"Start":"02:42.755 ","End":"02:46.415","Text":"They have the same span as these 2."},{"Start":"02:46.415 ","End":"02:53.220","Text":"Scroll. Now give it a name T prime."},{"Start":"02:53.220 ","End":"03:01.635","Text":"T prime now consists of the linearly independent set."},{"Start":"03:01.635 ","End":"03:05.930","Text":"The maximum we can get is 2 vectors."},{"Start":"03:05.930 ","End":"03:11.720","Text":"Now the last part was to extend this because we"},{"Start":"03:11.720 ","End":"03:18.380","Text":"know that we could get 3 linearly independent vectors in R3."},{"Start":"03:18.380 ","End":"03:28.535","Text":"So what I\u0027m going to do is suppose I replace this last row by 001,"},{"Start":"03:28.535 ","End":"03:30.200","Text":"just put a 1 here,"},{"Start":"03:30.200 ","End":"03:37.080","Text":"then these 2 with this 1 would now be linearly independent."},{"Start":"03:37.080 ","End":"03:38.445","Text":"Again, out of these 2,"},{"Start":"03:38.445 ","End":"03:41.500","Text":"I can replace them by these 2."},{"Start":"03:41.570 ","End":"03:48.480","Text":"Let\u0027s now call it T^* is these 2 with an extra 1."},{"Start":"03:48.480 ","End":"03:55.750","Text":"These are now 3 linearly independent vectors, and we\u0027re done."}],"ID":9963}],"Thumbnail":null,"ID":31676},{"Name":"Solution Space of Homogenous SLE","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction and Example","Duration":"18m 32s","ChapterTopicVideoID":9924,"CourseChapterTopicPlaylistID":31677,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9924.jpeg","UploadDate":"2019-02-07T19:40:04.1670000","DurationForVideoObject":"PT18M32S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.425","Text":"Now we come to a new topic in vector spaces."},{"Start":"00:04.425 ","End":"00:11.470","Text":"Basis and dimension for the solution space of a homogeneous system of linear equations."},{"Start":"00:11.470 ","End":"00:14.715","Text":"There are a lot technical words in there,"},{"Start":"00:14.715 ","End":"00:18.945","Text":"and it\u0027s pretty hard to understand what this says"},{"Start":"00:18.945 ","End":"00:21.195","Text":"but as we go along,"},{"Start":"00:21.195 ","End":"00:24.120","Text":"all these terms will become clear."},{"Start":"00:24.120 ","End":"00:26.970","Text":"I\u0027m going to start out more theoretically,"},{"Start":"00:26.970 ","End":"00:34.450","Text":"but we\u0027ll soon get into very practical examples of how to solve certain problems."},{"Start":"00:34.460 ","End":"00:38.460","Text":"Although the title says homogeneous,"},{"Start":"00:38.460 ","End":"00:44.480","Text":"I\u0027m actually going to start out by contrasting it with the non-homogeneous case."},{"Start":"00:44.480 ","End":"00:50.600","Text":"Remember, homogeneous with a system of equations means that it\u0027s all 0s on the right."},{"Start":"00:50.600 ","End":"00:53.810","Text":"A non-homogeneous means they\u0027re not all 0s."},{"Start":"00:53.810 ","End":"00:57.115","Text":"This is an example of a non-homogeneous,"},{"Start":"00:57.115 ","End":"01:00.000","Text":"2 equations and 3 unknowns."},{"Start":"01:00.000 ","End":"01:04.230","Text":"Turns out this system has infinitely many solutions."},{"Start":"01:04.230 ","End":"01:08.245","Text":"Let\u0027s do the computational go through this fairly quickly."},{"Start":"01:08.245 ","End":"01:10.710","Text":"Won\u0027t bother using matrices."},{"Start":"01:10.710 ","End":"01:15.200","Text":"Let\u0027s just subtract twice the first row from the second row."},{"Start":"01:15.200 ","End":"01:19.415","Text":"That gives us this system which is already in echelon form."},{"Start":"01:19.415 ","End":"01:22.100","Text":"The leading terms are"},{"Start":"01:22.100 ","End":"01:30.605","Text":"the dependent constrained variables and the other 1 is z would be the free variable."},{"Start":"01:30.605 ","End":"01:35.450","Text":"What we do is we let z equal a parameter, for example,"},{"Start":"01:35.450 ","End":"01:37.205","Text":"let z equals t,"},{"Start":"01:37.205 ","End":"01:40.975","Text":"and then we get y and x from substitution."},{"Start":"01:40.975 ","End":"01:44.450","Text":"We get y by plugging in z equals t here,"},{"Start":"01:44.450 ","End":"01:46.210","Text":"and then we\u0027ve got y and z here."},{"Start":"01:46.210 ","End":"01:50.419","Text":"In short, this is what we get for z,"},{"Start":"01:50.419 ","End":"01:52.960","Text":"y, and x,"},{"Start":"01:52.960 ","End":"01:55.940","Text":"and here it is again in vector form."},{"Start":"01:55.940 ","End":"01:57.770","Text":"Of course there are"},{"Start":"01:57.770 ","End":"02:01.790","Text":"infinitely many solutions because each value of t gives us a solution."},{"Start":"02:01.790 ","End":"02:03.260","Text":"Let\u0027s take 2 examples,"},{"Start":"02:03.260 ","End":"02:06.265","Text":"say t equals 0 and t equals 1."},{"Start":"02:06.265 ","End":"02:11.365","Text":"Plugging in t equals 0 gives us this solution, call it u,"},{"Start":"02:11.365 ","End":"02:13.670","Text":"and t equals 1 will give us this one,"},{"Start":"02:13.670 ","End":"02:18.545","Text":"call it v. We get a whole bunch of solutions that way."},{"Start":"02:18.545 ","End":"02:21.575","Text":"Mathematics, we call that a set."},{"Start":"02:21.575 ","End":"02:31.425","Text":"For each t we get a solution and the set of solutions is a subset of 3."},{"Start":"02:31.425 ","End":"02:39.200","Text":"The question is, is this set of solutions a subspace of R^3?"},{"Start":"02:39.200 ","End":"02:43.735","Text":"This might be the time to go and review the concept of subspace."},{"Start":"02:43.735 ","End":"02:46.695","Text":"Not every subset is a subspace,"},{"Start":"02:46.695 ","End":"02:50.780","Text":"it has to satisfy 3 axioms conditions."},{"Start":"02:50.780 ","End":"02:56.105","Text":"Now actually this set of S violates all 3 of the conditions"},{"Start":"02:56.105 ","End":"03:01.190","Text":"but it\u0027s enough to say no to one of them,"},{"Start":"03:01.190 ","End":"03:03.695","Text":"but let me go through all 3."},{"Start":"03:03.695 ","End":"03:08.420","Text":"The first axiom or condition is that to be a subspace,"},{"Start":"03:08.420 ","End":"03:10.355","Text":"it should be closed under addition."},{"Start":"03:10.355 ","End":"03:13.500","Text":"In other words, if 2 things are in S,"},{"Start":"03:13.500 ","End":"03:15.710","Text":"their sum should also be in S,"},{"Start":"03:15.710 ","End":"03:17.135","Text":"that\u0027s not the case."},{"Start":"03:17.135 ","End":"03:20.090","Text":"Let\u0027s take this u and v. U and v,"},{"Start":"03:20.090 ","End":"03:21.350","Text":"each of them is a solution,"},{"Start":"03:21.350 ","End":"03:24.470","Text":"so it\u0027s an S. But if I take u plus v,"},{"Start":"03:24.470 ","End":"03:26.520","Text":"I get 5, 2, 1,"},{"Start":"03:26.520 ","End":"03:31.819","Text":"and you can easily see that this does not satisfy the set of equations."},{"Start":"03:31.819 ","End":"03:36.300","Text":"It doesn\u0027t satisfy the 1 of the 2 equations, you should check that."},{"Start":"03:37.100 ","End":"03:39.590","Text":"Already S is not a subspace,"},{"Start":"03:39.590 ","End":"03:43.715","Text":"but just for the practice we\u0027ll continue with the other 2 axioms also."},{"Start":"03:43.715 ","End":"03:46.520","Text":"It\u0027s not closed under scalar multiplication,"},{"Start":"03:46.520 ","End":"03:50.225","Text":"meaning we have say that u is a solution,"},{"Start":"03:50.225 ","End":"03:53.990","Text":"but if I multiply it by a scalar say 2,"},{"Start":"03:53.990 ","End":"04:01.680","Text":"then that\u0027s not in S. It\u0027s not a solution because 2u is twice."},{"Start":"04:01.680 ","End":"04:03.600","Text":"This is 4, 4, 0,"},{"Start":"04:03.600 ","End":"04:05.030","Text":"and plug it in,"},{"Start":"04:05.030 ","End":"04:08.770","Text":"you\u0027ll see it does not satisfy the system of equations."},{"Start":"04:08.770 ","End":"04:13.680","Text":"Lastly, the final deathblow, if you wish."},{"Start":"04:13.680 ","End":"04:17.805","Text":"Usually this is the one I check first because it\u0027s easiest,"},{"Start":"04:17.805 ","End":"04:22.225","Text":"is that the 0 vector is not in the set."},{"Start":"04:22.225 ","End":"04:25.295","Text":"That\u0027s one of the basic conditions for a subspace."},{"Start":"04:25.295 ","End":"04:27.740","Text":"At 0 vector means x equals 0,"},{"Start":"04:27.740 ","End":"04:29.855","Text":"y equals 0, z equals 0."},{"Start":"04:29.855 ","End":"04:34.530","Text":"Plugging in to the SLE and you\u0027ll see it doesn\u0027t work."},{"Start":"04:35.830 ","End":"04:41.810","Text":"What I\u0027m concluding from all this and why looked at a non-homogeneous is just to show you"},{"Start":"04:41.810 ","End":"04:48.410","Text":"that set of solutions is not necessarily a vector subspace."},{"Start":"04:48.410 ","End":"04:51.090","Text":"That was for non-homogeneous."},{"Start":"04:51.090 ","End":"04:57.660","Text":"In a moment, we\u0027re going to contrast that with the behavior of homogeneous SLE."},{"Start":"04:57.950 ","End":"05:02.914","Text":"This time is considered the homogeneous system of linear equations."},{"Start":"05:02.914 ","End":"05:05.345","Text":"Took the same one as before,"},{"Start":"05:05.345 ","End":"05:07.955","Text":"except that I put 0s on the right."},{"Start":"05:07.955 ","End":"05:10.290","Text":"Now it\u0027s homogeneous."},{"Start":"05:10.290 ","End":"05:14.365","Text":"Let\u0027s see what the difference is in the behavior."},{"Start":"05:14.365 ","End":"05:16.470","Text":"Like the previous one,"},{"Start":"05:16.470 ","End":"05:19.895","Text":"this one also has infinitely many solutions."},{"Start":"05:19.895 ","End":"05:22.520","Text":"Going to do the same row operations,"},{"Start":"05:22.520 ","End":"05:26.425","Text":"subtract twice the first row from the second row,"},{"Start":"05:26.425 ","End":"05:28.845","Text":"and that gives us this."},{"Start":"05:28.845 ","End":"05:30.800","Text":"Once again, the x,"},{"Start":"05:30.800 ","End":"05:35.510","Text":"which is the leading term here and the y here are"},{"Start":"05:35.510 ","End":"05:43.800","Text":"the dependent variables and z is the free variable."},{"Start":"05:43.800 ","End":"05:46.395","Text":"These 2 are called constraint."},{"Start":"05:46.395 ","End":"05:49.810","Text":"We let z be a parameter t again."},{"Start":"05:49.810 ","End":"05:53.050","Text":"Just like before plugging it in."},{"Start":"05:53.050 ","End":"05:57.755","Text":"From here we get y and then from z and from y we get x,"},{"Start":"05:57.755 ","End":"06:00.200","Text":"and this is what it is."},{"Start":"06:00.200 ","End":"06:03.590","Text":"Put it in vector form."},{"Start":"06:03.590 ","End":"06:08.220","Text":"Like so. Just like before,"},{"Start":"06:08.220 ","End":"06:15.750","Text":"let\u0027s take 2 example solutions by setting t equals 0 and then t equals 1."},{"Start":"06:15.750 ","End":"06:20.555","Text":"I forget what we did before if it was 0 and 1 also."},{"Start":"06:20.555 ","End":"06:25.860","Text":"Anyway, we get u and v as these 2 vectors."},{"Start":"06:26.440 ","End":"06:31.430","Text":"Once again, we can consider not just these 2,"},{"Start":"06:31.430 ","End":"06:36.965","Text":"but all the solutions and call it S and ask the same question."},{"Start":"06:36.965 ","End":"06:42.335","Text":"This time, do the solutions form a subspace of R^3?"},{"Start":"06:42.335 ","End":"06:44.120","Text":"As you might expect,"},{"Start":"06:44.120 ","End":"06:46.770","Text":"the reason I chose ones"},{"Start":"06:46.770 ","End":"06:52.325","Text":"non-homogeneous and ones homogeneous is this time the answer is, yes."},{"Start":"06:52.325 ","End":"06:58.405","Text":"To prove it, we have to prove those 3 conditions or axioms."},{"Start":"06:58.405 ","End":"07:01.525","Text":"I brought them in again to remind you."},{"Start":"07:01.525 ","End":"07:04.190","Text":"I think I may have rearranged the order."},{"Start":"07:04.190 ","End":"07:05.825","Text":"I like to put this one first,"},{"Start":"07:05.825 ","End":"07:13.610","Text":"that the 0 vector belongs to the solution set and if 2 belong and so does their sum."},{"Start":"07:13.610 ","End":"07:18.310","Text":"Let\u0027s call closure under addition and closure under scalar multiplication."},{"Start":"07:18.310 ","End":"07:19.890","Text":"As for the first 1,"},{"Start":"07:19.890 ","End":"07:22.245","Text":"the 0 vector is just 0, 0, 0."},{"Start":"07:22.245 ","End":"07:23.340","Text":"Just plug it in,"},{"Start":"07:23.340 ","End":"07:25.545","Text":"check see it is the solution."},{"Start":"07:25.545 ","End":"07:29.040","Text":"I\u0027m not going to prove these 2,"},{"Start":"07:29.040 ","End":"07:31.805","Text":"let\u0027s just illustrate it with our particular u and"},{"Start":"07:31.805 ","End":"07:36.410","Text":"v. U and v have just disappeared off screen,"},{"Start":"07:36.410 ","End":"07:38.660","Text":"but check the sum,"},{"Start":"07:38.660 ","End":"07:39.935","Text":"you\u0027ll see that this is it."},{"Start":"07:39.935 ","End":"07:46.655","Text":"Plug it into the original system and you\u0027ll see that this is a solution."},{"Start":"07:46.655 ","End":"07:49.910","Text":"Scalar multiplication again, just an example,"},{"Start":"07:49.910 ","End":"07:53.715","Text":"twice u is this."},{"Start":"07:53.715 ","End":"07:55.805","Text":"This is also a solution,"},{"Start":"07:55.805 ","End":"07:58.150","Text":"again, leaving it to you to check."},{"Start":"07:58.150 ","End":"08:00.800","Text":"So this time in the homogeneous case,"},{"Start":"08:00.800 ","End":"08:04.895","Text":"the solution set is a vector subspace."},{"Start":"08:04.895 ","End":"08:07.340","Text":"Since it\u0027s a vector subspace,"},{"Start":"08:07.340 ","End":"08:10.700","Text":"which is also a vector space in its own right."},{"Start":"08:10.700 ","End":"08:13.835","Text":"Now we can try and find a basis for it,"},{"Start":"08:13.835 ","End":"08:18.695","Text":"or at least not now, but later on we\u0027ll show how to do this."},{"Start":"08:18.695 ","End":"08:21.245","Text":"It turns out that in general,"},{"Start":"08:21.245 ","End":"08:27.050","Text":"that when you take a solution set of a very important homogeneous system,"},{"Start":"08:27.050 ","End":"08:29.585","Text":"it is a vector space."},{"Start":"08:29.585 ","End":"08:32.405","Text":"It\u0027s like the subspace or a vector space."},{"Start":"08:32.405 ","End":"08:34.970","Text":"It\u0027s a subspace of R^n,"},{"Start":"08:34.970 ","End":"08:36.600","Text":"like we had R^3 before."},{"Start":"08:36.600 ","End":"08:39.700","Text":"Which n is the number of variables."},{"Start":"08:39.780 ","End":"08:43.195","Text":"In general, we\u0027ve seen this before,"},{"Start":"08:43.195 ","End":"08:51.280","Text":"we can write a general system in matrix form as A times x equals 0,"},{"Start":"08:51.280 ","End":"08:54.670","Text":"where x is x1 through xn, the end variables."},{"Start":"08:54.670 ","End":"08:59.695","Text":"Now, I was thinking of giving the proof at the end."},{"Start":"08:59.695 ","End":"09:06.670","Text":"I\u0027ll leave it up to you to skip,"},{"Start":"09:06.670 ","End":"09:08.770","Text":"or not to skip, or to return to this."},{"Start":"09:08.770 ","End":"09:13.690","Text":"At the end, I\u0027ll just briefly go through the proof,"},{"Start":"09:13.690 ","End":"09:17.240","Text":"but you certainly may skip this."},{"Start":"09:18.390 ","End":"09:25.390","Text":"I\u0027m going to prove that the solution set in the homogeneous case is actually a subspace."},{"Start":"09:25.390 ","End":"09:29.560","Text":"We\u0027re just going to show those 3 conditions are satisfied."},{"Start":"09:29.560 ","End":"09:35.590","Text":"The first condition was that the 0 vector is in the subspace,"},{"Start":"09:35.590 ","End":"09:45.550","Text":"and all we have to do is show it satisfies the equation that A times the 0 vector is 0,"},{"Start":"09:45.550 ","End":"09:50.410","Text":"which it certainly is because any matrix times 0 vector is 0,"},{"Start":"09:50.410 ","End":"09:52.405","Text":"so 0 is a solution."},{"Start":"09:52.405 ","End":"09:55.240","Text":"Secondly, you have to show closure under addition,"},{"Start":"09:55.240 ","End":"09:58.165","Text":"so if u and v are 2 solutions,"},{"Start":"09:58.165 ","End":"09:59.290","Text":"u is a solution,"},{"Start":"09:59.290 ","End":"10:01.015","Text":"so Au is 0,"},{"Start":"10:01.015 ","End":"10:04.225","Text":"v is a solution, so Av is 0."},{"Start":"10:04.225 ","End":"10:06.730","Text":"Since 0 plus 0 is 0,"},{"Start":"10:06.730 ","End":"10:08.530","Text":"I can add these 2,"},{"Start":"10:08.530 ","End":"10:12.310","Text":"then I could use the distributive law to take a out the brackets,"},{"Start":"10:12.310 ","End":"10:15.745","Text":"and if A times this thing is 0,"},{"Start":"10:15.745 ","End":"10:17.424","Text":"then this thing is a solution."},{"Start":"10:17.424 ","End":"10:20.815","Text":"In other words, u plus v is a solution also."},{"Start":"10:20.815 ","End":"10:26.875","Text":"The third condition is the scalar multiplication."},{"Start":"10:26.875 ","End":"10:29.395","Text":"If u is a solution,"},{"Start":"10:29.395 ","End":"10:32.230","Text":"then Au is 0,"},{"Start":"10:32.230 ","End":"10:35.080","Text":"which means that if I multiply this by a k,"},{"Start":"10:35.080 ","End":"10:38.620","Text":"constant scalar, it\u0027s still 0."},{"Start":"10:38.620 ","End":"10:42.535","Text":"The scalar can go inside by 1 of the rules,"},{"Start":"10:42.535 ","End":"10:45.640","Text":"and so A times ku is 0,"},{"Start":"10:45.640 ","End":"10:47.485","Text":"and if a times something is 0,"},{"Start":"10:47.485 ","End":"10:49.570","Text":"that something is a solution."},{"Start":"10:49.570 ","End":"10:52.510","Text":"That\u0027s the end of the optional proof."},{"Start":"10:52.510 ","End":"10:53.740","Text":"Maybe you skipped over this,"},{"Start":"10:53.740 ","End":"10:56.665","Text":"maybe not, let\u0027s continue."},{"Start":"10:56.665 ","End":"11:02.420","Text":"Now we come to the practical part I mentioned earlier about finding a basis."},{"Start":"11:02.420 ","End":"11:04.860","Text":"We mentioned that since the solution set of"},{"Start":"11:04.860 ","End":"11:08.715","Text":"homogeneous SLE is a vector space or vector subspace,"},{"Start":"11:08.715 ","End":"11:10.650","Text":"then it has a basis,"},{"Start":"11:10.650 ","End":"11:13.530","Text":"and here we\u0027re going to show how to find the basis,"},{"Start":"11:13.530 ","End":"11:17.085","Text":"and we\u0027re going to do it just by use of 1 main example."},{"Start":"11:17.085 ","End":"11:21.955","Text":"Let\u0027s start with this system,"},{"Start":"11:21.955 ","End":"11:25.900","Text":"and we\u0027ll let the solution space,"},{"Start":"11:25.900 ","End":"11:27.820","Text":"I\u0027ll use the letter W this time."},{"Start":"11:27.820 ","End":"11:35.035","Text":"We have to find a basis for W, and the dimension."},{"Start":"11:35.035 ","End":"11:40.510","Text":"Let\u0027s start, but I have to tell you there will be 2 different methods."},{"Start":"11:40.510 ","End":"11:43.960","Text":"Step 1 is common to both methods,"},{"Start":"11:43.960 ","End":"11:46.794","Text":"but then it splits up into 2 different methods."},{"Start":"11:46.794 ","End":"11:52.345","Text":"The first step is just to bring the system to row-echelon form."},{"Start":"11:52.345 ","End":"11:54.895","Text":"I\u0027ll go through this quickly."},{"Start":"11:54.895 ","End":"11:57.760","Text":"Bring it to matrix form."},{"Start":"11:57.760 ","End":"12:00.790","Text":"We don\u0027t need an augmented matrix in a homogeneous case,"},{"Start":"12:00.790 ","End":"12:03.250","Text":"we don\u0027t bother with the 0s."},{"Start":"12:03.250 ","End":"12:07.045","Text":"We want the 0 out what\u0027s below the 1 here."},{"Start":"12:07.045 ","End":"12:09.100","Text":"These are the 2 row operations."},{"Start":"12:09.100 ","End":"12:10.570","Text":"Subtract twice this from this,"},{"Start":"12:10.570 ","End":"12:12.790","Text":"and 4 times this from this."},{"Start":"12:12.790 ","End":"12:16.630","Text":"That gives us this matrix."},{"Start":"12:16.630 ","End":"12:18.610","Text":"Another [inaudible] form,"},{"Start":"12:18.610 ","End":"12:19.900","Text":"we want to make a 0 here."},{"Start":"12:19.900 ","End":"12:24.070","Text":"Subtract the 2nd from the 3rd,"},{"Start":"12:24.070 ","End":"12:27.910","Text":"and then we actually get echelon form,"},{"Start":"12:27.910 ","End":"12:30.880","Text":"but we even get a row of 0s,"},{"Start":"12:30.880 ","End":"12:36.220","Text":"which means that really there\u0027s only 2 equations,"},{"Start":"12:36.220 ","End":"12:39.505","Text":"because this 0 equals 0 we throw out."},{"Start":"12:39.505 ","End":"12:47.005","Text":"At this point, we split up into 2 different alternative continuations."},{"Start":"12:47.005 ","End":"12:53.290","Text":"The first method for the second step is especially what we did earlier in the examples."},{"Start":"12:53.290 ","End":"12:59.230","Text":"We identify the constrained variables, the non-free,"},{"Start":"12:59.230 ","End":"13:01.825","Text":"that\u0027s x_1 and x_2,"},{"Start":"13:01.825 ","End":"13:04.195","Text":"and the others, x_3, x_4,"},{"Start":"13:04.195 ","End":"13:06.355","Text":"x_5 are free variables,"},{"Start":"13:06.355 ","End":"13:10.030","Text":"and we assign parameters to the free variables."},{"Start":"13:10.030 ","End":"13:11.935","Text":"Previously we had just a t,"},{"Start":"13:11.935 ","End":"13:13.090","Text":"this time we have 3 of them,"},{"Start":"13:13.090 ","End":"13:16.090","Text":"so I\u0027ll still use rst."},{"Start":"13:16.590 ","End":"13:18.790","Text":"Once we have x_3,"},{"Start":"13:18.790 ","End":"13:19.990","Text":"x_4, x_5,"},{"Start":"13:19.990 ","End":"13:23.800","Text":"we compute x_2 and x_1 from these."},{"Start":"13:23.800 ","End":"13:28.900","Text":"First, we get x2 from the second equation as this,"},{"Start":"13:28.900 ","End":"13:34.585","Text":"and then we plug all these into the first 1 and we get x_1, which is this."},{"Start":"13:34.585 ","End":"13:36.760","Text":"Now in vector form,"},{"Start":"13:36.760 ","End":"13:41.510","Text":"we have all 5 variables in terms of rst,"},{"Start":"13:41.510 ","End":"13:46.050","Text":"and what we want to do is split this up into 3 pieces."},{"Start":"13:46.050 ","End":"13:48.630","Text":"We want to take the r bits separately,"},{"Start":"13:48.630 ","End":"13:50.625","Text":"the s bit separately,"},{"Start":"13:50.625 ","End":"13:53.670","Text":"and the t bits separately."},{"Start":"13:53.670 ","End":"13:55.680","Text":"For example, the r,"},{"Start":"13:55.680 ","End":"13:58.850","Text":"I look at the first component, there\u0027s nothing there."},{"Start":"13:58.850 ","End":"14:02.770","Text":"In the second component I have a minus 2r,"},{"Start":"14:02.770 ","End":"14:04.225","Text":"so it\u0027s minus 2,"},{"Start":"14:04.225 ","End":"14:05.740","Text":"here we have r,"},{"Start":"14:05.740 ","End":"14:08.365","Text":"so it\u0027s 1, there\u0027s no r\u0027s in this and this,"},{"Start":"14:08.365 ","End":"14:10.735","Text":"and similarly for s and t,"},{"Start":"14:10.735 ","End":"14:14.815","Text":"these 3 vectors,"},{"Start":"14:14.815 ","End":"14:23.150","Text":"and it is these 3 that form the basis for W, the solution space."},{"Start":"14:23.310 ","End":"14:31.120","Text":"They are a basis in the sense that each of them is a solution, you can check,"},{"Start":"14:31.120 ","End":"14:32.890","Text":"substitute each 1,"},{"Start":"14:32.890 ","End":"14:38.785","Text":"and every solution is the linear combination of these 3."},{"Start":"14:38.785 ","End":"14:42.025","Text":"What\u0027s more, they are also linearly independent,"},{"Start":"14:42.025 ","End":"14:46.640","Text":"which basically means that I can\u0027t do it in less than 3."},{"Start":"14:46.920 ","End":"14:50.740","Text":"To answer the second part of the question, that\u0027s the basis."},{"Start":"14:50.740 ","End":"14:52.750","Text":"The dimension is 3, of course,"},{"Start":"14:52.750 ","End":"14:57.894","Text":"just by counting the number of elements in the basis."},{"Start":"14:57.894 ","End":"15:04.030","Text":"I just repeated the original system in case you want"},{"Start":"15:04.030 ","End":"15:10.435","Text":"to here verify that these 3 really are solutions,"},{"Start":"15:10.435 ","End":"15:13.370","Text":"you can plug them in here."},{"Start":"15:13.650 ","End":"15:19.255","Text":"Now I\u0027d like to make an important note on the dimension."},{"Start":"15:19.255 ","End":"15:25.330","Text":"Notice that 3 was also the number of free variables."},{"Start":"15:25.330 ","End":"15:26.740","Text":"Remember x_3, x_4,"},{"Start":"15:26.740 ","End":"15:28.855","Text":"and x_5, were the free variables,"},{"Start":"15:28.855 ","End":"15:33.370","Text":"and it always works out that the number of free variables is the dimension,"},{"Start":"15:33.370 ","End":"15:35.230","Text":"but that\u0027s not all."},{"Start":"15:35.230 ","End":"15:37.255","Text":"You can also compute it,"},{"Start":"15:37.255 ","End":"15:39.925","Text":"that 3 is 5 minus 2."},{"Start":"15:39.925 ","End":"15:41.065","Text":"Now, what is 5?"},{"Start":"15:41.065 ","End":"15:44.905","Text":"5 is the number of variables altogether,"},{"Start":"15:44.905 ","End":"15:49.315","Text":"and 2 is the number of nonzero rows."},{"Start":"15:49.315 ","End":"15:50.680","Text":"It\u0027s scrolled off-screen,"},{"Start":"15:50.680 ","End":"15:53.800","Text":"but when we brought it into echelon form,"},{"Start":"15:53.800 ","End":"15:56.350","Text":"we got just 2 rows,"},{"Start":"15:56.350 ","End":"15:57.580","Text":"or 2 equations,"},{"Start":"15:57.580 ","End":"16:00.235","Text":"and 5 minus 2 is 3,"},{"Start":"16:00.235 ","End":"16:03.800","Text":"it\u0027s another way of getting to the dimension."},{"Start":"16:04.080 ","End":"16:09.820","Text":"What remains for me to show you is the other method of doing step 2."},{"Start":"16:09.820 ","End":"16:14.890","Text":"It doesn\u0027t have a name in the literature on the Internet,"},{"Start":"16:14.890 ","End":"16:17.080","Text":"but here, informally,"},{"Start":"16:17.080 ","End":"16:21.440","Text":"we call it the method of the Wandering 1."},{"Start":"16:21.630 ","End":"16:26.740","Text":"This method is relevant only for homogeneous systems."},{"Start":"16:26.740 ","End":"16:31.390","Text":"The previous step could have also been used for non-homogeneous,"},{"Start":"16:31.390 ","End":"16:33.760","Text":"but this technique, the Wandering 1,"},{"Start":"16:33.760 ","End":"16:35.875","Text":"is relevant only for homogeneous."},{"Start":"16:35.875 ","End":"16:41.950","Text":"What we do is to take the 3 free variables, x_3,"},{"Start":"16:41.950 ","End":"16:43.675","Text":"x_4, x_5,"},{"Start":"16:43.675 ","End":"16:49.060","Text":"and each time make 1 of them equal to 1 and the others equal to 0."},{"Start":"16:49.060 ","End":"16:52.660","Text":"Let\u0027s say I wrote them in the other order,"},{"Start":"16:52.660 ","End":"16:55.825","Text":"X5, let that equal 1,"},{"Start":"16:55.825 ","End":"16:58.915","Text":"and the other to be 0."},{"Start":"16:58.915 ","End":"17:01.570","Text":"If you plug those values in,"},{"Start":"17:01.570 ","End":"17:05.245","Text":"you\u0027ll see that x_2 is 1 and x_1 is minus 1."},{"Start":"17:05.245 ","End":"17:07.360","Text":"That\u0027s just the first case."},{"Start":"17:07.360 ","End":"17:11.680","Text":"Then I take the 3 free variables,"},{"Start":"17:11.680 ","End":"17:14.350","Text":"and this time, I\u0027ll let x_4 be 1,"},{"Start":"17:14.350 ","End":"17:17.600","Text":"and the other 2 free variables as 0."},{"Start":"17:18.690 ","End":"17:26.405","Text":"From these 3, compute the 2 constraint variables, x_2, x_1."},{"Start":"17:26.405 ","End":"17:28.410","Text":"Similarly, in the third case,"},{"Start":"17:28.410 ","End":"17:30.200","Text":"this time x_3 is 1,"},{"Start":"17:30.200 ","End":"17:32.105","Text":"and the other 2 is 0."},{"Start":"17:32.105 ","End":"17:34.690","Text":"That\u0027s what I mean by the Wandering 1."},{"Start":"17:34.690 ","End":"17:36.505","Text":"First of all, this is 1,"},{"Start":"17:36.505 ","End":"17:37.930","Text":"then this is 1,"},{"Start":"17:37.930 ","End":"17:39.090","Text":"then this is 1."},{"Start":"17:39.090 ","End":"17:42.415","Text":"Each of the free variables in turn is assigned 1,"},{"Start":"17:42.415 ","End":"17:44.500","Text":"and the other free variable 0,"},{"Start":"17:44.500 ","End":"17:50.640","Text":"and the other dependent constraint ones are computed from these."},{"Start":"17:50.640 ","End":"17:55.760","Text":"We get the basis simply by plugging these values in,"},{"Start":"17:55.760 ","End":"17:57.470","Text":"but you got to do them in the right order,"},{"Start":"17:57.470 ","End":"17:59.555","Text":"like x_1, x_2,"},{"Start":"17:59.555 ","End":"18:00.770","Text":"x_3, x_4, x_5."},{"Start":"18:00.770 ","End":"18:04.780","Text":"That would give us minus 1, 1, 0, 0,"},{"Start":"18:04.780 ","End":"18:09.670","Text":"1 here, and similarly for this 1,"},{"Start":"18:09.670 ","End":"18:11.140","Text":"1 minus 1, 0,"},{"Start":"18:11.140 ","End":"18:13.435","Text":"1, 0 and so on."},{"Start":"18:13.435 ","End":"18:15.020","Text":"If you look at these,"},{"Start":"18:15.020 ","End":"18:17.810","Text":"it\u0027s actually the same as the previous method,"},{"Start":"18:17.810 ","End":"18:20.280","Text":"just in a different order."},{"Start":"18:20.280 ","End":"18:24.890","Text":"Once again, the dimension is 3."},{"Start":"18:25.800 ","End":"18:29.785","Text":"That\u0027s it. Both methods,"},{"Start":"18:29.785 ","End":"18:33.080","Text":"and we are finally done."}],"ID":9970},{"Watched":false,"Name":"Finding a Basis and Dimension (Examples)","Duration":"14m 15s","ChapterTopicVideoID":9925,"CourseChapterTopicPlaylistID":31677,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9925.jpeg","UploadDate":"2017-08-07T11:36:26.7130000","DurationForVideoObject":"PT14M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.820","Text":"Continuing with this topic for"},{"Start":"00:02.820 ","End":"00:06.510","Text":"solution sets for homogeneous systems of linear equations,"},{"Start":"00:06.510 ","End":"00:13.335","Text":"this clip will just be devoted to solved exercises."},{"Start":"00:13.335 ","End":"00:20.100","Text":"Mostly we\u0027ll be finding a basis and also the dimension."},{"Start":"00:20.100 ","End":"00:21.780","Text":"Usually, you find the basis,"},{"Start":"00:21.780 ","End":"00:23.190","Text":"you also find the dimension."},{"Start":"00:23.190 ","End":"00:25.080","Text":"Just involves counting."},{"Start":"00:25.080 ","End":"00:30.345","Text":"The first one is the following exercise."},{"Start":"00:30.345 ","End":"00:32.520","Text":"Lets just read it first."},{"Start":"00:32.520 ","End":"00:38.010","Text":"We have W is a subspace of R^4,"},{"Start":"00:38.010 ","End":"00:39.915","Text":"in 4-dimensional space,"},{"Start":"00:39.915 ","End":"00:41.550","Text":"and it\u0027s defined as follows,"},{"Start":"00:41.550 ","End":"00:45.210","Text":"it\u0027s the set of all quadruplets a, b, c,"},{"Start":"00:45.210 ","End":"00:47.380","Text":"d from R^4,"},{"Start":"00:47.380 ","End":"00:50.480","Text":"subject to these conditions."},{"Start":"00:50.480 ","End":"00:54.290","Text":"The condition says that a plus b is equal to c,"},{"Start":"00:54.290 ","End":"00:57.110","Text":"and also that b plus c is d."},{"Start":"00:57.110 ","End":"01:04.650","Text":"Our task is to find a basis for W and the dimension."},{"Start":"01:04.650 ","End":"01:07.430","Text":"Now, it might occur to you to think,"},{"Start":"01:07.430 ","End":"01:10.640","Text":"where is the system of linear equations?"},{"Start":"01:10.640 ","End":"01:13.205","Text":"Homogeneous or otherwise."},{"Start":"01:13.205 ","End":"01:15.245","Text":"There is no SLE here."},{"Start":"01:15.245 ","End":"01:18.770","Text":"Coming to that. Usually the kind of exercises you"},{"Start":"01:18.770 ","End":"01:23.210","Text":"get are slightly disguised. We\u0027ll soon see."},{"Start":"01:23.210 ","End":"01:29.740","Text":"Meanwhile, let\u0027s try and understand what the subspace W is."},{"Start":"01:29.740 ","End":"01:35.775","Text":"Here are 3 examples of members of W,"},{"Start":"01:35.775 ","End":"01:38.820","Text":"1, 2, 3, 5 belongs."},{"Start":"01:38.820 ","End":"01:40.185","Text":"Why is that?"},{"Start":"01:40.185 ","End":"01:43.190","Text":"Because these two conditions are satisfied,"},{"Start":"01:43.190 ","End":"01:45.710","Text":"this means first plus second is the third,"},{"Start":"01:45.710 ","End":"01:48.700","Text":"and this plus this is equal to this."},{"Start":"01:48.700 ","End":"01:51.210","Text":"Also, b plus c equals d,"},{"Start":"01:51.210 ","End":"01:53.655","Text":"2 plus 3 equals 5."},{"Start":"01:53.655 ","End":"01:55.795","Text":"That\u0027s fine also."},{"Start":"01:55.795 ","End":"01:57.815","Text":"Another example is this."},{"Start":"01:57.815 ","End":"02:00.320","Text":"Well, you can see 0 plus 0 is 0,"},{"Start":"02:00.320 ","End":"02:02.705","Text":"and 0 plus 0 is 0."},{"Start":"02:02.705 ","End":"02:06.110","Text":"Similarly here, minus 1 and 2 is 1,"},{"Start":"02:06.110 ","End":"02:07.470","Text":"2 and 1 is 3."},{"Start":"02:07.470 ","End":"02:10.190","Text":"These are just some examples to give you"},{"Start":"02:10.190 ","End":"02:17.960","Text":"concrete feeling for what kind of entities live in this W space."},{"Start":"02:18.100 ","End":"02:22.985","Text":"But where is the SLE? Here it is."},{"Start":"02:22.985 ","End":"02:27.380","Text":"You should easily be able to translate from this to this."},{"Start":"02:27.380 ","End":"02:33.755","Text":"These equations can be with slight rewriting,"},{"Start":"02:33.755 ","End":"02:37.435","Text":"be brought into the form something equals 0."},{"Start":"02:37.435 ","End":"02:41.100","Text":"The first one will give us a plus b minus c is 0,"},{"Start":"02:41.100 ","End":"02:44.640","Text":"and here, b plus c minus d is 0."},{"Start":"02:44.640 ","End":"02:53.220","Text":"Maybe we\u0027re more used to x, y, z, and t."},{"Start":"02:53.220 ","End":"02:56.790","Text":"Put any 4 letters that will do,"},{"Start":"02:56.790 ","End":"03:00.270","Text":"and a, b, c, d are the 4 variables."},{"Start":"03:00.270 ","End":"03:03.600","Text":"We have two equations and four unknowns,"},{"Start":"03:03.600 ","End":"03:09.540","Text":"and it\u0027s homogeneous because we have a 0 here and we have a 0 here."},{"Start":"03:09.710 ","End":"03:14.860","Text":"What we have to do is just to solve this SLE."},{"Start":"03:14.860 ","End":"03:19.640","Text":"Now, we\u0027re in luck because normally I\u0027d go and bring this to echelon form."},{"Start":"03:19.640 ","End":"03:22.700","Text":"But this already is in row echelon form,"},{"Start":"03:22.700 ","End":"03:24.650","Text":"you can see the steps."},{"Start":"03:24.650 ","End":"03:26.690","Text":"If we look at it further,"},{"Start":"03:26.690 ","End":"03:30.050","Text":"we see that the leading terms,"},{"Start":"03:30.050 ","End":"03:32.570","Text":"that\u0027s the a and the b,"},{"Start":"03:32.570 ","End":"03:34.520","Text":"there\u0027s various names for these variables,"},{"Start":"03:34.520 ","End":"03:38.285","Text":"leading variables, pivot variables, basic variables."},{"Start":"03:38.285 ","End":"03:42.580","Text":"In any event, the other ones are called the free variables."},{"Start":"03:42.580 ","End":"03:44.550","Text":"That would be c and d,"},{"Start":"03:44.550 ","End":"03:46.085","Text":"these are the free variables."},{"Start":"03:46.085 ","End":"03:48.020","Text":"Now in the tutorial,"},{"Start":"03:48.020 ","End":"03:51.665","Text":"we had 2 different ways of proceeding from here."},{"Start":"03:51.665 ","End":"03:53.360","Text":"I\u0027m going to use,"},{"Start":"03:53.360 ","End":"03:58.700","Text":"I believe it was the second method and we informally called it the wandering ones."},{"Start":"03:58.700 ","End":"04:04.385","Text":"What we do is we set each of the free variables to 1 and the others to 0,"},{"Start":"04:04.385 ","End":"04:07.840","Text":"and we run through all the 3 variables."},{"Start":"04:07.840 ","End":"04:12.105","Text":"First time round I\u0027ll take d as 1 and c as 0,"},{"Start":"04:12.105 ","End":"04:15.830","Text":"and from these we compute the other variables that are constraint,"},{"Start":"04:15.830 ","End":"04:18.410","Text":"that are dependent on the free variables."},{"Start":"04:18.410 ","End":"04:20.180","Text":"Just substitution."},{"Start":"04:20.180 ","End":"04:22.100","Text":"I won\u0027t go into the boring details,"},{"Start":"04:22.100 ","End":"04:23.810","Text":"and then next we\u0027ll do the opposite."},{"Start":"04:23.810 ","End":"04:26.615","Text":"We\u0027ll let d is 0, c is 1."},{"Start":"04:26.615 ","End":"04:30.530","Text":"If we do that, plug in c and d here, we get b,"},{"Start":"04:30.530 ","End":"04:32.365","Text":"we plug in d,"},{"Start":"04:32.365 ","End":"04:35.090","Text":"well there is no d, but we plug in c and b here,"},{"Start":"04:35.090 ","End":"04:37.260","Text":"and we get a is 0."},{"Start":"04:37.510 ","End":"04:41.660","Text":"If I write these in the correct order and in vector form a,"},{"Start":"04:41.660 ","End":"04:42.710","Text":"b, c, d,"},{"Start":"04:42.710 ","End":"04:47.000","Text":"these are the 2 vectors we get and they are"},{"Start":"04:47.000 ","End":"04:53.940","Text":"the basis of the subspace W. If I count how many of these there are,"},{"Start":"04:54.080 ","End":"04:58.330","Text":"and there\u0027s 2 of them, so the dimension is 2,"},{"Start":"04:58.330 ","End":"05:01.800","Text":"and remember, there was also other techniques for computing this."},{"Start":"05:01.800 ","End":"05:05.540","Text":"One was to take the number of variables 4 and"},{"Start":"05:05.540 ","End":"05:09.800","Text":"to subtract the number of rows in the echelon form,"},{"Start":"05:09.800 ","End":"05:11.615","Text":"and you\u0027d also get 2."},{"Start":"05:11.615 ","End":"05:16.290","Text":"2 is also the number of the free variables."},{"Start":"05:16.290 ","End":"05:20.850","Text":"Let\u0027s just verify at least we can do that."},{"Start":"05:20.850 ","End":"05:25.280","Text":"That these two basis members really do"},{"Start":"05:25.280 ","End":"05:30.110","Text":"belong to W. I want to check that these two conditions are satisfied."},{"Start":"05:30.110 ","End":"05:31.355","Text":"I look at the first one,"},{"Start":"05:31.355 ","End":"05:36.010","Text":"a plus b equals c, minus 1 and 1 is 0,"},{"Start":"05:36.010 ","End":"05:42.075","Text":"and also b plus c is d because 1 plus 0 is 1,"},{"Start":"05:42.075 ","End":"05:48.760","Text":"and similarly here, and obviously these two are linearly independent."},{"Start":"05:50.330 ","End":"05:52.835","Text":"That\u0027s the answer."},{"Start":"05:52.835 ","End":"05:56.975","Text":"Here\u0027s the basis and here\u0027s the dimension."},{"Start":"05:56.975 ","End":"05:59.155","Text":"Onto the next question."},{"Start":"05:59.155 ","End":"06:01.350","Text":"Here\u0027s the next exercise,"},{"Start":"06:01.350 ","End":"06:05.150","Text":"and once again, it\u0027s slightly disguised."},{"Start":"06:05.150 ","End":"06:10.180","Text":"We don\u0027t see any homogeneous system of linear equations here."},{"Start":"06:10.180 ","End":"06:13.340","Text":"We have to work a little bit to get it to that form."},{"Start":"06:13.340 ","End":"06:15.890","Text":"That is typical an exam question,"},{"Start":"06:15.890 ","End":"06:21.725","Text":"is not to just give at you spoon-fed and say here is the system, slightly disguise it."},{"Start":"06:21.725 ","End":"06:26.790","Text":"Here we have a condition involving dot product of vectors."},{"Start":"06:26.790 ","End":"06:28.835","Text":"If you don\u0027t remember what dot product is,"},{"Start":"06:28.835 ","End":"06:31.535","Text":"perhaps now\u0027s the time to go and review it."},{"Start":"06:31.535 ","End":"06:36.450","Text":"We\u0027re in the space of 6, 6-dimensional space,"},{"Start":"06:36.450 ","End":"06:43.040","Text":"and we want all the vectors such that the dot product of u with this particular vector,"},{"Start":"06:43.040 ","End":"06:47.155","Text":"just six 1s is 0,"},{"Start":"06:47.155 ","End":"06:51.180","Text":"and we want to find,"},{"Start":"06:51.180 ","End":"06:54.155","Text":"as usual, basis and dimension."},{"Start":"06:54.155 ","End":"06:58.175","Text":"Of course as it is, we don\u0027t even know that W is a subspace,"},{"Start":"06:58.175 ","End":"07:04.860","Text":"but we\u0027ll soon show that this is a solution space of an SLE, a homogeneous one."},{"Start":"07:05.520 ","End":"07:10.315","Text":"Let\u0027s try the typical element like this, x_1 through x_6."},{"Start":"07:10.315 ","End":"07:13.120","Text":"I could have used a, b, c, d, e, f"},{"Start":"07:13.120 ","End":"07:17.770","Text":"but I want to generalize this to a general number, not just 6."},{"Start":"07:17.770 ","End":"07:19.870","Text":"I prefer the subscript,"},{"Start":"07:19.870 ","End":"07:23.590","Text":"6 variables just all x\u0027s with a different subscript."},{"Start":"07:23.590 ","End":"07:27.295","Text":"Let\u0027s see if I can interpret this condition."},{"Start":"07:27.295 ","End":"07:32.050","Text":"I just copied it here and that will replace u with this,"},{"Start":"07:32.050 ","End":"07:33.760","Text":"and this is what we get."},{"Start":"07:33.760 ","End":"07:35.785","Text":"If you remember how to do dot product,"},{"Start":"07:35.785 ","End":"07:40.704","Text":"just multiply component-wise in pairs and then add the products."},{"Start":"07:40.704 ","End":"07:45.760","Text":"It\u0027s x_1 times 1 plus x_2 times 1 plus x_3 times 1."},{"Start":"07:45.760 ","End":"07:51.205","Text":"In short, this is what the dot-product is and it\u0027s equal to 0."},{"Start":"07:51.205 ","End":"07:55.060","Text":"Actually this is a system of linear equations,"},{"Start":"07:55.060 ","End":"07:57.595","Text":"I could put a curly brace if that helps."},{"Start":"07:57.595 ","End":"08:01.570","Text":"It\u0027s just one equation and six unknowns."},{"Start":"08:01.570 ","End":"08:04.000","Text":"This is the leading term,"},{"Start":"08:04.000 ","End":"08:05.170","Text":"it\u0027s a leading variable,"},{"Start":"08:05.170 ","End":"08:06.250","Text":"a pivot variable,"},{"Start":"08:06.250 ","End":"08:08.935","Text":"a basic variables, various names for it."},{"Start":"08:08.935 ","End":"08:11.395","Text":"More importantly, the other ones,"},{"Start":"08:11.395 ","End":"08:16.970","Text":"the ones that are x_2 through x_6."},{"Start":"08:17.490 ","End":"08:22.690","Text":"What we have here is one equation in six unknowns."},{"Start":"08:22.690 ","End":"08:27.925","Text":"Notice that this system already is in row echelon form."},{"Start":"08:27.925 ","End":"08:31.555","Text":"The leading or pivot variable is x_1,"},{"Start":"08:31.555 ","End":"08:34.765","Text":"which means that x_2 through x_6,"},{"Start":"08:34.765 ","End":"08:38.455","Text":"that would be five of them are free variables."},{"Start":"08:38.455 ","End":"08:42.549","Text":"Once again, I\u0027m going to use the second method,"},{"Start":"08:42.549 ","End":"08:46.885","Text":"the one I called wandering ones informally."},{"Start":"08:46.885 ","End":"08:50.950","Text":"Remember it only good for homogeneous systems,"},{"Start":"08:50.950 ","End":"08:53.875","Text":"which of course it is because of the 0 here."},{"Start":"08:53.875 ","End":"08:56.830","Text":"For each of these five free variables,"},{"Start":"08:56.830 ","End":"09:02.260","Text":"we let one of them equal 1 and the rest equals 0 alternately."},{"Start":"09:02.260 ","End":"09:04.270","Text":"Let\u0027s say, first time round,"},{"Start":"09:04.270 ","End":"09:09.445","Text":"I say the x_2 is 1 all four others 0."},{"Start":"09:09.445 ","End":"09:11.545","Text":"From these I compute x_1,"},{"Start":"09:11.545 ","End":"09:13.705","Text":"x_1 is always computed from the others,"},{"Start":"09:13.705 ","End":"09:15.220","Text":"comes out to be this."},{"Start":"09:15.220 ","End":"09:16.750","Text":"I\u0027m just going to show you all of them."},{"Start":"09:16.750 ","End":"09:18.250","Text":"Why do them one at a time?"},{"Start":"09:18.250 ","End":"09:20.470","Text":"Notice that the next time round I\u0027ll let that x_3,"},{"Start":"09:20.470 ","End":"09:22.450","Text":"be 1 and the others 0,"},{"Start":"09:22.450 ","End":"09:23.830","Text":"then x_4 is 1,"},{"Start":"09:23.830 ","End":"09:26.680","Text":"and then x_5 is 1 and then x_6 is 1."},{"Start":"09:26.680 ","End":"09:34.400","Text":"In each case, so happens that x_1 comes out to be minus 1."},{"Start":"09:35.190 ","End":"09:38.740","Text":"If we collect these five vectors,"},{"Start":"09:38.740 ","End":"09:40.195","Text":"but you got to put them in order."},{"Start":"09:40.195 ","End":"09:42.280","Text":"First x_1, then x_2,"},{"Start":"09:42.280 ","End":"09:44.365","Text":"then x_3, x_4, x_5, x_6."},{"Start":"09:44.365 ","End":"09:47.560","Text":"The first row would give us the minus 1 is here."},{"Start":"09:47.560 ","End":"09:49.795","Text":"This one goes here,"},{"Start":"09:49.795 ","End":"09:52.220","Text":"and the other four, 0s."},{"Start":"09:52.220 ","End":"09:57.220","Text":"We get five vectors this way and these five vectors are"},{"Start":"09:57.220 ","End":"10:05.005","Text":"the basis B for the subspace W. If I count how many are here,"},{"Start":"10:05.005 ","End":"10:07.555","Text":"then we see that the dimension is 5,"},{"Start":"10:07.555 ","End":"10:12.250","Text":"and just want to remind you it can be useful,"},{"Start":"10:12.250 ","End":"10:14.650","Text":"that we can compute 5 in two other ways."},{"Start":"10:14.650 ","End":"10:17.830","Text":"One of them is to count the number of free variables,"},{"Start":"10:17.830 ","End":"10:19.240","Text":"so there were five of them."},{"Start":"10:19.240 ","End":"10:21.160","Text":"The other way was to say, okay,"},{"Start":"10:21.160 ","End":"10:23.530","Text":"I have six variables in total and"},{"Start":"10:23.530 ","End":"10:27.955","Text":"subtract the number of equations in echelon form, which is 1."},{"Start":"10:27.955 ","End":"10:31.610","Text":"So 6 minus 1 is 5 also."},{"Start":"10:31.860 ","End":"10:37.390","Text":"The next problem, here it is,"},{"Start":"10:37.390 ","End":"10:41.140","Text":"it\u0027s actually just a generalization of the previous one."},{"Start":"10:41.140 ","End":"10:42.715","Text":"In the previous one,"},{"Start":"10:42.715 ","End":"10:46.180","Text":"we had 6, n was 6."},{"Start":"10:46.180 ","End":"10:48.910","Text":"But nothing special about 6,"},{"Start":"10:48.910 ","End":"10:52.195","Text":"let\u0027s just generalize it to R^n."},{"Start":"10:52.195 ","End":"10:57.340","Text":"Only when you don\u0027t know what n is you\u0027ll have to use a lot of the dot,"},{"Start":"10:57.340 ","End":"11:00.640","Text":"dot, dot, also called the ellipsis,"},{"Start":"11:00.640 ","End":"11:03.620","Text":"meaning and so on and so on."},{"Start":"11:05.280 ","End":"11:14.035","Text":"As before, we\u0027ll take a typical member of W and the components,"},{"Start":"11:14.035 ","End":"11:16.420","Text":"we won\u0027t use letters, we don\u0027t know how many they are,"},{"Start":"11:16.420 ","End":"11:20.470","Text":"so we just use subscripts again, x_1 through x_n."},{"Start":"11:20.470 ","End":"11:23.245","Text":"These are the n variables."},{"Start":"11:23.245 ","End":"11:27.160","Text":"Now, we want to interpret this condition and hopefully get"},{"Start":"11:27.160 ","End":"11:32.365","Text":"a homogeneous system of linear equations from this,"},{"Start":"11:32.365 ","End":"11:36.310","Text":"so replace u by what it is."},{"Start":"11:36.310 ","End":"11:39.760","Text":"Then once again, we\u0027re going to use dot-product."},{"Start":"11:39.760 ","End":"11:44.915","Text":"Remember we just multiply pairs of components and add them."},{"Start":"11:44.915 ","End":"11:52.400","Text":"We get x_1 times 1 plus x_2 times 1 and so on up to x_n times 1."},{"Start":"11:52.400 ","End":"11:56.830","Text":"This is what we get and it\u0027s a system,"},{"Start":"11:56.830 ","End":"11:59.380","Text":"I\u0027ll just put a curly brace to show you that I can look at it as"},{"Start":"11:59.380 ","End":"12:04.270","Text":"a system of one equation in n unknowns."},{"Start":"12:04.270 ","End":"12:06.850","Text":"It is already in echelon form,"},{"Start":"12:06.850 ","End":"12:09.850","Text":"so at least we can skip that part."},{"Start":"12:09.850 ","End":"12:15.475","Text":"The pivotal leading term is x_1,"},{"Start":"12:15.475 ","End":"12:18.460","Text":"x_1 is going to be the variable that depends on the others"},{"Start":"12:18.460 ","End":"12:22.510","Text":"and x_2 through x_n will be the free variables."},{"Start":"12:22.510 ","End":"12:30.400","Text":"Actually, there will be n minus 1 free variables because n altogether but excluding x_1."},{"Start":"12:30.400 ","End":"12:34.720","Text":"N once again, will use that Method two which I informally"},{"Start":"12:34.720 ","End":"12:41.470","Text":"called the wandering ones or just the wondering one."},{"Start":"12:41.470 ","End":"12:49.210","Text":"First-time round I let x_2 equal 1 and the other free variables 0."},{"Start":"12:49.210 ","End":"12:52.870","Text":"We always compute x_1 from the remaining in"},{"Start":"12:52.870 ","End":"12:56.155","Text":"this case because they all add up to 0 and x_2 is 1,"},{"Start":"12:56.155 ","End":"12:59.410","Text":"that makes x_1 be minus 1."},{"Start":"12:59.410 ","End":"13:04.840","Text":"Next time round I let x_3 be 1,"},{"Start":"13:04.840 ","End":"13:07.570","Text":"and x_2 is 0 and the others are 0."},{"Start":"13:07.570 ","End":"13:12.010","Text":"Again you get x_1 is minus 1, and so on,"},{"Start":"13:12.010 ","End":"13:21.520","Text":"until we get to the last one where we put x_n is 1 and x_2 up to x_n minus 1 is 0."},{"Start":"13:21.520 ","End":"13:24.655","Text":"Again, x_1 is minus 1."},{"Start":"13:24.655 ","End":"13:34.450","Text":"We want to collect these n minus 1 vectors together into a basis."},{"Start":"13:34.450 ","End":"13:37.240","Text":"You just have to make sure to put things in the right order."},{"Start":"13:37.240 ","End":"13:39.610","Text":"X_1 is first and then x_2,"},{"Start":"13:39.610 ","End":"13:41.230","Text":"x_3 up to x_n."},{"Start":"13:41.230 ","End":"13:43.840","Text":"This first line gives us the minus 1,"},{"Start":"13:43.840 ","End":"13:47.570","Text":"then the 1 and then the rest of them, 0s and so on."},{"Start":"13:47.700 ","End":"13:51.170","Text":"This is the basis."},{"Start":"13:51.180 ","End":"13:54.520","Text":"The dimension is n minus 1,"},{"Start":"13:54.520 ","End":"13:55.855","Text":"which like we said,"},{"Start":"13:55.855 ","End":"13:59.320","Text":"is also equal to the number of free variables,"},{"Start":"13:59.320 ","End":"14:00.955","Text":"which we said is n minus 1."},{"Start":"14:00.955 ","End":"14:04.750","Text":"It\u0027s also equal to the total number of variables n"},{"Start":"14:04.750 ","End":"14:08.770","Text":"minus the number of equations in the echelon form,"},{"Start":"14:08.770 ","End":"14:11.540","Text":"so again n minus 1."},{"Start":"14:11.670 ","End":"14:15.200","Text":"We are done with this clip."}],"ID":9971},{"Watched":false,"Name":"Exercise 1","Duration":"14m 20s","ChapterTopicVideoID":9927,"CourseChapterTopicPlaylistID":31677,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9927.jpeg","UploadDate":"2017-08-07T11:37:23.2470000","DurationForVideoObject":"PT14M20S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:03.640","Text":"In this exercise, which is fairly lengthy,"},{"Start":"00:03.640 ","End":"00:06.355","Text":"doesn\u0027t even fit in on the screen. That\u0027s okay."},{"Start":"00:06.355 ","End":"00:10.690","Text":"We have 3 systems of linear equations, 1, 2,"},{"Start":"00:10.690 ","End":"00:14.770","Text":"and 3, but they\u0027re not just any old systems, they are homogeneous."},{"Start":"00:14.770 ","End":"00:19.660","Text":"I perhaps should mention that it\u0027s important enough."},{"Start":"00:19.660 ","End":"00:24.220","Text":"Now one of the things about homogeneous linear equations is that,"},{"Start":"00:24.220 ","End":"00:28.060","Text":"the set of solutions is a vector subspace."},{"Start":"00:28.060 ","End":"00:33.100","Text":"In this case, we are in our 4 because we have x, y, z,"},{"Start":"00:33.100 ","End":"00:38.425","Text":"and w. We always have 0 as a solution."},{"Start":"00:38.425 ","End":"00:39.760","Text":"I mean, if I put x, y, z,"},{"Start":"00:39.760 ","End":"00:42.340","Text":"and w or 0, certainly that will work."},{"Start":"00:42.340 ","End":"00:45.320","Text":"In general, we get a subspace."},{"Start":"00:45.320 ","End":"00:51.050","Text":"Now, the space of solutions here is going to be called w,"},{"Start":"00:51.050 ","End":"00:53.765","Text":"here we\u0027ll call it u,"},{"Start":"00:53.765 ","End":"01:02.780","Text":"and here we\u0027ll call it v and we want to find a basis for each of these."},{"Start":"01:02.780 ","End":"01:03.860","Text":"That\u0027s the first part."},{"Start":"01:03.860 ","End":"01:08.810","Text":"Find the basis for w for u and for v, and the dimension."},{"Start":"01:08.810 ","End":"01:11.780","Text":"Once you have a basis the dimension is easy."},{"Start":"01:11.780 ","End":"01:21.305","Text":"Then after that, we want to find the basis for the sum of these 2 sub-spaces."},{"Start":"01:21.305 ","End":"01:25.400","Text":"Hope you remember what a sum of 2 spaces is."},{"Start":"01:25.400 ","End":"01:30.800","Text":"If not, maybe I\u0027ll remind you what u plus v is?"},{"Start":"01:30.800 ","End":"01:34.850","Text":"Is the set of all little u plus little v,"},{"Start":"01:34.850 ","End":"01:40.185","Text":"where u comes from u and v comes from v,"},{"Start":"01:40.185 ","End":"01:42.295","Text":"in case you\u0027ve forgotten."},{"Start":"01:42.295 ","End":"01:48.050","Text":"We also want the dimension of the intersection."},{"Start":"01:48.050 ","End":"01:53.825","Text":"The sum of 2 vector spaces is a vector space and so is the intersection."},{"Start":"01:53.825 ","End":"01:58.190","Text":"Then finally, we\u0027re going to find a basis"},{"Start":"01:58.190 ","End":"02:02.660","Text":"for this intersection of u and v. That\u0027s a lot of work."},{"Start":"02:02.660 ","End":"02:06.295","Text":"Let\u0027s get started on a fresh page."},{"Start":"02:06.295 ","End":"02:08.630","Text":"In a part 1,"},{"Start":"02:08.630 ","End":"02:12.094","Text":"we want to find the solution space of this,"},{"Start":"02:12.094 ","End":"02:15.230","Text":"and we\u0027ll do it with matrices."},{"Start":"02:15.230 ","End":"02:17.660","Text":"This is the corresponding matrix."},{"Start":"02:17.660 ","End":"02:20.480","Text":"Remember with homogeneous, we don\u0027t need the augmented matrix."},{"Start":"02:20.480 ","End":"02:23.330","Text":"We don\u0027t need a column of zeros."},{"Start":"02:23.330 ","End":"02:25.970","Text":"Let\u0027s do row operations."},{"Start":"02:25.970 ","End":"02:28.930","Text":"Let\u0027s make 0 here and here."},{"Start":"02:28.930 ","End":"02:31.880","Text":"We subtract 3 times this row from"},{"Start":"02:31.880 ","End":"02:35.900","Text":"the second row and we add 5 times this row to the last row,"},{"Start":"02:35.900 ","End":"02:38.080","Text":"and then we\u0027ll get this,"},{"Start":"02:38.080 ","End":"02:43.415","Text":"still not echelon form I need a 0 where the 8 is."},{"Start":"02:43.415 ","End":"02:48.170","Text":"We just have to add twice this row to this row, we get a 0 here."},{"Start":"02:48.170 ","End":"02:52.270","Text":"As a matter of fact, the whole row comes out to be 0."},{"Start":"02:52.270 ","End":"02:58.955","Text":"Now, what I want to do is go back from the matrix form to the equation form."},{"Start":"02:58.955 ","End":"03:04.540","Text":"Here they are, 2 equations in 4 unknowns."},{"Start":"03:04.540 ","End":"03:08.270","Text":"Z and w are the free variables,"},{"Start":"03:08.270 ","End":"03:14.810","Text":"and then x and y are called pivot leading variables."},{"Start":"03:14.810 ","End":"03:18.280","Text":"Anyway, they\u0027re constrained, they depend on z and w,"},{"Start":"03:18.280 ","End":"03:21.335","Text":"so we let z and w be a parameter."},{"Start":"03:21.335 ","End":"03:24.050","Text":"No, change of mind, we don\u0027t want the general solution,"},{"Start":"03:24.050 ","End":"03:25.205","Text":"we just want the basis."},{"Start":"03:25.205 ","End":"03:29.975","Text":"We\u0027ll use the method of the and I called it the wandering ones."},{"Start":"03:29.975 ","End":"03:35.695","Text":"Each time we let one of the free variables be 1 and the others 0."},{"Start":"03:35.695 ","End":"03:37.170","Text":"First-time around then let,"},{"Start":"03:37.170 ","End":"03:40.080","Text":"w be 1 and z be 0."},{"Start":"03:40.080 ","End":"03:42.410","Text":"Once I have w and z,"},{"Start":"03:42.410 ","End":"03:46.460","Text":"everything else follows from back substitution from z and w here,"},{"Start":"03:46.460 ","End":"03:48.460","Text":"we get y here,"},{"Start":"03:48.460 ","End":"03:50.150","Text":"and then once we have y, z,"},{"Start":"03:50.150 ","End":"03:52.655","Text":"and w, we plug in here and we get x."},{"Start":"03:52.655 ","End":"03:57.180","Text":"Anyway, I\u0027ll leave you to pause and check if these are the calculations."},{"Start":"03:57.950 ","End":"04:01.420","Text":"Remember the solution space we called it w,"},{"Start":"04:01.420 ","End":"04:04.310","Text":"the basis would be these,"},{"Start":"04:04.310 ","End":"04:06.050","Text":"but you got to get them in the right order."},{"Start":"04:06.050 ","End":"04:11.955","Text":"We want x, y, z, w. This one gives us this,"},{"Start":"04:11.955 ","End":"04:16.410","Text":"and this one gives us the x, y, z, w in that order."},{"Start":"04:17.650 ","End":"04:22.860","Text":"Oh sorry, I might gotten the wrong way around."},{"Start":"04:22.860 ","End":"04:28.220","Text":"This row belongs here and this one belongs here."},{"Start":"04:28.220 ","End":"04:32.390","Text":"That\u0027s right. Like you can see a 2.5 here from the y. Yeah,"},{"Start":"04:32.390 ","End":"04:34.650","Text":"that\u0027s the right way around."},{"Start":"04:34.850 ","End":"04:40.070","Text":"The other dimension that we have to do is count how many elements in the basis,"},{"Start":"04:40.070 ","End":"04:42.590","Text":"and that is 2."},{"Start":"04:42.590 ","End":"04:46.655","Text":"That\u0027s that. That\u0027s the first of 3."},{"Start":"04:46.655 ","End":"04:50.295","Text":"The second homogeneous system was this,"},{"Start":"04:50.295 ","End":"04:54.070","Text":"and here\u0027s the matrix."},{"Start":"04:54.070 ","End":"05:01.280","Text":"We want to subtract the first row from both the second and from the third."},{"Start":"05:01.280 ","End":"05:04.700","Text":"Then we get this still multinational on form."},{"Start":"05:04.700 ","End":"05:06.820","Text":"We need a 0 here."},{"Start":"05:06.820 ","End":"05:11.750","Text":"We subtract twice the second row from the third row,"},{"Start":"05:11.750 ","End":"05:16.295","Text":"and we get this, and once again we have a row of zeros."},{"Start":"05:16.295 ","End":"05:25.655","Text":"This means that once again we have just 2 equations in 4 unknowns."},{"Start":"05:25.655 ","End":"05:27.830","Text":"Again, we\u0027re going to use this method of"},{"Start":"05:27.830 ","End":"05:32.215","Text":"the wandering ones to each time let some think,"},{"Start":"05:32.215 ","End":"05:36.315","Text":"one of the free variables be 1 and the others 0."},{"Start":"05:36.315 ","End":"05:41.880","Text":"In this case we have z and w. Just like before at onetime I let w be 1,"},{"Start":"05:41.880 ","End":"05:45.775","Text":"z is 0, and then the other way around w is 0, z is 1."},{"Start":"05:45.775 ","End":"05:48.760","Text":"I first compute the y from the second equation,"},{"Start":"05:48.760 ","End":"05:52.300","Text":"and then I have all 3 of them, y,"},{"Start":"05:52.300 ","End":"05:54.280","Text":"z, and w to compute x from,"},{"Start":"05:54.280 ","End":"05:55.930","Text":"and this is what it comes out."},{"Start":"05:55.930 ","End":"05:58.345","Text":"Now I just have to get them in the proper order,"},{"Start":"05:58.345 ","End":"06:01.255","Text":"which gives us the basis for u."},{"Start":"06:01.255 ","End":"06:04.180","Text":"We want 1, 2, 0, 1."},{"Start":"06:04.180 ","End":"06:06.835","Text":"I guess that 1 is this one,"},{"Start":"06:06.835 ","End":"06:09.610","Text":"and this one is that one,"},{"Start":"06:09.610 ","End":"06:12.190","Text":"and that\u0027s the basis for u."},{"Start":"06:12.190 ","End":"06:15.190","Text":"Again dimension is 2."},{"Start":"06:15.190 ","End":"06:21.565","Text":"Now the third SLE homogeneous, Here\u0027s its matrix."},{"Start":"06:21.565 ","End":"06:26.750","Text":"I said we don\u0027t need the 0s with homogeneous and we want to bring it to echelon form,"},{"Start":"06:26.750 ","End":"06:30.995","Text":"so I would subtract twice the first row from the second row."},{"Start":"06:30.995 ","End":"06:35.620","Text":"If we do that, what we end up with is this, which is,"},{"Start":"06:35.620 ","End":"06:37.375","Text":"I mean there\u0027s a 0 row,"},{"Start":"06:37.375 ","End":"06:43.195","Text":"so there\u0027s only going to be one equation in 4 unknowns."},{"Start":"06:43.195 ","End":"06:45.985","Text":"This is the equation from the top row,"},{"Start":"06:45.985 ","End":"06:49.720","Text":"another 3 free variables, y, z, and w."},{"Start":"06:49.720 ","End":"06:52.690","Text":"We going to have 3 members of"},{"Start":"06:52.690 ","End":"06:58.130","Text":"the basis each time we\u0027ll let one of these be 1 and the others 0."},{"Start":"06:58.130 ","End":"07:01.145","Text":"Here we are. I\u0027ve colored the 1,"},{"Start":"07:01.145 ","End":"07:06.575","Text":"you have a 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, for w, z, and y."},{"Start":"07:06.575 ","End":"07:08.240","Text":"Once we have w, z and y,"},{"Start":"07:08.240 ","End":"07:10.730","Text":"we can plug in here and get the x,"},{"Start":"07:10.730 ","End":"07:13.575","Text":"and then gather these together."},{"Start":"07:13.575 ","End":"07:15.610","Text":"That gives us a basis,"},{"Start":"07:15.610 ","End":"07:20.034","Text":"V was the name of the solution subspace for this."},{"Start":"07:20.034 ","End":"07:23.695","Text":"If we put them in the right order, let\u0027s see,"},{"Start":"07:23.695 ","End":"07:25.330","Text":"minus 1, 0, 0, 1,"},{"Start":"07:25.330 ","End":"07:27.100","Text":"that would be this one."},{"Start":"07:27.100 ","End":"07:30.940","Text":"I guess the middle one would be this one,"},{"Start":"07:30.940 ","End":"07:33.040","Text":"and finally 1, 1, 0, 0,"},{"Start":"07:33.040 ","End":"07:35.125","Text":"yeah, the last one,"},{"Start":"07:35.125 ","End":"07:37.870","Text":"and the dimension this time is of course 3,"},{"Start":"07:37.870 ","End":"07:39.950","Text":"as we can count."},{"Start":"07:40.050 ","End":"07:43.225","Text":"Now on to part b, b1."},{"Start":"07:43.225 ","End":"07:47.750","Text":"Here again,"},{"Start":"07:57.690 ","End":"08:03.400","Text":"if I want to compute a basis for U plus V,"},{"Start":"08:03.400 ","End":"08:12.085","Text":"all we do is throw in together the union of these 2 basis,"},{"Start":"08:12.085 ","End":"08:18.265","Text":"and then just do row operations to cut it down to size."},{"Start":"08:18.265 ","End":"08:21.820","Text":"I\u0027ll show you. I put these,"},{"Start":"08:21.820 ","End":"08:24.100","Text":"2 plus 3, 5 vectors in here."},{"Start":"08:24.100 ","End":"08:27.100","Text":"For some reason I did it from right to left."},{"Start":"08:27.100 ","End":"08:28.495","Text":"This, then this,"},{"Start":"08:28.495 ","End":"08:31.370","Text":"and then this, this, this."},{"Start":"08:33.090 ","End":"08:37.180","Text":"I even put the Vs first."},{"Start":"08:37.180 ","End":"08:40.015","Text":"This one is here,"},{"Start":"08:40.015 ","End":"08:41.275","Text":"this one is here."},{"Start":"08:41.275 ","End":"08:46.660","Text":"I don\u0027t know why I did V. These 3 from the V,"},{"Start":"08:46.660 ","End":"08:49.915","Text":"these are from U."},{"Start":"08:49.915 ","End":"08:52.270","Text":"That\u0027s not a U, that\u0027s the union."},{"Start":"08:52.270 ","End":"08:54.040","Text":"Maybe I\u0027ll erase it."},{"Start":"08:54.040 ","End":"08:57.475","Text":"Here I just get the order, it doesn\u0027t really matter."},{"Start":"08:57.475 ","End":"08:59.260","Text":"First this, this, this,"},{"Start":"08:59.260 ","End":"09:01.660","Text":"this, this, that\u0027s how the order is."},{"Start":"09:01.660 ","End":"09:09.730","Text":"Now, something you know so well to do is to bring to row echelon form."},{"Start":"09:09.730 ","End":"09:12.400","Text":"I would just skim through that quickly."},{"Start":"09:12.400 ","End":"09:18.670","Text":"Add multiples of the top row to all the rest to make these 0."},{"Start":"09:18.670 ","End":"09:20.860","Text":"Okay. I\u0027ll spare you the details,"},{"Start":"09:20.860 ","End":"09:23.125","Text":"this is what we get."},{"Start":"09:23.125 ","End":"09:27.655","Text":"Next, you want 0s in these places."},{"Start":"09:27.655 ","End":"09:32.590","Text":"This is what we get after we add multiples of the second row."},{"Start":"09:32.590 ","End":"09:35.530","Text":"If I could just subtract the second row from each of the third, fourth,"},{"Start":"09:35.530 ","End":"09:39.045","Text":"and fifth and get this."},{"Start":"09:39.045 ","End":"09:43.020","Text":"This, just a second,"},{"Start":"09:43.020 ","End":"09:47.955","Text":"can finally be brought to row echelon form by subtracting"},{"Start":"09:47.955 ","End":"09:57.620","Text":"the third row from the fourth row."},{"Start":"09:57.620 ","End":"10:00.175","Text":"Yeah. Anyway, we do have"},{"Start":"10:00.175 ","End":"10:09.895","Text":"the echelon form and 2 rows of 0s."},{"Start":"10:09.895 ","End":"10:15.655","Text":"The basis for the sum of the vector subspaces,"},{"Start":"10:15.655 ","End":"10:21.860","Text":"U plus V is just what we got here without the 0s."},{"Start":"10:22.590 ","End":"10:27.040","Text":"Again, I don\u0027t know why I\u0027m doing it in the wrong order."},{"Start":"10:27.040 ","End":"10:30.730","Text":"This one, 1, 1, 0, 0 is here,"},{"Start":"10:30.730 ","End":"10:33.025","Text":"guess I\u0027m going in this direction."},{"Start":"10:33.025 ","End":"10:36.160","Text":"Then 0, 1, 1, 0 here,"},{"Start":"10:36.160 ","End":"10:37.900","Text":"and then 0, 0, minus 1, 1"},{"Start":"10:37.900 ","End":"10:41.395","Text":"is here, all accounted for."},{"Start":"10:41.395 ","End":"10:47.560","Text":"The dimension is 3 by counting the basis."},{"Start":"10:47.560 ","End":"10:56.770","Text":"We\u0027re still in b, b2, let us expose it all,"},{"Start":"10:56.770 ","End":"10:59.365","Text":"there\u0027s no point showing you a row at a time."},{"Start":"10:59.365 ","End":"11:06.775","Text":"After the dimension of U plus V, the subspace,"},{"Start":"11:06.775 ","End":"11:14.845","Text":"there\u0027s a formula that this is equal to the dimension of each one of them added,"},{"Start":"11:14.845 ","End":"11:19.105","Text":"and then we subtract the dimension of the intersection."},{"Start":"11:19.105 ","End":"11:22.060","Text":"Now, since we know this and this,"},{"Start":"11:22.060 ","End":"11:24.565","Text":"we can use it to find what this is."},{"Start":"11:24.565 ","End":"11:26.680","Text":"This is 3, this was 2,"},{"Start":"11:26.680 ","End":"11:31.795","Text":"this was 3 from the previous exercises."},{"Start":"11:31.795 ","End":"11:35.695","Text":"We have 3 equals 2 plus 3 minus something."},{"Start":"11:35.695 ","End":"11:37.765","Text":"That something is 2."},{"Start":"11:37.765 ","End":"11:45.040","Text":"That\u0027s the answer for the dimension of the intersection subspace."},{"Start":"11:45.040 ","End":"11:51.760","Text":"Now for part c, we want the intersection of the solution sets."},{"Start":"11:51.760 ","End":"11:57.730","Text":"We want a U intersection with V. That was the 2 systems of equations,"},{"Start":"11:57.730 ","End":"11:59.620","Text":"we call them 2 and 3."},{"Start":"11:59.620 ","End":"12:03.580","Text":"What we do is if you want something to belong to both,"},{"Start":"12:03.580 ","End":"12:09.620","Text":"it has to satisfy the equations of 2 and 3."},{"Start":"12:09.810 ","End":"12:18.025","Text":"Here we are. The first 3 equations belong to U,"},{"Start":"12:18.025 ","End":"12:23.965","Text":"that\u0027s part 2, and the second 2 belong to the system number 3."},{"Start":"12:23.965 ","End":"12:26.935","Text":"We take them all together and that will give us the and,"},{"Start":"12:26.935 ","End":"12:29.660","Text":"the intersection of both."},{"Start":"12:29.790 ","End":"12:33.940","Text":"Okay. Now, here\u0027s the matrix corresponding to this,"},{"Start":"12:33.940 ","End":"12:39.805","Text":"we just take the coefficients and we\u0027re going to do row operations on this."},{"Start":"12:39.805 ","End":"12:44.335","Text":"First step was to get all 0s here."},{"Start":"12:44.335 ","End":"12:51.475","Text":"Just subtracted this top row from these 3 and twice the top row from this,"},{"Start":"12:51.475 ","End":"12:53.755","Text":"and we get this."},{"Start":"12:53.755 ","End":"13:03.055","Text":"Then after we subtract twice this second row from the third row, we get this."},{"Start":"13:03.055 ","End":"13:06.310","Text":"Notice that there are only 2 rows,"},{"Start":"13:06.310 ","End":"13:09.880","Text":"and the rest of it is 0s."},{"Start":"13:09.880 ","End":"13:15.530","Text":"If I go back to the equation form,"},{"Start":"13:15.660 ","End":"13:17.875","Text":"here it is,"},{"Start":"13:17.875 ","End":"13:22.600","Text":"and w and z are free variables,"},{"Start":"13:22.600 ","End":"13:24.730","Text":"and x and y depend on these,"},{"Start":"13:24.730 ","End":"13:29.980","Text":"then we use back substitution and we\u0027ll use the wandering 1s method again."},{"Start":"13:29.980 ","End":"13:33.490","Text":"Here\u0027s where w is 1 and z is 0."},{"Start":"13:33.490 ","End":"13:34.990","Text":"First we compute y,"},{"Start":"13:34.990 ","End":"13:37.510","Text":"and then we use that to compute x."},{"Start":"13:37.510 ","End":"13:40.480","Text":"Similarly, the other way around w is 0,"},{"Start":"13:40.480 ","End":"13:42.010","Text":"z is 1,"},{"Start":"13:42.010 ","End":"13:51.850","Text":"and that gives us a basis for U intersection with V as being these 2 vectors."},{"Start":"13:51.850 ","End":"13:53.170","Text":"Once you have them in the right order,"},{"Start":"13:53.170 ","End":"13:55.030","Text":"Let\u0027s see, 1, 2, 0,1,"},{"Start":"13:55.030 ","End":"13:58.730","Text":"that\u0027s this 1 here."},{"Start":"13:59.010 ","End":"14:01.225","Text":"This 1, let\u0027s see,"},{"Start":"14:01.225 ","End":"14:04.030","Text":"minus 2, minus 1, 1, 0,"},{"Start":"14:04.030 ","End":"14:07.460","Text":"that would be this one here."},{"Start":"14:07.830 ","End":"14:10.945","Text":"I\u0027m just repeating what we saw earlier that"},{"Start":"14:10.945 ","End":"14:13.690","Text":"the dimension of U intersection V really is 2."},{"Start":"14:13.690 ","End":"14:16.735","Text":"We got it previously from a computation."},{"Start":"14:16.735 ","End":"14:20.660","Text":"Okay. That\u0027s it for this exercise."}],"ID":9972},{"Watched":false,"Name":"Exercise 2","Duration":"2m 32s","ChapterTopicVideoID":9928,"CourseChapterTopicPlaylistID":31677,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9928.jpeg","UploadDate":"2017-08-07T11:37:31.5900000","DurationForVideoObject":"PT2M32S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.610","Text":"In this exercise, we\u0027re looking at a subspace of 4,"},{"Start":"00:05.610 ","End":"00:07.800","Text":"4 dimensional real space."},{"Start":"00:07.800 ","End":"00:11.610","Text":"You want all the vectors a, b, c, d,"},{"Start":"00:11.610 ","End":"00:14.295","Text":"which satisfy these 2 conditions,"},{"Start":"00:14.295 ","End":"00:19.320","Text":"a equals c and b equals d. I\u0027ll give you an example."},{"Start":"00:19.320 ","End":"00:20.850","Text":"Let\u0027s take, I don\u0027t know,"},{"Start":"00:20.850 ","End":"00:26.580","Text":"1, 4, 1, 4. a equals c,"},{"Start":"00:26.580 ","End":"00:28.785","Text":"the first and the third are the same."},{"Start":"00:28.785 ","End":"00:32.925","Text":"b equals d, means the second and the fourth are the same."},{"Start":"00:32.925 ","End":"00:35.650","Text":"That\u0027s a subspace."},{"Start":"00:37.310 ","End":"00:39.710","Text":"It\u0027s a subset anyway,"},{"Start":"00:39.710 ","End":"00:41.180","Text":"we\u0027ll see that it is a subspace."},{"Start":"00:41.180 ","End":"00:48.415","Text":"We want to find a basis for this subspace and its dimension."},{"Start":"00:48.415 ","End":"00:56.090","Text":"Now, notice that I can rewrite these 2 conditions as a system of linear equations,"},{"Start":"00:56.090 ","End":"00:58.700","Text":"2 equations and 4 unknowns, a, b, c,"},{"Start":"00:58.700 ","End":"01:00.350","Text":"d. a equals c,"},{"Start":"01:00.350 ","End":"01:02.780","Text":"then a minus c is 0. b equal d,"},{"Start":"01:02.780 ","End":"01:04.355","Text":"b minus d is 0."},{"Start":"01:04.355 ","End":"01:06.520","Text":"Just arrange them nicely."},{"Start":"01:06.520 ","End":"01:10.415","Text":"Notice that it already is in echelon form."},{"Start":"01:10.415 ","End":"01:12.890","Text":"You just look for the free variables,"},{"Start":"01:12.890 ","End":"01:14.210","Text":"those would be c and d,"},{"Start":"01:14.210 ","End":"01:16.340","Text":"and a and b depend on them."},{"Start":"01:16.340 ","End":"01:18.230","Text":"To get the basis,"},{"Start":"01:18.230 ","End":"01:22.320","Text":"we use what I call the wandering 1 \u0027s method."},{"Start":"01:22.340 ","End":"01:27.285","Text":"Once we let d be 1 and c be 0,"},{"Start":"01:27.285 ","End":"01:33.010","Text":"and then the other way around we take d as 0 and c as 1."},{"Start":"01:33.220 ","End":"01:37.730","Text":"From this equation, we compute b from this 1 we compute a."},{"Start":"01:37.730 ","End":"01:42.015","Text":"Well, because a equals c and b equals d, that\u0027s pretty easy."},{"Start":"01:42.015 ","End":"01:44.145","Text":"d is 0, b is 0."},{"Start":"01:44.145 ","End":"01:50.235","Text":"Whatever c is a [inaudible] we\u0027ve got these solutions."},{"Start":"01:50.235 ","End":"01:54.150","Text":"Then these give us the basis,"},{"Start":"01:54.150 ","End":"01:57.615","Text":"the basis for the solution base u."},{"Start":"01:57.615 ","End":"02:00.540","Text":"The first 1 gives us 0, 1,"},{"Start":"02:00.540 ","End":"02:03.060","Text":"0, 1, which is this."},{"Start":"02:03.060 ","End":"02:04.280","Text":"Second 1 gives us 1,"},{"Start":"02:04.280 ","End":"02:06.485","Text":"0, 1, 0, which is this."},{"Start":"02:06.485 ","End":"02:12.920","Text":"Certainly these satisfy the condition that a equals c and b equals d. These 2 form"},{"Start":"02:12.920 ","End":"02:19.595","Text":"a basis for all solutions to this condition."},{"Start":"02:19.595 ","End":"02:24.710","Text":"They all are linear combination of these 2.The dimension of u,"},{"Start":"02:24.710 ","End":"02:33.270","Text":"of course, just by counting the number of elements in the basis is 2, and we\u0027re done."}],"ID":9973},{"Watched":false,"Name":"Exercise 3","Duration":"2m 11s","ChapterTopicVideoID":9929,"CourseChapterTopicPlaylistID":31677,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9929.jpeg","UploadDate":"2017-08-07T11:37:38.2100000","DurationForVideoObject":"PT2M11S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.190","Text":"We start out with a vector space R^4."},{"Start":"00:08.190 ","End":"00:12.690","Text":"We let U be the subset of R^4 all those a,"},{"Start":"00:12.690 ","End":"00:17.610","Text":"b, c, d, such that these 2 conditions are satisfied,"},{"Start":"00:17.610 ","End":"00:19.875","Text":"that c is the sum of a and b,"},{"Start":"00:19.875 ","End":"00:24.660","Text":"and d is the sum of b and c. We want to"},{"Start":"00:24.660 ","End":"00:30.500","Text":"find a basis for this subspace."},{"Start":"00:30.500 ","End":"00:33.950","Text":"Well, we\u0027ll show that it is a subspace much just a subset."},{"Start":"00:33.950 ","End":"00:35.480","Text":"If we have the basis,"},{"Start":"00:35.480 ","End":"00:38.400","Text":"then also the dimension."},{"Start":"00:38.600 ","End":"00:46.045","Text":"What we can do is write these 2 equations in this form."},{"Start":"00:46.045 ","End":"00:49.520","Text":"Then we can see that it\u0027s a system of linear equations,"},{"Start":"00:49.520 ","End":"00:53.660","Text":"but it\u0027s homogeneous because of the 0s on the right."},{"Start":"00:53.660 ","End":"01:00.220","Text":"Since it\u0027s homogeneous, the solution space is a subspace."},{"Start":"01:00.220 ","End":"01:02.930","Text":"What we have to do now is solve it."},{"Start":"01:02.930 ","End":"01:05.405","Text":"It already is in echelon form,"},{"Start":"01:05.405 ","End":"01:09.545","Text":"and clearly c and d are the free variables,"},{"Start":"01:09.545 ","End":"01:14.149","Text":"but a and b are dependent on those."},{"Start":"01:14.170 ","End":"01:18.620","Text":"We\u0027re going to use the method we just call it in this course wandering ones,"},{"Start":"01:18.620 ","End":"01:20.360","Text":"it\u0027s not a general name,"},{"Start":"01:20.360 ","End":"01:23.330","Text":"where from the free variables in this case d and c,"},{"Start":"01:23.330 ","End":"01:26.599","Text":"each time we let 1 of them be 1 and the rest 0."},{"Start":"01:26.599 ","End":"01:32.325","Text":"First we\u0027ll let d equal 1 and c be 0 and then the other way around."},{"Start":"01:32.325 ","End":"01:33.930","Text":"Once we have c and d,"},{"Start":"01:33.930 ","End":"01:38.800","Text":"we can compute b from here and once we have b, we compute a."},{"Start":"01:39.470 ","End":"01:42.960","Text":"We collect these in the right order, a, b,"},{"Start":"01:42.960 ","End":"01:46.470","Text":"c, d, this is minus 1, 1,"},{"Start":"01:46.470 ","End":"01:51.620","Text":"0, 1, That\u0027s this vector and this 1 gives us 2 minus 1,"},{"Start":"01:51.620 ","End":"01:53.615","Text":"1, 0, that\u0027s here."},{"Start":"01:53.615 ","End":"02:00.475","Text":"These 2 vectors will give us a basis of U."},{"Start":"02:00.475 ","End":"02:05.270","Text":"By counting the number of elements in the set 1,"},{"Start":"02:05.270 ","End":"02:08.540","Text":"2, that\u0027s the dimension of U."},{"Start":"02:08.540 ","End":"02:12.280","Text":"We\u0027re done."}],"ID":9974},{"Watched":false,"Name":"Exercise 4","Duration":"2m 43s","ChapterTopicVideoID":9926,"CourseChapterTopicPlaylistID":31677,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9926.jpeg","UploadDate":"2017-08-07T11:36:38.7700000","DurationForVideoObject":"PT2M43S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.990","Text":"In this exercise, we\u0027ll once again with the vector space R^4 dimensional space."},{"Start":"00:08.030 ","End":"00:12.465","Text":"This time we have an unusual condition."},{"Start":"00:12.465 ","End":"00:17.580","Text":"We take a subset which will be a subspace of all the vectors V"},{"Start":"00:17.580 ","End":"00:23.610","Text":"and R^4 such that the dot product of v with this particular vector,"},{"Start":"00:23.610 ","End":"00:28.200","Text":"1 minus 1, 1 minus 1 is 0."},{"Start":"00:28.200 ","End":"00:33.150","Text":"Our task is to find a basis and the dimension of"},{"Start":"00:33.150 ","End":"00:38.475","Text":"U with the understanding that it will turn out to be a subspace."},{"Start":"00:38.475 ","End":"00:42.360","Text":"So let\u0027s see, I hope you remember the dot product."},{"Start":"00:42.360 ","End":"00:45.490","Text":"Let\u0027s give V some coordinates."},{"Start":"00:45.490 ","End":"00:47.460","Text":"We\u0027ll call them X, Y, Z,"},{"Start":"00:47.460 ","End":"00:56.060","Text":"and T. I want to interpret what it means to say that V dot product with this vector is 0."},{"Start":"00:56.060 ","End":"00:58.890","Text":"Well, it just means that since V is X,"},{"Start":"00:58.890 ","End":"01:02.495","Text":"Y, Z, T, that the dot product of these 2 is 0."},{"Start":"01:02.495 ","End":"01:05.710","Text":"But we know how to do a dot product of 2 vectors."},{"Start":"01:05.710 ","End":"01:09.995","Text":"You multiply component wise and add the products."},{"Start":"01:09.995 ","End":"01:19.390","Text":"So we will get X times 1 plus Y times minus 1 plus C times 1 plus T times minus 1 is 0."},{"Start":"01:19.390 ","End":"01:21.730","Text":"In short, this is what we get."},{"Start":"01:21.730 ","End":"01:25.550","Text":"It\u0027s a homogeneous system of linear equations."},{"Start":"01:25.550 ","End":"01:28.800","Text":"1 equation, 4 unknowns."},{"Start":"01:28.800 ","End":"01:32.460","Text":"Only x is not free."},{"Start":"01:32.460 ","End":"01:34.500","Text":"The other 3 Y, Z,"},{"Start":"01:34.500 ","End":"01:37.210","Text":"and T are free variables."},{"Start":"01:37.210 ","End":"01:45.640","Text":"X is a pivot variable leading wherever you\u0027re really sure what it\u0027s called."},{"Start":"01:45.640 ","End":"01:47.870","Text":"It\u0027s constrained."},{"Start":"01:47.870 ","End":"01:52.270","Text":"We\u0027re going to use our system, our trick,"},{"Start":"01:52.270 ","End":"01:56.230","Text":"The wondering 1s for finding a basis from"},{"Start":"01:56.230 ","End":"02:02.805","Text":"the 3 free variables each time we let 1 of them be 1 and the other to 0."},{"Start":"02:02.805 ","End":"02:06.705","Text":"If I let T equal 1 and Y and Z are 0,"},{"Start":"02:06.705 ","End":"02:09.600","Text":"then X minus T is 0,"},{"Start":"02:09.600 ","End":"02:12.155","Text":"but T is 1, so X is 1."},{"Start":"02:12.155 ","End":"02:14.375","Text":"Similarly with the other 2,"},{"Start":"02:14.375 ","End":"02:18.200","Text":"then I\u0027ll have to do is arrange these in the order X, Y, Z,"},{"Start":"02:18.200 ","End":"02:23.740","Text":"T. The first 1 is 1,0,0,1 is here,"},{"Start":"02:23.740 ","End":"02:29.760","Text":"minus 1,0,1,0 that\u0027s here and 1,1,0,0 is here."},{"Start":"02:29.760 ","End":"02:33.460","Text":"That\u0027s the basis for U."},{"Start":"02:33.590 ","End":"02:40.445","Text":"The dimension is 3 just by counting how many members there are in the basis."},{"Start":"02:40.445 ","End":"02:43.200","Text":"Okay. Done."}],"ID":9975}],"Thumbnail":null,"ID":31677},{"Name":"Subspaces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Vector Subspaces","Duration":"14m 17s","ChapterTopicVideoID":25069,"CourseChapterTopicPlaylistID":31678,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/25069.jpeg","UploadDate":"2021-06-27T08:08:39.8800000","DurationForVideoObject":"PT14M17S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.140","Text":"Continuing with the topic of vector spaces."},{"Start":"00:04.140 ","End":"00:08.955","Text":"In this clip, we\u0027ll learn a new concept, vector subspaces."},{"Start":"00:08.955 ","End":"00:14.610","Text":"Let me start by returning to a familiar vector space R^3,"},{"Start":"00:14.610 ","End":"00:20.685","Text":"which is the set of all triplets x,y,z of real numbers."},{"Start":"00:20.685 ","End":"00:25.155","Text":"A couple of examples of members of R^3,"},{"Start":"00:25.155 ","End":"00:30.570","Text":"4 1 0, root 2 0 pi."},{"Start":"00:30.570 ","End":"00:34.380","Text":"Mainly I wrote this to remind you of the set theory notation."},{"Start":"00:34.380 ","End":"00:36.680","Text":"This little e means,"},{"Start":"00:36.680 ","End":"00:39.280","Text":"belongs to, in case you\u0027ve forgotten."},{"Start":"00:39.280 ","End":"00:45.710","Text":"This belongs to R^3 and this belongs to R^3 and an infinite number of others."},{"Start":"00:45.710 ","End":"00:48.395","Text":"Now, I\u0027m going to define a set."},{"Start":"00:48.395 ","End":"00:50.510","Text":"I\u0027ll call it W,"},{"Start":"00:50.510 ","End":"00:55.605","Text":"which is all the triplets a, b, c,"},{"Start":"00:55.605 ","End":"00:59.740","Text":"that belong to R^3 but with a condition,"},{"Start":"00:59.740 ","End":"01:04.940","Text":"this vertical line means such that a plus b plus c equals 1."},{"Start":"01:04.940 ","End":"01:07.885","Text":"The sum of the components is 1."},{"Start":"01:07.885 ","End":"01:10.705","Text":"Some examples of elements of"},{"Start":"01:10.705 ","End":"01:17.290","Text":"W,0,0,1 because if you add the components 0 plus 0 plus 1, it is 1."},{"Start":"01:17.290 ","End":"01:19.935","Text":"Another example, 1/2, 1/2,"},{"Start":"01:19.935 ","End":"01:23.580","Text":"0 also the sum of these 3 is 1."},{"Start":"01:23.580 ","End":"01:26.805","Text":"Third example, 4 and 8,"},{"Start":"01:26.805 ","End":"01:28.305","Text":"that brings me up to 12."},{"Start":"01:28.305 ","End":"01:30.319","Text":"If I write here minus 11,"},{"Start":"01:30.319 ","End":"01:35.785","Text":"then the sum is 1 and countless others."},{"Start":"01:35.785 ","End":"01:41.475","Text":"Now, notice that W is a subset of R^3."},{"Start":"01:41.475 ","End":"01:45.740","Text":"The reason it\u0027s a subset is by definition,"},{"Start":"01:45.740 ","End":"01:49.650","Text":"I only took triplets that belong to R^3."},{"Start":"01:49.650 ","End":"01:50.810","Text":"In the first place,"},{"Start":"01:50.810 ","End":"01:53.605","Text":"I just attached an extra condition."},{"Start":"01:53.605 ","End":"01:56.720","Text":"Like diagrammatically in set theory,"},{"Start":"01:56.720 ","End":"02:00.515","Text":"if this was the set R^3,"},{"Start":"02:00.515 ","End":"02:05.395","Text":"then W would just be a part of this,"},{"Start":"02:05.395 ","End":"02:10.000","Text":"label this W. Now,"},{"Start":"02:10.000 ","End":"02:13.160","Text":"the question is that although it\u0027s a subset,"},{"Start":"02:13.160 ","End":"02:16.415","Text":"is it also a subspace of R^3?"},{"Start":"02:16.415 ","End":"02:19.820","Text":"Of course, you can\u0027t tell me because I haven\u0027t defined it yet."},{"Start":"02:19.820 ","End":"02:21.200","Text":"That\u0027s what I\u0027m about to do."},{"Start":"02:21.200 ","End":"02:24.595","Text":"This was an introduction to defining a subspace."},{"Start":"02:24.595 ","End":"02:31.955","Text":"We start with a subset W of a vector space V. In our case,"},{"Start":"02:31.955 ","End":"02:39.860","Text":"this would be my V is the vector space and W is a subset of it."},{"Start":"02:39.860 ","End":"02:46.595","Text":"It\u0027s going to be called a subspace if 3 conditions are satisfied."},{"Start":"02:46.595 ","End":"02:50.705","Text":"When we say V, you can just think of R^n"},{"Start":"02:50.705 ","End":"02:55.160","Text":"because so far these are the only vector spaces that we\u0027ve encountered."},{"Start":"02:55.160 ","End":"02:57.545","Text":"Let\u0027s see what the conditions are."},{"Start":"02:57.545 ","End":"03:02.210","Text":"The first condition is that 0 and when I say 0,"},{"Start":"03:02.210 ","End":"03:05.480","Text":"I mean the 0 vector maybe put a little underline."},{"Start":"03:05.480 ","End":"03:11.135","Text":"The 0 vector must be inside the subset W. Now,"},{"Start":"03:11.135 ","End":"03:13.520","Text":"before I even show you points 2 and 3,"},{"Start":"03:13.520 ","End":"03:17.360","Text":"let\u0027s see if our example satisfies that."},{"Start":"03:17.360 ","End":"03:22.795","Text":"Now, in our case, the 0 is 0,0,0,"},{"Start":"03:22.795 ","End":"03:25.780","Text":"and does it belong to W?"},{"Start":"03:25.780 ","End":"03:30.815","Text":"Well, if it did, we would have to have a plus b plus c equals 1."},{"Start":"03:30.815 ","End":"03:34.460","Text":"In other words, 0 plus 0 plus 0 equals 1,"},{"Start":"03:34.460 ","End":"03:36.815","Text":"which is not true."},{"Start":"03:36.815 ","End":"03:42.690","Text":"Already, we\u0027ve fallen flat with the first condition we failed."},{"Start":"03:42.690 ","End":"03:45.110","Text":"It doesn\u0027t matter what the other 2 conditions are."},{"Start":"03:45.110 ","End":"03:48.070","Text":"For this example, it\u0027s been ruled out."},{"Start":"03:48.070 ","End":"03:51.590","Text":"Still, we want to know the general definition."},{"Start":"03:51.590 ","End":"03:55.715","Text":"Let\u0027s go on to requirement number 2."},{"Start":"03:55.715 ","End":"04:04.160","Text":"The second condition says that if u and v are in W,"},{"Start":"04:04.160 ","End":"04:08.060","Text":"then the sum of them, u plus v,"},{"Start":"04:08.060 ","End":"04:14.850","Text":"has also got to be in W. That\u0027s the second requirement."},{"Start":"04:14.930 ","End":"04:20.885","Text":"It also has a name called closure under addition,"},{"Start":"04:20.885 ","End":"04:23.480","Text":"don\u0027t worry about that name."},{"Start":"04:23.480 ","End":"04:30.395","Text":"The third condition it\u0027s that"},{"Start":"04:30.395 ","End":"04:37.070","Text":"if we have a vector in W and any scalar,"},{"Start":"04:37.070 ","End":"04:42.855","Text":"the way I write scalar is I write that k belongs to R the real numbers."},{"Start":"04:42.855 ","End":"04:45.170","Text":"I take a vector and take a scalar."},{"Start":"04:45.170 ","End":"04:50.075","Text":"Then the scalar times the vector also has to be inside"},{"Start":"04:50.075 ","End":"04:55.890","Text":"W. This rule also has a name."},{"Start":"04:55.890 ","End":"04:58.640","Text":"Won\u0027t even read it, I\u0027m just putting it here for reference in"},{"Start":"04:58.640 ","End":"05:01.340","Text":"case you encounter the concept of closure."},{"Start":"05:01.340 ","End":"05:07.485","Text":"Let\u0027s see if these conditions 2 and 3 are satisfied in our example."},{"Start":"05:07.485 ","End":"05:10.150","Text":"Now, I know it\u0027s already failed 1,"},{"Start":"05:10.150 ","End":"05:12.620","Text":"but just for educational purposes,"},{"Start":"05:12.620 ","End":"05:16.800","Text":"we\u0027ll just see if it satisfies conditions 2 and 3."},{"Start":"05:16.800 ","End":"05:20.420","Text":"Now, I claim it satisfies none of these conditions."},{"Start":"05:20.420 ","End":"05:21.995","Text":"Let\u0027s look at number 2,"},{"Start":"05:21.995 ","End":"05:23.975","Text":"we see 2 vectors here."},{"Start":"05:23.975 ","End":"05:25.955","Text":"If I take, for example,"},{"Start":"05:25.955 ","End":"05:28.750","Text":"the 0, 0, 1,"},{"Start":"05:28.750 ","End":"05:30.495","Text":"which is in W,"},{"Start":"05:30.495 ","End":"05:36.415","Text":"and I add to it 1/2, 1/2, 0."},{"Start":"05:36.415 ","End":"05:38.479","Text":"Then what I get,"},{"Start":"05:38.479 ","End":"05:40.250","Text":"we just add component-wise,"},{"Start":"05:40.250 ","End":"05:45.300","Text":"we get 0 and 1/2 is 1/2 and then 1/2 and then 1."},{"Start":"05:45.300 ","End":"05:49.480","Text":"This plus this plus this is not equal to 1."},{"Start":"05:49.480 ","End":"05:53.595","Text":"1/2 plus 1/2 plus 1 is equal to 2,"},{"Start":"05:53.595 ","End":"05:55.020","Text":"not equal to 1."},{"Start":"05:55.020 ","End":"05:59.140","Text":"It\u0027s failed in that condition also."},{"Start":"05:59.810 ","End":"06:09.485","Text":"The last condition, it actually also fails that because let\u0027s take the 0, 0,"},{"Start":"06:09.485 ","End":"06:13.535","Text":"1 as my u and as the scalar,"},{"Start":"06:13.535 ","End":"06:18.545","Text":"I don\u0027t know, I\u0027ll take 3 as the scalar."},{"Start":"06:18.545 ","End":"06:22.315","Text":"I take 3 times the vector."},{"Start":"06:22.315 ","End":"06:27.970","Text":"This is equal to 0, 0, 3."},{"Start":"06:28.040 ","End":"06:30.660","Text":"Is this in our W?"},{"Start":"06:30.660 ","End":"06:35.300","Text":"No, because 0 plus 0 plus 3 is equal to 3,"},{"Start":"06:35.300 ","End":"06:37.600","Text":"which is not equal to 1."},{"Start":"06:37.600 ","End":"06:45.500","Text":"Actually, the example I gave satisfies none of the 3 conditions,"},{"Start":"06:45.500 ","End":"06:50.965","Text":"but it\u0027s enough for it to fail 1 and it\u0027s not a subspace."},{"Start":"06:50.965 ","End":"06:57.155","Text":"Now, I mentioned that I\u0027m also going to give a little diagram. Let\u0027s see."},{"Start":"06:57.155 ","End":"07:02.345","Text":"Here we have our vector space R^n."},{"Start":"07:02.345 ","End":"07:10.180","Text":"Well in general, let\u0027s just call it V. Then we have our subspace W."},{"Start":"07:10.260 ","End":"07:17.665","Text":"The first condition says that the 0 vector must be in there."},{"Start":"07:17.665 ","End":"07:22.255","Text":"The second condition says that if I have u in there,"},{"Start":"07:22.255 ","End":"07:25.205","Text":"and if I have v in here,"},{"Start":"07:25.205 ","End":"07:31.380","Text":"then also u plus v that vector"},{"Start":"07:31.380 ","End":"07:38.420","Text":"is also inside W. Also k times v,"},{"Start":"07:38.420 ","End":"07:44.440","Text":"the vector than just representing as dots points, k,"},{"Start":"07:44.440 ","End":"07:50.425","Text":"v is also inside W. Now another remark,"},{"Start":"07:50.425 ","End":"07:58.750","Text":"some books or some teachers have a slightly different set of requirements."},{"Start":"07:58.750 ","End":"08:05.275","Text":"Number 1, instead of saying that 0 is a member of W,"},{"Start":"08:05.275 ","End":"08:11.620","Text":"instead of that, the requirement is that W is not equal to the empty set."},{"Start":"08:11.620 ","End":"08:15.025","Text":"This is the empty set symbol in set theory."},{"Start":"08:15.025 ","End":"08:19.390","Text":"To say that W is not empty means that there is some vector,"},{"Start":"08:19.390 ","End":"08:25.495","Text":"at least 1 vector inside W. It turns out that these are equivalent,"},{"Start":"08:25.495 ","End":"08:29.845","Text":"but I\u0027m mentioning it in case you encounter this."},{"Start":"08:29.845 ","End":"08:36.310","Text":"I\u0027m going to give another example of a subset and we\u0027ll see if it\u0027s a subspace or not."},{"Start":"08:36.310 ","End":"08:39.910","Text":"We\u0027ll take R^3 as our vector space."},{"Start":"08:39.910 ","End":"08:46.600","Text":"Again, this is the V. And we\u0027ll take all the a, b,"},{"Start":"08:46.600 ","End":"08:52.540","Text":"c in V such that a different condition this time,"},{"Start":"08:52.540 ","End":"08:55.885","Text":"last time we had a condition a plus b plus c equals 1."},{"Start":"08:55.885 ","End":"08:59.965","Text":"This time, the condition is that a is less than or equal to b,"},{"Start":"08:59.965 ","End":"09:04.330","Text":"less than or equal to c. For example, 1,"},{"Start":"09:04.330 ","End":"09:09.310","Text":"4, 7 is in W because 1 is less than or equal to 4 and 4 is less than or equal to 7."},{"Start":"09:09.310 ","End":"09:11.875","Text":"Another example, 4, 4, 4,"},{"Start":"09:11.875 ","End":"09:16.600","Text":"which is okay because notice it\u0027s less than or equal to so 4 is less than or equal to 4,"},{"Start":"09:16.600 ","End":"09:18.490","Text":"which is less than or equal to 4."},{"Start":"09:18.490 ","End":"09:21.220","Text":"Third example, less than or equal to."},{"Start":"09:21.220 ","End":"09:23.530","Text":"It could be equal to and it could be less than."},{"Start":"09:23.530 ","End":"09:26.305","Text":"Anyway, there\u0027s millions of other examples."},{"Start":"09:26.305 ","End":"09:30.985","Text":"Let\u0027s see which of these conditions is satisfied."},{"Start":"09:30.985 ","End":"09:34.555","Text":"Now, notice that we\u0027re okay with number 1,"},{"Start":"09:34.555 ","End":"09:39.540","Text":"because 0, 0, 0 is okay."},{"Start":"09:39.540 ","End":"09:44.070","Text":"It is in W because 0 is less than or equal to 0,"},{"Start":"09:44.070 ","End":"09:46.260","Text":"which is less than or equal to 0."},{"Start":"09:46.260 ","End":"09:48.555","Text":"Condition 1 is satisfied."},{"Start":"09:48.555 ","End":"09:51.770","Text":"Now how about Condition 2?"},{"Start":"09:51.770 ","End":"09:55.255","Text":"I claim that\u0027s also satisfied."},{"Start":"09:55.255 ","End":"09:58.390","Text":"Let\u0027s just take first an example."},{"Start":"09:58.390 ","End":"10:00.850","Text":"Let\u0027s add these 2."},{"Start":"10:00.850 ","End":"10:03.490","Text":"If I take 1, 4,"},{"Start":"10:03.490 ","End":"10:10.840","Text":"7 and add it to 4, 4,"},{"Start":"10:10.840 ","End":"10:16.210","Text":"4, what I get is 1 and 4 is 5,"},{"Start":"10:16.210 ","End":"10:17.500","Text":"4 and 4 is 8,"},{"Start":"10:17.500 ","End":"10:19.315","Text":"7 and 4 is 11,"},{"Start":"10:19.315 ","End":"10:22.180","Text":"and that\u0027s also inside."},{"Start":"10:22.180 ","End":"10:24.085","Text":"Of course this is just 1 example."},{"Start":"10:24.085 ","End":"10:28.585","Text":"In general, for these exercises you would have to prove it."},{"Start":"10:28.585 ","End":"10:31.120","Text":"You would say, for example,"},{"Start":"10:31.120 ","End":"10:33.909","Text":"that if a, b,"},{"Start":"10:33.909 ","End":"10:37.495","Text":"and c is in W,"},{"Start":"10:37.495 ","End":"10:39.070","Text":"and another 1,"},{"Start":"10:39.070 ","End":"10:41.470","Text":"let\u0027s call it big A, big B,"},{"Start":"10:41.470 ","End":"10:46.555","Text":"big C is in W. Then we have 2 things,"},{"Start":"10:46.555 ","End":"10:48.655","Text":"that a is less than or equal to b,"},{"Start":"10:48.655 ","End":"10:55.780","Text":"less than or equal to C. Also for the other vector big A less than big B less than or"},{"Start":"10:55.780 ","End":"11:04.130","Text":"equal to big C. We have to answer the question."},{"Start":"11:04.380 ","End":"11:06.520","Text":"What is the sum?"},{"Start":"11:06.520 ","End":"11:11.470","Text":"The sum is a plus A, b plus B."},{"Start":"11:11.470 ","End":"11:14.125","Text":"That\u0027s how we define the addition component-wise."},{"Start":"11:14.125 ","End":"11:19.164","Text":"Little c, and this is big C,"},{"Start":"11:19.164 ","End":"11:23.950","Text":"little c plus big C. The question is,"},{"Start":"11:23.950 ","End":"11:26.705","Text":"is this also in W?"},{"Start":"11:26.705 ","End":"11:28.200","Text":"I\u0027m claiming that yes,"},{"Start":"11:28.200 ","End":"11:30.705","Text":"because you can add inequalities."},{"Start":"11:30.705 ","End":"11:33.930","Text":"I can put a equal sign even if it\u0027s a double inequality."},{"Start":"11:33.930 ","End":"11:37.650","Text":"If a is less than or equal to b and big A is less than or equal to big B,"},{"Start":"11:37.650 ","End":"11:44.080","Text":"then this plus this a plus A is got to be less than or equal to b plus B."},{"Start":"11:44.080 ","End":"11:47.920","Text":"Similarly, this is going to be less than or equal to c plus"},{"Start":"11:47.920 ","End":"11:53.425","Text":"C. The sum has the property of less than or equal to, less than or equal to."},{"Start":"11:53.425 ","End":"11:55.405","Text":"It\u0027s also in W?"},{"Start":"11:55.405 ","End":"12:04.710","Text":"Yes. We already have yes for requirement Condition 1,"},{"Start":"12:04.710 ","End":"12:06.705","Text":"we\u0027ve got yes, for Condition 2,"},{"Start":"12:06.705 ","End":"12:08.999","Text":"what about Condition 3?"},{"Start":"12:08.999 ","End":"12:11.840","Text":"Multiplication with a scalar?"},{"Start":"12:11.840 ","End":"12:16.779","Text":"Well, let\u0027s take our example of 1, 4, 7."},{"Start":"12:16.779 ","End":"12:18.940","Text":"Now if I take 1, 4,"},{"Start":"12:18.940 ","End":"12:21.520","Text":"7 and multiply it by a scalar,"},{"Start":"12:21.520 ","End":"12:22.840","Text":"just as an example,"},{"Start":"12:22.840 ","End":"12:25.585","Text":"say 5 times this,"},{"Start":"12:25.585 ","End":"12:30.250","Text":"then this would equal 5,"},{"Start":"12:30.250 ","End":"12:31.780","Text":"5 times 4 is 20,"},{"Start":"12:31.780 ","End":"12:34.525","Text":"5 times 7 is 35."},{"Start":"12:34.525 ","End":"12:39.684","Text":"It looks like indeed still 5 is less than or equal to 20,"},{"Start":"12:39.684 ","End":"12:42.415","Text":"less than or equal to 35."},{"Start":"12:42.415 ","End":"12:45.070","Text":"This is an example. It\u0027s not a proof."},{"Start":"12:45.070 ","End":"12:48.685","Text":"Is it possible that a scalar times"},{"Start":"12:48.685 ","End":"12:55.615","Text":"such a vector will violate the condition of the subspace?"},{"Start":"12:55.615 ","End":"12:57.685","Text":"The answer is, yes."},{"Start":"12:57.685 ","End":"12:59.260","Text":"Just think about it in a moment."},{"Start":"12:59.260 ","End":"13:01.539","Text":"What if I took a negative number?"},{"Start":"13:01.539 ","End":"13:04.180","Text":"Suppose I took instead of 5,"},{"Start":"13:04.180 ","End":"13:06.190","Text":"I took minus 5."},{"Start":"13:06.190 ","End":"13:10.390","Text":"If I took minus 5 times 1, 4,"},{"Start":"13:10.390 ","End":"13:17.440","Text":"7, then I would get minus 5,"},{"Start":"13:17.440 ","End":"13:21.325","Text":"minus 20, minus 35."},{"Start":"13:21.325 ","End":"13:30.685","Text":"This is no longer true because actually minus 5 is bigger than minus 20"},{"Start":"13:30.685 ","End":"13:40.135","Text":"and it\u0027s bigger than minus 35 so to want less than or equal to here is wrong."},{"Start":"13:40.135 ","End":"13:46.010","Text":"We almost made it to being a subspace but, no."},{"Start":"13:47.430 ","End":"13:54.175","Text":"Condition 3 fails when k is a negative scalar."},{"Start":"13:54.175 ","End":"13:57.535","Text":"Not good enough, not a subspace."},{"Start":"13:57.535 ","End":"14:01.675","Text":"Now you might say you want an example of something that is a subspace."},{"Start":"14:01.675 ","End":"14:04.330","Text":"Well, not here, not in this clip."},{"Start":"14:04.330 ","End":"14:10.945","Text":"There are lots of example clips on subspaces following the tutorial."},{"Start":"14:10.945 ","End":"14:14.800","Text":"I\u0027ll leave you to look at the examples there."},{"Start":"14:14.800 ","End":"14:18.290","Text":"I think we\u0027re done for this clip."}],"ID":25837},{"Watched":false,"Name":"Basis for a Subspace","Duration":"3m 57s","ChapterTopicVideoID":9916,"CourseChapterTopicPlaylistID":31678,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9916.jpeg","UploadDate":"2017-08-07T11:35:29.8270000","DurationForVideoObject":"PT3M57S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:07.425","Text":"In this clip, we\u0027ll learn how to find a basis for a subspace of R^n."},{"Start":"00:07.425 ","End":"00:10.110","Text":"We\u0027ve done something similar before when"},{"Start":"00:10.110 ","End":"00:14.580","Text":"that subspace was the solution set for a homogeneous SLE."},{"Start":"00:14.580 ","End":"00:20.535","Text":"This time the subspace will be given as the span of a certain set of vectors."},{"Start":"00:20.535 ","End":"00:23.830","Text":"I think it\u0027s best to start with an example."},{"Start":"00:24.200 ","End":"00:30.690","Text":"Here we are. The subspace W is in 3D space R^3,"},{"Start":"00:30.690 ","End":"00:36.520","Text":"and it\u0027s given us the span of these 3 vectors."},{"Start":"00:36.640 ","End":"00:42.830","Text":"We have to find the basis for W. Once we have the basis then might as well,"},{"Start":"00:42.830 ","End":"00:45.200","Text":"give the dimensional also."},{"Start":"00:45.200 ","End":"00:49.380","Text":"There will be a part b. I\u0027ll reveal this later on."},{"Start":"00:49.760 ","End":"00:53.300","Text":"We\u0027ll solve this using a matrix."},{"Start":"00:53.300 ","End":"00:59.415","Text":"We take these 3 vectors and just put them as rows of a matrix."},{"Start":"00:59.415 ","End":"01:04.800","Text":"We want to bring this matrix to row echelon form."},{"Start":"01:04.910 ","End":"01:06.980","Text":"We have a 1 here."},{"Start":"01:06.980 ","End":"01:10.580","Text":"We can 0 out the first column by subtracting"},{"Start":"01:10.580 ","End":"01:15.300","Text":"4 times this from this and 7 times this from this,"},{"Start":"01:15.590 ","End":"01:18.675","Text":"and we end up with this matrix."},{"Start":"01:18.675 ","End":"01:24.025","Text":"Again, we need in a row operation to make a 0 under here."},{"Start":"01:24.025 ","End":"01:29.120","Text":"If I subtract twice the second row from the third row,"},{"Start":"01:29.120 ","End":"01:31.875","Text":"then this is what we get."},{"Start":"01:31.875 ","End":"01:38.680","Text":"Notice that there is a 0 in the last row,"},{"Start":"01:38.680 ","End":"01:43.135","Text":"which means that we only need 2 vectors."},{"Start":"01:43.135 ","End":"01:47.680","Text":"The basis for W consists of 2 vectors,"},{"Start":"01:47.680 ","End":"01:49.840","Text":"this 1 and this 1."},{"Start":"01:49.840 ","End":"01:58.310","Text":"If it was me, I\u0027d probably divide this by minus 3 and go for 0,1,2."},{"Start":"01:58.310 ","End":"02:01.700","Text":"It\u0027s more aesthetic to work with smaller numbers,"},{"Start":"02:01.700 ","End":"02:02.810","Text":"but you don\u0027t have to."},{"Start":"02:02.810 ","End":"02:05.580","Text":"This is optional."},{"Start":"02:06.080 ","End":"02:11.160","Text":"The dimension is 2 just by counting the number of elements."},{"Start":"02:11.160 ","End":"02:15.150","Text":"Now I said that be a continuation."},{"Start":"02:15.150 ","End":"02:20.185","Text":"Let\u0027s go back to the question and part b,"},{"Start":"02:20.185 ","End":"02:24.770","Text":"very common type of question on an exam is to complete"},{"Start":"02:24.770 ","End":"02:30.260","Text":"the basis that we found for W to a basis of all of our 3."},{"Start":"02:30.260 ","End":"02:37.920","Text":"I other words, I need a third vector scope down here."},{"Start":"02:37.920 ","End":"02:39.780","Text":"Here we have 2 vectors."},{"Start":"02:39.780 ","End":"02:43.630","Text":"We need a third 1 to make it a basis for all of our 3."},{"Start":"02:43.630 ","End":"02:49.400","Text":"All we need is to find something that\u0027s linearly independent of these 2."},{"Start":"02:49.400 ","End":"02:53.000","Text":"Because there\u0027s a theorem when we have 3 linearly independent vectors,"},{"Start":"02:53.000 ","End":"02:55.525","Text":"that will be a basis."},{"Start":"02:55.525 ","End":"03:00.800","Text":"I\u0027m just going to give it to you a 1 example would be 0,0,1,"},{"Start":"03:00.800 ","End":"03:04.595","Text":"and you can check that these are linearly independent."},{"Start":"03:04.595 ","End":"03:09.170","Text":"1 way to check that these are linearly independent is to write them in a matrix."},{"Start":"03:09.170 ","End":"03:13.455","Text":"Notice that if we have 1, 2, 3, 0,"},{"Start":"03:13.455 ","End":"03:18.795","Text":"I\u0027ll use the 0,1,2 form and 0,0,1."},{"Start":"03:18.795 ","End":"03:27.645","Text":"Then it\u0027s actually a matrix in echelon form already."},{"Start":"03:27.645 ","End":"03:32.415","Text":"It has 3 rows, not less."},{"Start":"03:32.415 ","End":"03:36.005","Text":"These really are linearly independent."},{"Start":"03:36.005 ","End":"03:39.660","Text":"Otherwise we\u0027d get a row with 0s in it."},{"Start":"03:40.090 ","End":"03:46.970","Text":"There are some methodologies for how to find this but I don\u0027t want to get into that now."},{"Start":"03:46.970 ","End":"03:52.110","Text":"We just like a trial and error found 0,0,1."},{"Start":"03:52.840 ","End":"03:55.355","Text":"That is all there is to it."},{"Start":"03:55.355 ","End":"03:57.570","Text":"We are done."}],"ID":9981},{"Watched":false,"Name":"Exercise 1","Duration":"4m 26s","ChapterTopicVideoID":9918,"CourseChapterTopicPlaylistID":31678,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9918.jpeg","UploadDate":"2017-08-07T11:35:57.2600000","DurationForVideoObject":"PT4M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:08.055","Text":"In this exercise, we\u0027re considering a subspace of R^4 for dimensional real space,"},{"Start":"00:08.055 ","End":"00:14.190","Text":"which is defined to be the span of these 3 vectors."},{"Start":"00:14.190 ","End":"00:15.750","Text":"We give it a name,"},{"Start":"00:15.750 ","End":"00:17.220","Text":"we call it U."},{"Start":"00:17.220 ","End":"00:19.830","Text":"The span is always a subspace."},{"Start":"00:19.830 ","End":"00:23.309","Text":"Also notice that for consistency,"},{"Start":"00:23.309 ","End":"00:28.875","Text":"I just make sure that they all have 4 components so we really are in R^4."},{"Start":"00:28.875 ","End":"00:35.355","Text":"If it\u0027s a subspace, we can find the basis for it and also the dimension of U."},{"Start":"00:35.355 ","End":"00:39.230","Text":"Part b is a more unusual question."},{"Start":"00:39.230 ","End":"00:44.690","Text":"We usually have the reverse question where we give a system of"},{"Start":"00:44.690 ","End":"00:50.540","Text":"linear equations homogeneous and we look for it\u0027s solution space."},{"Start":"00:50.540 ","End":"00:54.620","Text":"This time we start with the solution space and we have to"},{"Start":"00:54.620 ","End":"01:00.060","Text":"reverse engineer it to get a system of homogeneous equations."},{"Start":"01:00.580 ","End":"01:04.170","Text":"If it wasn\u0027t for b,"},{"Start":"01:04.210 ","End":"01:07.880","Text":"we\u0027d start off with matrices and we\u0027d have a matrix with"},{"Start":"01:07.880 ","End":"01:12.330","Text":"3 rows with these 3 and because of part b,"},{"Start":"01:14.110 ","End":"01:17.045","Text":"we put in an extra row,"},{"Start":"01:17.045 ","End":"01:18.710","Text":"x, y, z, t,"},{"Start":"01:18.710 ","End":"01:24.890","Text":"whatever 4 variables we would use for a system with 4 unknowns."},{"Start":"01:24.890 ","End":"01:28.880","Text":"Basically, I\u0027m teaching this technique through this example."},{"Start":"01:28.880 ","End":"01:32.280","Text":"We haven\u0027t seen this before. Maybe you have."},{"Start":"01:32.360 ","End":"01:36.680","Text":"As you might have guessed, we probably want to bring this to row echelon form."},{"Start":"01:36.680 ","End":"01:40.015","Text":"That\u0027s pretty much what we do with matrices."},{"Start":"01:40.015 ","End":"01:44.625","Text":"As usual, we want to get 0s below this"},{"Start":"01:44.625 ","End":"01:49.920","Text":"1 here and we subtract multiples of this first row from the others."},{"Start":"01:50.110 ","End":"01:52.700","Text":"Especially the last 1,"},{"Start":"01:52.700 ","End":"01:54.110","Text":"I\u0027ll just show you the last 1."},{"Start":"01:54.110 ","End":"01:57.905","Text":"We\u0027re going to subtract x times this row from this row."},{"Start":"01:57.905 ","End":"02:00.740","Text":"That would be this row operation."},{"Start":"02:00.740 ","End":"02:07.085","Text":"When we\u0027ve done those row operations, we get this."},{"Start":"02:07.085 ","End":"02:11.630","Text":"Of course, you can always pause at any stage and verify the computations."},{"Start":"02:11.630 ","End":"02:13.265","Text":"I\u0027m not going to dwell on it."},{"Start":"02:13.265 ","End":"02:15.725","Text":"Now we have 0s here."},{"Start":"02:15.725 ","End":"02:19.950","Text":"Next we want a 0 here and here."},{"Start":"02:20.030 ","End":"02:24.080","Text":"What we\u0027ll do is to get a 0 here."},{"Start":"02:24.080 ","End":"02:29.030","Text":"We\u0027ll just add twice the second row to the third row."},{"Start":"02:29.030 ","End":"02:31.595","Text":"To get a 0 here,"},{"Start":"02:31.595 ","End":"02:33.725","Text":"we\u0027ll take 4 times this,"},{"Start":"02:33.725 ","End":"02:37.710","Text":"y minus x times this and add them."},{"Start":"02:37.850 ","End":"02:43.260","Text":"After we do all that, now look,"},{"Start":"02:43.260 ","End":"02:47.870","Text":"we\u0027ve got a row of 0s here and also note that if it didn\u0027t have the row of 0s,"},{"Start":"02:47.870 ","End":"02:52.130","Text":"we\u0027d already be in row echelon form so we can stop by row operations."},{"Start":"02:52.130 ","End":"02:54.120","Text":"If we didn\u0027t have part b,"},{"Start":"02:54.120 ","End":"02:56.070","Text":"we\u0027d be done and we\u0027d say,"},{"Start":"02:56.070 ","End":"03:02.220","Text":"okay, these 2 vectors are the basis for U."},{"Start":"03:02.220 ","End":"03:04.170","Text":"Then we count them 1,"},{"Start":"03:04.170 ","End":"03:06.165","Text":"2 and say the dimension is 2."},{"Start":"03:06.165 ","End":"03:12.215","Text":"We\u0027ll do that in a moment but what do we do with this last row?"},{"Start":"03:12.215 ","End":"03:18.525","Text":"The idea is to take each of these and set it to 0."},{"Start":"03:18.525 ","End":"03:19.970","Text":"We\u0027ll get 2 equations."},{"Start":"03:19.970 ","End":"03:22.655","Text":"This equals 0 and this equals 0,"},{"Start":"03:22.655 ","End":"03:26.195","Text":"and that will be our part b."},{"Start":"03:26.195 ","End":"03:29.690","Text":"That\u0027s how we get the answer to part b,"},{"Start":"03:29.690 ","End":"03:31.850","Text":"the system of homogeneous linear equations"},{"Start":"03:31.850 ","End":"03:34.310","Text":"and perhaps I shouldn\u0027t have put them in boxes,"},{"Start":"03:34.310 ","End":"03:36.740","Text":"but instead put a curly brace."},{"Start":"03:36.740 ","End":"03:38.105","Text":"That\u0027s not important."},{"Start":"03:38.105 ","End":"03:40.265","Text":"This is the system."},{"Start":"03:40.265 ","End":"03:44.460","Text":"As I said, the basis for you,"},{"Start":"03:44.460 ","End":"03:46.920","Text":"these top 2 like 1, 1 minus 1,"},{"Start":"03:46.920 ","End":"03:48.690","Text":"2, 1, 1 minus 1, 2,"},{"Start":"03:48.690 ","End":"03:54.240","Text":"and then the other 1 also and the dimension just by counting is 2."},{"Start":"03:54.430 ","End":"03:59.720","Text":"It\u0027s also a good idea to verify if you have"},{"Start":"03:59.720 ","End":"04:05.090","Text":"time that these actually satisfy these equations."},{"Start":"04:05.090 ","End":"04:10.880","Text":"Let\u0027s just do 1 example that the first vector satisfies the first equation."},{"Start":"04:10.880 ","End":"04:14.090","Text":"If I stick in 1, 1 minus 1, 2,"},{"Start":"04:14.090 ","End":"04:19.095","Text":"I\u0027ll get minus 3 plus 5, minus 2."},{"Start":"04:19.095 ","End":"04:23.120","Text":"It is 0 and you can mentally check the others also."},{"Start":"04:23.120 ","End":"04:26.880","Text":"That\u0027s it. We\u0027re done."}],"ID":9982},{"Watched":false,"Name":"Exercise 2","Duration":"3m 23s","ChapterTopicVideoID":9919,"CourseChapterTopicPlaylistID":31678,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9919.jpeg","UploadDate":"2017-08-07T11:36:12.7830000","DurationForVideoObject":"PT3M23S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.050","Text":"This exercise is a continuation of the previous 1."},{"Start":"00:04.050 ","End":"00:06.720","Text":"It\u0027s actually the second in a 4-part series of"},{"Start":"00:06.720 ","End":"00:11.430","Text":"exercises and it\u0027s just like the previous exercise."},{"Start":"00:11.430 ","End":"00:13.790","Text":"We have again a subspace of R^4."},{"Start":"00:13.790 ","End":"00:16.470","Text":"The only thing that\u0027s different is the set of vectors is"},{"Start":"00:16.470 ","End":"00:19.620","Text":"changed and the subspace is called V,"},{"Start":"00:19.620 ","End":"00:22.200","Text":"whereas it was U before."},{"Start":"00:22.200 ","End":"00:30.120","Text":"We want to find for this V a basis for the subspace and"},{"Start":"00:30.120 ","End":"00:34.260","Text":"its dimension and also a homogeneous system"},{"Start":"00:34.260 ","End":"00:39.120","Text":"of linear equations for which V is the solution space."},{"Start":"00:39.120 ","End":"00:41.750","Text":"We\u0027ll be using the same technique as before."},{"Start":"00:41.750 ","End":"00:47.425","Text":"We\u0027ll start off by putting these 4 rows into a matrix."},{"Start":"00:47.425 ","End":"00:53.810","Text":"Here they are the 4 vectors and because of the second part of the question,"},{"Start":"00:53.810 ","End":"00:56.300","Text":"we also put x, y,"},{"Start":"00:56.300 ","End":"00:59.585","Text":"z, t, just like we did in the previous."},{"Start":"00:59.585 ","End":"01:05.315","Text":"We want to start row operations to bring it into row echelon form."},{"Start":"01:05.315 ","End":"01:08.510","Text":"First we want 0s beneath this 1,"},{"Start":"01:08.510 ","End":"01:11.900","Text":"so we subtract multiples of this row from the other."},{"Start":"01:11.900 ","End":"01:15.200","Text":"In the last case it\u0027s a variable number."},{"Start":"01:15.200 ","End":"01:18.860","Text":"We subtract x times row 1 from row 5."},{"Start":"01:18.860 ","End":"01:20.870","Text":"Everything else is obvious."},{"Start":"01:20.870 ","End":"01:22.700","Text":"What we get is this,"},{"Start":"01:22.700 ","End":"01:25.055","Text":"I\u0027ll leave you to verify and the 0s are here."},{"Start":"01:25.055 ","End":"01:27.050","Text":"Now we want to have 0s onto"},{"Start":"01:27.050 ","End":"01:33.455","Text":"this 1 here so similar set of row operations and here they are."},{"Start":"01:33.455 ","End":"01:35.510","Text":"As a result of which,"},{"Start":"01:35.510 ","End":"01:38.025","Text":"we end up with this."},{"Start":"01:38.025 ","End":"01:41.225","Text":"But now, I want to do 2 things at once."},{"Start":"01:41.225 ","End":"01:44.420","Text":"Notice that there\u0027s a row of 0s,"},{"Start":"01:44.420 ","End":"01:45.980","Text":"and I\u0027ll get rid of that,"},{"Start":"01:45.980 ","End":"01:48.710","Text":"but also divide this by 3."},{"Start":"01:48.710 ","End":"01:51.370","Text":"I like to reduce numbers."},{"Start":"01:51.370 ","End":"01:56.780","Text":"Now this is what we have and we\u0027re still not in row echelon form."},{"Start":"01:56.780 ","End":"02:00.325","Text":"I can still get a 0 here."},{"Start":"02:00.325 ","End":"02:04.610","Text":"I\u0027ll take twice this and z minus 2x minus y"},{"Start":"02:04.610 ","End":"02:08.645","Text":"times this and add them and put that into the last row."},{"Start":"02:08.645 ","End":"02:11.580","Text":"If we do that,"},{"Start":"02:12.110 ","End":"02:14.865","Text":"then this is what we have."},{"Start":"02:14.865 ","End":"02:18.140","Text":"Of course this is echelon form."},{"Start":"02:18.140 ","End":"02:20.375","Text":"Now that last extra row,"},{"Start":"02:20.375 ","End":"02:21.740","Text":"ignore that for the moment,"},{"Start":"02:21.740 ","End":"02:26.255","Text":"for the first part of the question and then we would say that these 3,"},{"Start":"02:26.255 ","End":"02:27.995","Text":"and I\u0027ll write them again in a moment,"},{"Start":"02:27.995 ","End":"02:30.440","Text":"are the basis vectors."},{"Start":"02:30.440 ","End":"02:33.530","Text":"We\u0027ll see that the dimension will be 3,"},{"Start":"02:33.530 ","End":"02:35.465","Text":"those 1, 2, 3."},{"Start":"02:35.465 ","End":"02:37.865","Text":"For the last part,"},{"Start":"02:37.865 ","End":"02:41.780","Text":"I mean the bit about finding the homogeneous SLA,"},{"Start":"02:41.780 ","End":"02:44.630","Text":"we just set this to be 0."},{"Start":"02:44.630 ","End":"02:46.100","Text":"Previously we had 2 equations,"},{"Start":"02:46.100 ","End":"02:49.520","Text":"this time we just get 1 equation, and this is it."},{"Start":"02:49.520 ","End":"02:56.520","Text":"It\u0027s a system with just 1 equation in 4 unknowns."},{"Start":"02:56.520 ","End":"02:59.610","Text":"Of course we want to tidy that up a bit."},{"Start":"02:59.610 ","End":"03:02.300","Text":"Open brackets, collect like terms,"},{"Start":"03:02.300 ","End":"03:03.935","Text":"put it in a nice box."},{"Start":"03:03.935 ","End":"03:07.405","Text":"This is the answer to the second part."},{"Start":"03:07.405 ","End":"03:09.140","Text":"Here, like I said,"},{"Start":"03:09.140 ","End":"03:12.260","Text":"is the answer to the first part where we just take"},{"Start":"03:12.260 ","End":"03:19.505","Text":"these 3 vectors and put them here and we already said that the dimension is 3."},{"Start":"03:19.505 ","End":"03:23.790","Text":"We\u0027re done with this second part out of 4."}],"ID":9983},{"Watched":false,"Name":"Exercise 3","Duration":"4m 10s","ChapterTopicVideoID":9917,"CourseChapterTopicPlaylistID":31678,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9917.jpeg","UploadDate":"2017-08-07T11:35:43.2730000","DurationForVideoObject":"PT4M10S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:05.070","Text":"This exercise is the third of a 4 part series."},{"Start":"00:05.070 ","End":"00:07.470","Text":"In the first exercise,"},{"Start":"00:07.470 ","End":"00:08.900","Text":"we dealt with U,"},{"Start":"00:08.900 ","End":"00:11.040","Text":"this was the same U from there."},{"Start":"00:11.040 ","End":"00:14.130","Text":"This is the same V from Part 2."},{"Start":"00:14.130 ","End":"00:21.375","Text":"This time Part 3 concerns U plus V,"},{"Start":"00:21.375 ","End":"00:23.805","Text":"the sum of the 2 subspaces."},{"Start":"00:23.805 ","End":"00:29.340","Text":"I want to remind you what it means for the sum of 2 subspaces."},{"Start":"00:29.340 ","End":"00:37.800","Text":"U plus V is simply the set of all sums where I take 1 part of"},{"Start":"00:37.800 ","End":"00:44.240","Text":"the sum from the subspace U and the"},{"Start":"00:44.240 ","End":"00:52.645","Text":"other from the subspace V. Set of all possible sums I can get 1 from here, 1 from here."},{"Start":"00:52.645 ","End":"00:55.910","Text":"If you think about it, this U plus V is going to be"},{"Start":"00:55.910 ","End":"01:00.020","Text":"a linear combination of these plus a linear combination of these."},{"Start":"01:00.020 ","End":"01:04.595","Text":"Together we\u0027ve got a linear combination of all of these together."},{"Start":"01:04.595 ","End":"01:07.690","Text":"How many of them 3 and 4, 7 of them."},{"Start":"01:07.690 ","End":"01:11.910","Text":"Yeah, so U plus V is actually the span and"},{"Start":"01:11.910 ","End":"01:15.480","Text":"I could write all these 7 in a row and that would be it."},{"Start":"01:15.480 ","End":"01:18.990","Text":"Here they are 1,2,3,4,5,6,7 of them."},{"Start":"01:18.990 ","End":"01:23.410","Text":"These 3 at the top, then these 4."},{"Start":"01:23.690 ","End":"01:30.950","Text":"We start bringing it into row echelon form by a series of row operations."},{"Start":"01:30.950 ","End":"01:35.410","Text":"I\u0027m going to do this fairly quickly because you already know the routine."},{"Start":"01:35.410 ","End":"01:39.160","Text":"We\u0027re going to subtract multiples of the top row,"},{"Start":"01:39.160 ","End":"01:42.430","Text":"subtract, or add to the rest of them."},{"Start":"01:42.430 ","End":"01:44.170","Text":"If we do that,"},{"Start":"01:44.170 ","End":"01:47.035","Text":"then I\u0027ll leave you to check the computation."},{"Start":"01:47.035 ","End":"01:48.804","Text":"We get this."},{"Start":"01:48.804 ","End":"01:50.695","Text":"We\u0027ve got 0s here."},{"Start":"01:50.695 ","End":"01:56.300","Text":"Next, we want to get 0s below the minus 4."},{"Start":"01:56.300 ","End":"01:59.380","Text":"Note that to get a 0 here,"},{"Start":"01:59.380 ","End":"02:02.110","Text":"I would add twice a second row to the third row."},{"Start":"02:02.110 ","End":"02:06.425","Text":"That actually gives me a whole row of 0s."},{"Start":"02:06.425 ","End":"02:12.595","Text":"Also, if I subtract this row from the last row since they\u0027re the same,"},{"Start":"02:12.595 ","End":"02:15.710","Text":"I get also 0s here."},{"Start":"02:16.200 ","End":"02:18.760","Text":"Perhaps I should have waited with that because we\u0027re"},{"Start":"02:18.760 ","End":"02:20.740","Text":"still not finished with the row operations,"},{"Start":"02:20.740 ","End":"02:22.210","Text":"you might get more 0s even."},{"Start":"02:22.210 ","End":"02:26.320","Text":"Anyway, here I have a 6 and"},{"Start":"02:26.320 ","End":"02:31.810","Text":"then I need the entries below it to be 0 so I need to make 0 here and here."},{"Start":"02:31.810 ","End":"02:38.065","Text":"For example, I could subtract 1/3 of this row."},{"Start":"02:38.065 ","End":"02:44.005","Text":"No, I would add 1/3 of this row to this row and I would get a 0 here."},{"Start":"02:44.005 ","End":"02:46.390","Text":"Here I could take, I don\u0027t know,"},{"Start":"02:46.390 ","End":"02:50.155","Text":"6 times this minus 4 times this or something."},{"Start":"02:50.155 ","End":"02:52.745","Text":"This is what we get."},{"Start":"02:52.745 ","End":"02:55.190","Text":"We\u0027re still not quite done,"},{"Start":"02:55.190 ","End":"03:00.935","Text":"still not row echelon because I have this minus 30."},{"Start":"03:00.935 ","End":"03:03.080","Text":"I think I have a missing step here,"},{"Start":"03:03.080 ","End":"03:04.350","Text":"never mind, we don\u0027t need it."},{"Start":"03:04.350 ","End":"03:06.110","Text":"To get rid of the minus 30,"},{"Start":"03:06.110 ","End":"03:10.085","Text":"I would subtract 3 times this row from this row,"},{"Start":"03:10.085 ","End":"03:12.200","Text":"and that would make that into a 0."},{"Start":"03:12.200 ","End":"03:15.800","Text":"Then I would get rid of all the 0 rows,"},{"Start":"03:15.800 ","End":"03:18.370","Text":"which is here, here, and here."},{"Start":"03:18.370 ","End":"03:21.254","Text":"We\u0027re just left with 4 rows."},{"Start":"03:21.254 ","End":"03:25.190","Text":"Also, I would divide this row by 10,"},{"Start":"03:25.190 ","End":"03:28.130","Text":"so that would make it just a 1 here."},{"Start":"03:28.130 ","End":"03:32.155","Text":"Then assembling all the bits together."},{"Start":"03:32.155 ","End":"03:38.010","Text":"Here also, I would divide this by 6 and get a 1 here."},{"Start":"03:38.480 ","End":"03:40.840","Text":"Collecting it all together,"},{"Start":"03:40.840 ","End":"03:42.800","Text":"this is what we get."},{"Start":"03:42.800 ","End":"03:48.270","Text":"This is now in row echelon form."},{"Start":"03:48.590 ","End":"03:53.610","Text":"We conclude that these 4 vectors,"},{"Start":"03:53.610 ","End":"03:55.050","Text":"I\u0027ve written them out here,"},{"Start":"03:55.050 ","End":"03:57.120","Text":"1,1 minus 1,2 is here,"},{"Start":"03:57.120 ","End":"03:59.625","Text":"and so on, 0,0,0,1 is here,"},{"Start":"03:59.625 ","End":"04:07.979","Text":"but these 4 are a basis for U plus V. The dimension is 4 just by counting."},{"Start":"04:07.979 ","End":"04:10.930","Text":"We\u0027re done with part 3."}],"ID":9984},{"Watched":false,"Name":"Exercise 4","Duration":"6m 2s","ChapterTopicVideoID":9920,"CourseChapterTopicPlaylistID":31678,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9920.jpeg","UploadDate":"2017-08-07T11:36:35.4770000","DurationForVideoObject":"PT6M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.520","Text":"This exercise is the 4th and last in a 4-part series."},{"Start":"00:05.520 ","End":"00:10.140","Text":"In the first 2 parts we considered U and then"},{"Start":"00:10.140 ","End":"00:17.325","Text":"V. The previous exercise dealt with U plus V. This time,"},{"Start":"00:17.325 ","End":"00:21.330","Text":"there\u0027s another operation we can do on vector spaces,"},{"Start":"00:21.330 ","End":"00:23.205","Text":"and that is take their intersection."},{"Start":"00:23.205 ","End":"00:26.775","Text":"The intersection of 2 vector spaces is also a vector space."},{"Start":"00:26.775 ","End":"00:31.965","Text":"It\u0027s just all the elements of vectors that they have in common."},{"Start":"00:31.965 ","End":"00:36.950","Text":"We now have to figure out the basis and dimension of U intersection with"},{"Start":"00:36.950 ","End":"00:43.600","Text":"V. Now if I just have to find the dimension and not the basis,"},{"Start":"00:43.600 ","End":"00:45.975","Text":"then there\u0027s a formula."},{"Start":"00:45.975 ","End":"00:54.620","Text":"This is the formula that ties in the only 4 spaces, the U, V,"},{"Start":"00:54.620 ","End":"00:56.705","Text":"the sum U plus V,"},{"Start":"00:56.705 ","End":"01:03.550","Text":"and the intersection U intersection with V. We already know 3 of these 4."},{"Start":"01:03.550 ","End":"01:12.100","Text":"We know that the dimension of U from exercise 1 came out to be 2,"},{"Start":"01:13.670 ","End":"01:18.530","Text":"the second part of the series came out to be 3,"},{"Start":"01:18.530 ","End":"01:20.330","Text":"and the third part of the series,"},{"Start":"01:20.330 ","End":"01:22.430","Text":"we computed this as 4,"},{"Start":"01:22.430 ","End":"01:25.530","Text":"so this will come out to be,"},{"Start":"01:25.530 ","End":"01:28.020","Text":"if we bring this to the left everything else to the right,"},{"Start":"01:28.020 ","End":"01:32.775","Text":"will be 2 plus 3 minus 4, which equals 1."},{"Start":"01:32.775 ","End":"01:35.860","Text":"We can already get the dimension."},{"Start":"01:36.760 ","End":"01:41.930","Text":"We could have saved a lot of work if the dimension had come out to be 0,"},{"Start":"01:41.930 ","End":"01:52.065","Text":"then we\u0027d be done because the basis for would then be the empty set,"},{"Start":"01:52.065 ","End":"01:55.685","Text":"and the space contains just 0."},{"Start":"01:55.685 ","End":"01:58.985","Text":"It\u0027s a trivial vector space,"},{"Start":"01:58.985 ","End":"02:03.570","Text":"and the dimension here would be 0."},{"Start":"02:03.770 ","End":"02:07.475","Text":"But it\u0027s not so we have to do some work."},{"Start":"02:07.475 ","End":"02:12.570","Text":"Now here\u0027s how the strategy is for the intersection."},{"Start":"02:14.170 ","End":"02:22.330","Text":"In part 1 and part 2 we found the equations for which these were solution spaces."},{"Start":"02:22.330 ","End":"02:24.240","Text":"In the first part,"},{"Start":"02:24.240 ","End":"02:29.150","Text":"we got these 2 equations that define the solution space U,"},{"Start":"02:29.150 ","End":"02:30.920","Text":"and in the second part,"},{"Start":"02:30.920 ","End":"02:37.130","Text":"we had just a single equation in the system and it defined V. In other words,"},{"Start":"02:37.130 ","End":"02:40.040","Text":"U are those x, y, z, t,"},{"Start":"02:40.040 ","End":"02:44.570","Text":"which satisfies these U and V when they satisfy this,"},{"Start":"02:44.570 ","End":"02:49.190","Text":"and to belong to the intersection is the and of the conditions,"},{"Start":"02:49.190 ","End":"02:52.880","Text":"which means that we take all 3 equations to belong"},{"Start":"02:52.880 ","End":"02:56.720","Text":"to both U and V has to satisfy these 2, and this 1."},{"Start":"02:56.720 ","End":"02:59.635","Text":"We have a system of 3 equations,"},{"Start":"02:59.635 ","End":"03:02.235","Text":"and that\u0027s the strategy."},{"Start":"03:02.235 ","End":"03:06.405","Text":"We\u0027ll proceed using matrices."},{"Start":"03:06.405 ","End":"03:09.770","Text":"Three equations in 4 unknown, but it\u0027s homogeneous,"},{"Start":"03:09.770 ","End":"03:12.005","Text":"so we don\u0027t need the augmented matrix,"},{"Start":"03:12.005 ","End":"03:16.000","Text":"what we need are the coefficients of the left-hand side, which are here."},{"Start":"03:16.000 ","End":"03:23.330","Text":"Here are the row operations that will get 0s below the minus 3."},{"Start":"03:23.330 ","End":"03:32.595","Text":"I just subtract the first row from the second and 3 times this minus 8 times this,"},{"Start":"03:32.595 ","End":"03:34.830","Text":"and that will give us this."},{"Start":"03:34.830 ","End":"03:41.065","Text":"Now we have to have a 0 below the minus 6."},{"Start":"03:41.065 ","End":"03:44.870","Text":"Oops, I got the arrow the wrong way around."},{"Start":"03:44.870 ","End":"03:50.220","Text":"This is what I put into here, of course."},{"Start":"03:50.480 ","End":"03:53.595","Text":"Six times this,"},{"Start":"03:53.595 ","End":"03:57.405","Text":"minus 43 times this, put it into here."},{"Start":"03:57.405 ","End":"04:01.500","Text":"That will give us Row Echelon form,"},{"Start":"04:01.500 ","End":"04:04.490","Text":"but the numbers are a bit large."},{"Start":"04:04.490 ","End":"04:07.100","Text":"We could do some simplification."},{"Start":"04:07.100 ","End":"04:11.450","Text":"Divided this row by minus 3,"},{"Start":"04:11.450 ","End":"04:13.430","Text":"sorry, by minus 2,"},{"Start":"04:13.430 ","End":"04:14.765","Text":"and got this row,"},{"Start":"04:14.765 ","End":"04:17.615","Text":"and here just divide by 10,"},{"Start":"04:17.615 ","End":"04:21.720","Text":"and now we have this matrix."},{"Start":"04:22.610 ","End":"04:27.965","Text":"Next, back from a matrix to a system of equations,"},{"Start":"04:27.965 ","End":"04:30.335","Text":"this is what we get."},{"Start":"04:30.335 ","End":"04:36.260","Text":"Notice that t is a free variable and that the x,"},{"Start":"04:36.260 ","End":"04:39.455","Text":"the y, and the z,"},{"Start":"04:39.455 ","End":"04:46.415","Text":"not the pivot elements or the constrained variables,"},{"Start":"04:46.415 ","End":"04:51.470","Text":"and if we use the Wondering One\u0027s method to get a basis,"},{"Start":"04:51.470 ","End":"04:56.160","Text":"we would let t equals 1 and the other 3 variables 0,"},{"Start":"04:56.160 ","End":"04:58.230","Text":"but there are no other 3 variables,"},{"Start":"04:58.230 ","End":"04:59.790","Text":"I\u0027ll do it from there."},{"Start":"04:59.790 ","End":"05:05.030","Text":"But they\u0027ll get messy numbers with fractions."},{"Start":"05:05.030 ","End":"05:07.460","Text":"I mean, if I have a vector,"},{"Start":"05:07.460 ","End":"05:12.620","Text":"I can take 8 times that vector also and it will still be the basis."},{"Start":"05:12.620 ","End":"05:17.540","Text":"Let\u0027s take t equals 8 just to get nicer numbers to work with."},{"Start":"05:17.540 ","End":"05:19.825","Text":"You could use t equals 1."},{"Start":"05:19.825 ","End":"05:22.360","Text":"With back substitution here,"},{"Start":"05:22.360 ","End":"05:24.380","Text":"if you put t equals 8,"},{"Start":"05:24.380 ","End":"05:26.530","Text":"you get that z equals 5."},{"Start":"05:26.530 ","End":"05:29.610","Text":"Then you put t and z in here,"},{"Start":"05:29.610 ","End":"05:31.740","Text":"and you\u0027ll get that y is 1,"},{"Start":"05:31.740 ","End":"05:33.390","Text":"and then put those 3 in here,"},{"Start":"05:33.390 ","End":"05:36.225","Text":"you\u0027ll get that x is 5,"},{"Start":"05:36.225 ","End":"05:38.370","Text":"and so here\u0027s the basis."},{"Start":"05:38.370 ","End":"05:42.405","Text":"I mean, we use the dimension is 1 and it works out."},{"Start":"05:42.405 ","End":"05:45.510","Text":"We\u0027ll just put them in order, x, y, z,"},{"Start":"05:45.510 ","End":"05:47.280","Text":"t; 5, 1,"},{"Start":"05:47.280 ","End":"05:49.465","Text":"5, 8, that\u0027s here."},{"Start":"05:49.465 ","End":"05:51.790","Text":"If you would have taken t equals 1,"},{"Start":"05:51.790 ","End":"05:55.475","Text":"you just would have got this thing divided by 8,"},{"Start":"05:55.475 ","End":"05:57.975","Text":"would still be a basis."},{"Start":"05:57.975 ","End":"06:02.650","Text":"That concludes the 4 part series and we\u0027re done."}],"ID":9985}],"Thumbnail":null,"ID":31678},{"Name":"Row and Column Spaces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Row and Column Spaces","Duration":"6m 48s","ChapterTopicVideoID":9921,"CourseChapterTopicPlaylistID":31679,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9921.jpeg","UploadDate":"2017-08-07T11:36:39.4300000","DurationForVideoObject":"PT6M48S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Starting a new topic."},{"Start":"00:01.890 ","End":"00:06.195","Text":"The basis and dimension for the row space of a matrix."},{"Start":"00:06.195 ","End":"00:10.485","Text":"Likewise, there\u0027ll be a column space of a matrix."},{"Start":"00:10.485 ","End":"00:13.035","Text":"In order to understand what\u0027s going on here,"},{"Start":"00:13.035 ","End":"00:15.495","Text":"best to start with an example."},{"Start":"00:15.495 ","End":"00:19.410","Text":"Here is our example of 4 by 5 matrix."},{"Start":"00:19.410 ","End":"00:25.785","Text":"I\u0027m going to start with the row space of a matrix later we\u0027ll do the column."},{"Start":"00:25.785 ","End":"00:28.755","Text":"What is a row space?"},{"Start":"00:28.755 ","End":"00:30.975","Text":"We just take the rows,"},{"Start":"00:30.975 ","End":"00:33.195","Text":"consider them as vectors,"},{"Start":"00:33.195 ","End":"00:39.400","Text":"and we take the subspace spanned by these rows."},{"Start":"00:39.490 ","End":"00:43.640","Text":"Each of these is a vector in R_5,"},{"Start":"00:43.640 ","End":"00:48.590","Text":"there are 5 components and there\u0027s 4 of them, and span,"},{"Start":"00:48.590 ","End":"00:52.190","Text":"we sometimes called it just sp without the"},{"Start":"00:52.190 ","End":"00:57.640","Text":"A-N. Go back and review the concept of span if you\u0027ve forgotten."},{"Start":"00:57.640 ","End":"01:01.445","Text":"It\u0027s just the set of all linear combinations of these basically."},{"Start":"01:01.445 ","End":"01:04.190","Text":"It is a subspace of R_5."},{"Start":"01:04.190 ","End":"01:07.340","Text":"As a subspace, we can ask what the dimension is"},{"Start":"01:07.340 ","End":"01:12.010","Text":"and we can try and find a basis for this subspace."},{"Start":"01:12.010 ","End":"01:17.900","Text":"This subspace is called the row space of the matrix A,"},{"Start":"01:17.900 ","End":"01:23.420","Text":"and it\u0027s usually denoted like this;"},{"Start":"01:23.420 ","End":"01:26.565","Text":"rowsp, row space of A."},{"Start":"01:26.565 ","End":"01:35.360","Text":"We\u0027ll do this the usual way by just taking these as rows of a matrix,"},{"Start":"01:35.360 ","End":"01:36.845","Text":"which, in this case,"},{"Start":"01:36.845 ","End":"01:38.480","Text":"is our original matrix A,"},{"Start":"01:38.480 ","End":"01:40.775","Text":"and we want to bring it to row echelon form."},{"Start":"01:40.775 ","End":"01:43.220","Text":"We\u0027ll do this briskly because you know how to"},{"Start":"01:43.220 ","End":"01:46.480","Text":"do this and that\u0027s not the point of the lesson."},{"Start":"01:46.480 ","End":"01:51.540","Text":"The first thing I did was I subtracted multiples of the top row from the other row."},{"Start":"01:51.540 ","End":"01:54.905","Text":"I subtracted 5 times top row from the second row,"},{"Start":"01:54.905 ","End":"01:58.475","Text":"6 times from this third row, and so on."},{"Start":"01:58.475 ","End":"02:00.670","Text":"End up with this,"},{"Start":"02:00.670 ","End":"02:05.330","Text":"and then we just subtract the second row from the third and the fourth,"},{"Start":"02:05.330 ","End":"02:07.640","Text":"and we end up with this."},{"Start":"02:07.640 ","End":"02:11.570","Text":"Notice that there are 2 rows of 0s and we don\u0027t care about those."},{"Start":"02:11.570 ","End":"02:15.175","Text":"We only want the non-0 rows."},{"Start":"02:15.175 ","End":"02:20.760","Text":"That gives us the basis, these 2 vectors."},{"Start":"02:20.760 ","End":"02:25.215","Text":"I called it basis of row just for short."},{"Start":"02:25.215 ","End":"02:28.725","Text":"Just a short for the row space of A."},{"Start":"02:28.725 ","End":"02:33.130","Text":"The dimension we get by counting 1, 2."},{"Start":"02:33.340 ","End":"02:37.130","Text":"If it was me, I would have also"},{"Start":"02:37.130 ","End":"02:42.420","Text":"continued a bit because these are all divisible by 4 or minus 4."}],"ID":9976},{"Watched":false,"Name":"Exercise 1","Duration":"4m 11s","ChapterTopicVideoID":9923,"CourseChapterTopicPlaylistID":31679,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9923.jpeg","UploadDate":"2017-08-07T11:37:07.7170000","DurationForVideoObject":"PT4M11S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.010","Text":"In this exercise,"},{"Start":"00:02.010 ","End":"00:04.080","Text":"we\u0027re given a matrix,"},{"Start":"00:04.080 ","End":"00:05.700","Text":"I see it\u0027s 4 by 4,"},{"Start":"00:05.700 ","End":"00:08.290","Text":"it\u0027s 4 rows, 4 columns."},{"Start":"00:08.720 ","End":"00:12.390","Text":"This is concerned with the row space and column space,"},{"Start":"00:12.390 ","End":"00:15.705","Text":"so review those concept if you\u0027ve forgotten them."},{"Start":"00:15.705 ","End":"00:19.740","Text":"We\u0027re going to find the basis and the dimension for the row space,"},{"Start":"00:19.740 ","End":"00:22.320","Text":"and then the same thing for the column space."},{"Start":"00:22.320 ","End":"00:23.555","Text":"After we\u0027ve done that,"},{"Start":"00:23.555 ","End":"00:26.765","Text":"question is what is the rank of this matrix?"},{"Start":"00:26.765 ","End":"00:29.990","Text":"We begin with the row space."},{"Start":"00:29.990 ","End":"00:34.035","Text":"I just copied the matrix,"},{"Start":"00:34.035 ","End":"00:39.505","Text":"and we\u0027re going to do row operations to try and bring it to echelon form."},{"Start":"00:39.505 ","End":"00:41.450","Text":"Let\u0027s start with the first column."},{"Start":"00:41.450 ","End":"00:42.500","Text":"We have a 0 here,"},{"Start":"00:42.500 ","End":"00:43.865","Text":"we want 0 here,"},{"Start":"00:43.865 ","End":"00:46.790","Text":"so we\u0027ll do twice this minus this, and here,"},{"Start":"00:46.790 ","End":"00:50.260","Text":"4 times this minus this, as written here."},{"Start":"00:50.260 ","End":"00:52.545","Text":"That brings us to this,"},{"Start":"00:52.545 ","End":"00:57.020","Text":"and now we want 0s below this 11."},{"Start":"00:57.020 ","End":"00:59.120","Text":"Well, we\u0027ll just add this row to this,"},{"Start":"00:59.120 ","End":"01:01.805","Text":"and subtract it from this."},{"Start":"01:01.805 ","End":"01:04.450","Text":"After we do that,"},{"Start":"01:04.450 ","End":"01:07.380","Text":"we have this with 2 rows of 0 \u0027s."},{"Start":"01:07.380 ","End":"01:11.650","Text":"Just the first 2 rows are non-0,"},{"Start":"01:11.650 ","End":"01:16.700","Text":"and these are the 2 that provide us with the basis for the row spaces,"},{"Start":"01:16.700 ","End":"01:19.100","Text":"4,1,1,5, just copy them."},{"Start":"01:19.100 ","End":"01:23.925","Text":"Yes, the dimension is just by counting 1,2."},{"Start":"01:23.925 ","End":"01:30.040","Text":"Next, we move on to the column space."},{"Start":"01:30.200 ","End":"01:32.960","Text":"I don\u0027t have the original matrix with me,"},{"Start":"01:32.960 ","End":"01:34.340","Text":"but if you go back and look,"},{"Start":"01:34.340 ","End":"01:36.170","Text":"you see this is the transpose."},{"Start":"01:36.170 ","End":"01:39.760","Text":"The original matrix, the first row was 4,1,1,5,"},{"Start":"01:39.760 ","End":"01:41.610","Text":"and then the 2nd row is this."},{"Start":"01:41.610 ","End":"01:44.270","Text":"Reverse rows and columns, get the transpose."},{"Start":"01:44.270 ","End":"01:47.405","Text":"That\u0027s how we deal with column space."},{"Start":"01:47.405 ","End":"01:50.220","Text":"Transpose."},{"Start":"01:50.630 ","End":"01:53.830","Text":"It seems like we\u0027re forever doing row operations,"},{"Start":"01:53.830 ","End":"01:56.090","Text":"that\u0027s what we do with matrices mostly."},{"Start":"01:56.090 ","End":"01:58.790","Text":"You want to get 0 \u0027s here, here,"},{"Start":"01:58.790 ","End":"02:01.100","Text":"and here, 4 times this minus this,"},{"Start":"02:01.100 ","End":"02:02.915","Text":"4 times this minus this,"},{"Start":"02:02.915 ","End":"02:08.220","Text":"and in last 1, 4 times this minus 5 times this."},{"Start":"02:08.220 ","End":"02:11.250","Text":"These are the row operations,"},{"Start":"02:11.250 ","End":"02:12.740","Text":"and this is the result."},{"Start":"02:12.740 ","End":"02:15.395","Text":"Feel free at any time to pause, check the computations."},{"Start":"02:15.395 ","End":"02:16.910","Text":"You\u0027ve got 3 0s here."},{"Start":"02:16.910 ","End":"02:18.785","Text":"Now we\u0027ve got to continue."},{"Start":"02:18.785 ","End":"02:23.110","Text":"We want 0 \u0027s under this entry."},{"Start":"02:23.110 ","End":"02:29.145","Text":"Yes, but just notice this 1 can be divided by 11,"},{"Start":"02:29.145 ","End":"02:32.130","Text":"we can divide this by minus 5,"},{"Start":"02:32.130 ","End":"02:34.815","Text":"we can divide this row by 3,"},{"Start":"02:34.815 ","End":"02:36.525","Text":"is role divisions,"},{"Start":"02:36.525 ","End":"02:39.200","Text":"and then we have something much simpler."},{"Start":"02:39.200 ","End":"02:43.100","Text":"In fact, at this point we have 3 rows the same,"},{"Start":"02:43.100 ","End":"02:44.900","Text":"so we know what\u0027s going to happen."},{"Start":"02:44.900 ","End":"02:48.030","Text":"We\u0027re going to get a couple of 0 \u0027s"},{"Start":"02:48.030 ","End":"02:51.465","Text":"here after we subtracted the 2nd from the 3rd and the 4th,"},{"Start":"02:51.465 ","End":"02:53.520","Text":"and sure enough, yes,"},{"Start":"02:53.520 ","End":"02:55.365","Text":"we have 2 rows of 0 \u0027s,"},{"Start":"02:55.365 ","End":"03:01.840","Text":"which means that these 2 would be a basis."},{"Start":"03:01.840 ","End":"03:06.680","Text":"But remember we were dealing in the column space and we did the transpose,"},{"Start":"03:06.680 ","End":"03:09.005","Text":"where we reversed rows and columns."},{"Start":"03:09.005 ","End":"03:11.750","Text":"This is really a column,"},{"Start":"03:11.750 ","End":"03:13.115","Text":"and so is this."},{"Start":"03:13.115 ","End":"03:18.925","Text":"The basis for the column space are these 2 column vectors,"},{"Start":"03:18.925 ","End":"03:23.825","Text":"and the dimension of the column space is therefore 2."},{"Start":"03:23.825 ","End":"03:29.065","Text":"Now, the last part was a question about the rank of the matrix."},{"Start":"03:29.065 ","End":"03:32.650","Text":"Now, you\u0027ll notice that we\u0027ve got the same number"},{"Start":"03:32.650 ","End":"03:36.040","Text":"2 in the column space as in the row space,"},{"Start":"03:36.040 ","End":"03:43.345","Text":"which is not a coincidence because it\u0027s a definition and theorem combined."},{"Start":"03:43.345 ","End":"03:49.510","Text":"The theorem says that the column space and the row space have the same dimension."},{"Start":"03:49.510 ","End":"03:52.975","Text":"Since these 2 have the same number,"},{"Start":"03:52.975 ","End":"03:56.510","Text":"that common value, we call it the rank of the matrix,"},{"Start":"03:56.510 ","End":"03:57.790","Text":"that\u0027s the definition part."},{"Start":"03:57.790 ","End":"03:59.720","Text":"This equality is the theorem,"},{"Start":"03:59.720 ","End":"04:02.170","Text":"this equality is definition."},{"Start":"04:02.170 ","End":"04:04.580","Text":"In this case, we have 2 equals 2,"},{"Start":"04:04.580 ","End":"04:11.720","Text":"and therefore the rank is also equal to 2. We\u0027re done."}],"ID":9977},{"Watched":false,"Name":"Exercise 2","Duration":"3m 53s","ChapterTopicVideoID":9922,"CourseChapterTopicPlaylistID":31679,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9922.jpeg","UploadDate":"2017-08-07T11:36:52.4800000","DurationForVideoObject":"PT3M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.309","Text":"In this exercise, we have this matrix here 4 rows, 5 columns."},{"Start":"00:05.309 ","End":"00:09.420","Text":"We have to find a basis and"},{"Start":"00:09.420 ","End":"00:11.790","Text":"the dimension of the row space for"},{"Start":"00:11.790 ","End":"00:15.375","Text":"this matrix and then the same thing for the column space."},{"Start":"00:15.375 ","End":"00:19.250","Text":"Finally, to state what is the rank of this matrix."},{"Start":"00:19.250 ","End":"00:21.755","Text":"We start off with the row space"},{"Start":"00:21.755 ","End":"00:27.425","Text":"and we start doing row operations to try and get it into echelon form."},{"Start":"00:27.425 ","End":"00:30.475","Text":"Let\u0027s see, subtract twice this from this,"},{"Start":"00:30.475 ","End":"00:33.585","Text":"subtract this from this add twice this to this."},{"Start":"00:33.585 ","End":"00:36.120","Text":"Then we have 0 \u0027s here."},{"Start":"00:36.120 ","End":"00:41.130","Text":"Now continuing, we want to"},{"Start":"00:41.130 ","End":"00:46.820","Text":"subtract multiples of this row or add 3 times this row to this row,"},{"Start":"00:46.820 ","End":"00:49.000","Text":"subtract 7 times from this row."},{"Start":"00:49.000 ","End":"00:54.090","Text":"Then we get 0 \u0027s here and also 0 \u0027s here."},{"Start":"00:54.090 ","End":"00:55.985","Text":"We continue to this column."},{"Start":"00:55.985 ","End":"01:00.240","Text":"We want 0 \u0027s below the minus 5."},{"Start":"01:00.260 ","End":"01:05.400","Text":"The first step, divide this by minus 16 and this by"},{"Start":"01:05.400 ","End":"01:11.340","Text":"37 and then we have 1 \u0027s here and that\u0027s going to make it a lot easier."},{"Start":"01:12.560 ","End":"01:15.470","Text":"We wanted 0 \u0027s below this 1,"},{"Start":"01:15.470 ","End":"01:22.730","Text":"so we just subtract this row from this row that gives us the 0 and we also have a 0 here."},{"Start":"01:22.730 ","End":"01:25.060","Text":"We\u0027ve got a 0 row."},{"Start":"01:25.060 ","End":"01:31.630","Text":"It already looks like the dimension is going to be 3 and these 3 are going to be a basis."},{"Start":"01:31.630 ","End":"01:34.620","Text":"Yeah. That\u0027s the row space."},{"Start":"01:34.620 ","End":"01:36.300","Text":"Just copied like 1, 2, 1,"},{"Start":"01:36.300 ","End":"01:38.270","Text":"3, 5, 1, 2, 1, 3, 5."},{"Start":"01:38.270 ","End":"01:40.225","Text":"Similarly for this 1 and for this 1,"},{"Start":"01:40.225 ","End":"01:42.620","Text":"and dimension is 3."},{"Start":"01:42.620 ","End":"01:45.760","Text":"Let\u0027s move on to the column space."},{"Start":"01:45.760 ","End":"01:49.800","Text":"What you see here is not a 4 by 5 matrix,"},{"Start":"01:49.800 ","End":"01:51.250","Text":"but a 5 by 4."},{"Start":"01:51.250 ","End":"01:54.005","Text":"I took the transpose of the original matrix."},{"Start":"01:54.005 ","End":"01:55.400","Text":"In the original matrix,"},{"Start":"01:55.400 ","End":"01:57.920","Text":"the first row was 1, 2, 1, 3, 5."},{"Start":"01:57.920 ","End":"01:59.255","Text":"Here\u0027s the first column,"},{"Start":"01:59.255 ","End":"02:01.920","Text":"swap rows and columns."},{"Start":"02:01.970 ","End":"02:07.500","Text":"Again, we\u0027re going to do row operations to bring it to echelon form."},{"Start":"02:07.910 ","End":"02:15.990","Text":"Subtracting multiples of the first row here from the others twice this row from this row,"},{"Start":"02:15.990 ","End":"02:18.395","Text":"1 \u0027s from here so on."},{"Start":"02:18.395 ","End":"02:20.940","Text":"You know how to do this."},{"Start":"02:20.940 ","End":"02:23.269","Text":"Pause and verify the computations."},{"Start":"02:23.269 ","End":"02:26.915","Text":"You get all 0 \u0027s here and this is what we have."},{"Start":"02:26.915 ","End":"02:30.965","Text":"Now continuing with the process,"},{"Start":"02:30.965 ","End":"02:35.165","Text":"we want to 0 out what\u0027s below the 1 \u0027s here."},{"Start":"02:35.165 ","End":"02:40.405","Text":"Again, by subtracting multiples of this from this, this and this."},{"Start":"02:40.405 ","End":"02:45.230","Text":"We\u0027ve got 0 \u0027s here and already we\u0027ve got a 0 row."},{"Start":"02:45.230 ","End":"02:54.740","Text":"These 2 are the same and so doing 2 steps in 1 after we subtracted this from this,"},{"Start":"02:54.740 ","End":"02:57.695","Text":"this becomes a 0, eliminate the 0 \u0027s,"},{"Start":"02:57.695 ","End":"02:59.810","Text":"and we just have this first,"},{"Start":"02:59.810 ","End":"03:02.390","Text":"second, and the fourth 1 \u0027s here."},{"Start":"03:02.390 ","End":"03:06.600","Text":"This is what we have. There are 3 non 0 rows and"},{"Start":"03:06.600 ","End":"03:11.815","Text":"so the dimension will be 3 and these 3 will form a basis."},{"Start":"03:11.815 ","End":"03:16.940","Text":"Now remember we\u0027re working in the column space or we have to transpose again."},{"Start":"03:16.940 ","End":"03:18.800","Text":"This row is really at the column."},{"Start":"03:18.800 ","End":"03:25.650","Text":"These 3 columns are a basis for the column space and the dimension,"},{"Start":"03:25.650 ","End":"03:28.125","Text":"of course is, 3 just by counting."},{"Start":"03:28.125 ","End":"03:35.270","Text":"Then finally, we were asked about the rank of the matrix and just like before,"},{"Start":"03:35.270 ","End":"03:38.644","Text":"we have this definition theorem,"},{"Start":"03:38.644 ","End":"03:42.530","Text":"the dimension of the column is equal to the dimension of the row is a theorem and"},{"Start":"03:42.530 ","End":"03:46.430","Text":"it works out here because we got 3 for both of these."},{"Start":"03:46.430 ","End":"03:52.025","Text":"This common value, 3 will be the rank of the matrix so the rank is 3,"},{"Start":"03:52.025 ","End":"03:54.480","Text":"and we are done."}],"ID":9978}],"Thumbnail":null,"ID":31679},{"Name":"Change of Basis","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Coordinate Vectors","Duration":"9m 26s","ChapterTopicVideoID":9931,"CourseChapterTopicPlaylistID":31680,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9931.jpeg","UploadDate":"2017-08-07T11:37:53.1470000","DurationForVideoObject":"PT9M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.720","Text":"Starting a new topic in vector spaces,"},{"Start":"00:03.720 ","End":"00:06.765","Text":"something called Coordinate Vectors."},{"Start":"00:06.765 ","End":"00:09.630","Text":"I\u0027ll explain through an example and then I\u0027ll also"},{"Start":"00:09.630 ","End":"00:12.705","Text":"explain what I mean by an ordered basis."},{"Start":"00:12.705 ","End":"00:18.490","Text":"We\u0027ll consider the vector space R^3, 3-dimensional real space."},{"Start":"00:19.010 ","End":"00:23.070","Text":"We\u0027ll take the basis B and I\u0027ll"},{"Start":"00:23.070 ","End":"00:26.730","Text":"just ask you to take my word for it that this is indeed a basis."},{"Start":"00:26.730 ","End":"00:29.905","Text":"I don\u0027t want to spend time verifying that."},{"Start":"00:29.905 ","End":"00:32.780","Text":"But all you would have to do would be to show that these"},{"Start":"00:32.780 ","End":"00:35.150","Text":"3 are linearly independent because the 3 of them."},{"Start":"00:35.150 ","End":"00:39.270","Text":"Anyway, take my word for this is a basis."},{"Start":"00:39.270 ","End":"00:45.180","Text":"Now, in this section makes a difference the order."},{"Start":"00:45.180 ","End":"00:46.950","Text":"The basis is in a specific order,"},{"Start":"00:46.950 ","End":"00:47.960","Text":"so this is the 1st,"},{"Start":"00:47.960 ","End":"00:50.700","Text":"this is the 2nd, this is the 3rd."},{"Start":"00:51.260 ","End":"00:55.820","Text":"It\u0027s not quite a set because in a set the order doesn\u0027t make a difference,"},{"Start":"00:55.820 ","End":"00:57.155","Text":"but here it does."},{"Start":"00:57.155 ","End":"00:59.705","Text":"That\u0027s as far as audit goes."},{"Start":"00:59.705 ","End":"01:05.195","Text":"Now, because it\u0027s a basis of R^3, of course,"},{"Start":"01:05.195 ","End":"01:08.975","Text":"then any vector in R^3 can be written as"},{"Start":"01:08.975 ","End":"01:15.365","Text":"a linear combination of these 3 and in only 1 way,"},{"Start":"01:15.365 ","End":"01:19.670","Text":"it\u0027ll be something times this plus something times this plus something times this."},{"Start":"01:19.670 ","End":"01:22.640","Text":"I\u0027ll take just some random example."},{"Start":"01:22.640 ","End":"01:24.320","Text":"Let\u0027s take the vector 2,"},{"Start":"01:24.320 ","End":"01:27.005","Text":"8, 12 and R^3."},{"Start":"01:27.005 ","End":"01:32.630","Text":"You could check that this is equal to minus"},{"Start":"01:32.630 ","End":"01:37.745","Text":"2 times the first element of the basis and then plus 4 times the 2nd,"},{"Start":"01:37.745 ","End":"01:40.070","Text":"plus 10 times the 3rd."},{"Start":"01:40.070 ","End":"01:42.740","Text":"How I got to this minus 2, 4,"},{"Start":"01:42.740 ","End":"01:45.320","Text":"10 is not important at this moment."},{"Start":"01:45.320 ","End":"01:47.600","Text":"In fact, I might have even cooked it up backwards."},{"Start":"01:47.600 ","End":"01:50.915","Text":"I might\u0027ve started with this and given you this."},{"Start":"01:50.915 ","End":"01:55.770","Text":"Just to verify that this is so with computation."},{"Start":"01:57.080 ","End":"02:00.235","Text":"Now that we have these numbers,"},{"Start":"02:00.235 ","End":"02:05.125","Text":"these coefficients of the basis,"},{"Start":"02:05.125 ","End":"02:12.240","Text":"we say that the coordinate vector of the vector u which is 2, 8,"},{"Start":"02:12.240 ","End":"02:16.000","Text":"12 with respect to the basis B,"},{"Start":"02:16.000 ","End":"02:18.065","Text":"it\u0027s going to be an ordered basis,"},{"Start":"02:18.065 ","End":"02:25.925","Text":"is we write it as u in a square brackets with the name of the basis here."},{"Start":"02:25.925 ","End":"02:30.150","Text":"These 3 components are these 3 scalars from"},{"Start":"02:30.150 ","End":"02:34.320","Text":"here and that\u0027s what gives us the coordinate vector."},{"Start":"02:34.320 ","End":"02:41.019","Text":"Remember the basis, you have to keep the order of the 3 basis elements."},{"Start":"02:41.019 ","End":"02:46.185","Text":"If we didn\u0027t give the name u,"},{"Start":"02:46.185 ","End":"02:50.990","Text":"we could just write it like this to put the vector itself in the square brackets to"},{"Start":"02:50.990 ","End":"02:55.745","Text":"indicate the name of the basis and this is the coordinate vector."},{"Start":"02:55.745 ","End":"02:59.275","Text":"Here\u0027s the second example."},{"Start":"02:59.275 ","End":"03:04.080","Text":"Another random vector 7, 2, 7."},{"Start":"03:04.080 ","End":"03:07.230","Text":"Check that this holds."},{"Start":"03:07.230 ","End":"03:10.050","Text":"It doesn\u0027t matter where I got the 2,"},{"Start":"03:10.050 ","End":"03:11.985","Text":"5, 0 from."},{"Start":"03:11.985 ","End":"03:15.710","Text":"Later, we\u0027ll learn how to find them but I might have just cooked it up"},{"Start":"03:15.710 ","End":"03:19.115","Text":"by starting from the right-hand side and getting to the left."},{"Start":"03:19.115 ","End":"03:24.125","Text":"Anyway, just verify this and when you\u0027ve checked that,"},{"Start":"03:24.125 ","End":"03:33.290","Text":"then we\u0027ll be able to say that the coordinate vector of this vector 7,"},{"Start":"03:33.290 ","End":"03:38.655","Text":"2, 7 v is the vector 2,"},{"Start":"03:38.655 ","End":"03:39.720","Text":"5, 0,"},{"Start":"03:39.720 ","End":"03:41.850","Text":"which I get from these 3 here."},{"Start":"03:41.850 ","End":"03:46.430","Text":"Again, we put it in square brackets with the name of the basis,"},{"Start":"03:46.430 ","End":"03:53.070","Text":"the audit basis here and if we haven\u0027t given this a name v,"},{"Start":"03:53.070 ","End":"03:57.655","Text":"we can just write like so."},{"Start":"03:57.655 ","End":"04:02.285","Text":"Now it\u0027s time to discuss how we can actually compute"},{"Start":"04:02.285 ","End":"04:09.230","Text":"a coordinate vector of a given vector relative to a given basis."},{"Start":"04:09.230 ","End":"04:12.560","Text":"As usual we\u0027ll do it with examples."},{"Start":"04:12.560 ","End":"04:18.770","Text":"I\u0027ll use the example we had earlier for this ordered basis of"},{"Start":"04:18.770 ","End":"04:25.435","Text":"R3 and go back and check and see if this was the example we had."},{"Start":"04:25.435 ","End":"04:28.020","Text":"We actually had 2 examples."},{"Start":"04:28.020 ","End":"04:32.030","Text":"Remember, we had u which was this and we had v which was"},{"Start":"04:32.030 ","End":"04:36.860","Text":"this and in each case I pulled these numbers out of a hat,"},{"Start":"04:36.860 ","End":"04:38.840","Text":"so to speak, the minus 2, 4, 10."},{"Start":"04:38.840 ","End":"04:40.900","Text":"I just asked you to verify."},{"Start":"04:40.900 ","End":"04:42.285","Text":"Once we verified,"},{"Start":"04:42.285 ","End":"04:44.135","Text":"then we said, okay,"},{"Start":"04:44.135 ","End":"04:54.170","Text":"the coordinate vector of u with respect to the basis B is this from these numbers and we"},{"Start":"04:54.170 ","End":"04:58.440","Text":"said that the coordinate vector of"},{"Start":"04:58.440 ","End":"05:06.275","Text":"this v with respect to this same ordered basis was 2, 5, 0."},{"Start":"05:06.275 ","End":"05:07.790","Text":"All this we had."},{"Start":"05:07.790 ","End":"05:09.110","Text":"But the question is,"},{"Start":"05:09.110 ","End":"05:14.660","Text":"how would we compute these numbers if I didn\u0027t just produce them?"},{"Start":"05:14.660 ","End":"05:18.320","Text":"The idea is instead of taking say,"},{"Start":"05:18.320 ","End":"05:21.125","Text":"2, 8, 12 or 7, 2, 7,"},{"Start":"05:21.125 ","End":"05:26.670","Text":"is to take a general vector like x, y,"},{"Start":"05:26.670 ","End":"05:32.210","Text":"z and if I do the computation for a general x, y, z,"},{"Start":"05:32.210 ","End":"05:36.860","Text":"then I can afterwards plug in whatever I want 2,"},{"Start":"05:36.860 ","End":"05:38.210","Text":"8, 12, 7, 2,"},{"Start":"05:38.210 ","End":"05:40.355","Text":"7 or anything else."},{"Start":"05:40.355 ","End":"05:42.750","Text":"Here is our vector x, y, z."},{"Start":"05:42.750 ","End":"05:50.495","Text":"Give it a name w. You want to find it as the linear combination of the basis vectors."},{"Start":"05:50.495 ","End":"05:52.850","Text":"These scalars a, b,"},{"Start":"05:52.850 ","End":"05:55.880","Text":"c are the unknowns."},{"Start":"05:55.880 ","End":"05:58.160","Text":"In other words, we want to find a, b,"},{"Start":"05:58.160 ","End":"06:00.170","Text":"and c in terms of x,"},{"Start":"06:00.170 ","End":"06:01.310","Text":"y, z as if x,"},{"Start":"06:01.310 ","End":"06:03.490","Text":"y, z were known."},{"Start":"06:03.490 ","End":"06:07.910","Text":"I multiply the scalars and then I do the addition,"},{"Start":"06:07.910 ","End":"06:10.225","Text":"then we get this."},{"Start":"06:10.225 ","End":"06:17.670","Text":"This gives us a system of 3 linear equations and 3 unknowns,"},{"Start":"06:17.670 ","End":"06:21.985","Text":"a, b, c. We\u0027ll do it using matrices."},{"Start":"06:21.985 ","End":"06:27.075","Text":"This is the corresponding augmented matrix."},{"Start":"06:27.075 ","End":"06:31.850","Text":"What we\u0027re going to do is not just bring it into row echelon form,"},{"Start":"06:31.850 ","End":"06:38.780","Text":"let\u0027s make this part the identity matrix and then it will be easier,"},{"Start":"06:38.780 ","End":"06:40.760","Text":"at least that\u0027s 1 way of doing it."},{"Start":"06:40.760 ","End":"06:47.750","Text":"I\u0027ll start out by subtracting this top row from the other 2."},{"Start":"06:47.750 ","End":"06:49.790","Text":"In row notation,"},{"Start":"06:49.790 ","End":"06:55.430","Text":"this is what I mean and if we do that, we get this."},{"Start":"06:55.430 ","End":"06:59.435","Text":"This is already in row echelon form, but we\u0027re continuing."},{"Start":"06:59.435 ","End":"07:03.755","Text":"Let\u0027s subtract this row from this row."},{"Start":"07:03.755 ","End":"07:09.390","Text":"This is the notation and then we get this and"},{"Start":"07:09.390 ","End":"07:16.610","Text":"the next thing is to add this row to this row and that will make this 0."},{"Start":"07:16.610 ","End":"07:25.420","Text":"Then finally we get this and all we have to do is multiply the middle row by minus 1."},{"Start":"07:25.420 ","End":"07:34.130","Text":"We have the identity matrix here which means that we have exactly what a,"},{"Start":"07:34.130 ","End":"07:37.630","Text":"b, and c are in terms of x, y, z."},{"Start":"07:37.630 ","End":"07:40.985","Text":"If we go back to where we came from,"},{"Start":"07:40.985 ","End":"07:46.880","Text":"we had that w was a times this vector plus b times this,"},{"Start":"07:46.880 ","End":"07:48.410","Text":"plus c times this."},{"Start":"07:48.410 ","End":"07:50.540","Text":"I just replaced a, b,"},{"Start":"07:50.540 ","End":"07:54.300","Text":"and c with what they are from here."},{"Start":"07:54.380 ","End":"07:58.469","Text":"We have this formula and now we can convert"},{"Start":"07:58.469 ","End":"08:08.100","Text":"any vector into it\u0027s coordinate vector."},{"Start":"08:08.100 ","End":"08:11.805","Text":"For instance, let\u0027s take 1, 2, 3."},{"Start":"08:11.805 ","End":"08:14.400","Text":"I just follow this recipe."},{"Start":"08:14.400 ","End":"08:18.830","Text":"These 3 basis vectors stay and all I\u0027ll do is these computation."},{"Start":"08:18.830 ","End":"08:21.155","Text":"X plus y minus z,"},{"Start":"08:21.155 ","End":"08:25.320","Text":"1 plus 2 minus 3, and so on."},{"Start":"08:26.090 ","End":"08:30.545","Text":"Here we get 0 here 1 and here 2,"},{"Start":"08:30.545 ","End":"08:35.260","Text":"which means that the coordinate vector of this 1, 2,"},{"Start":"08:35.260 ","End":"08:38.920","Text":"3 with respect to that basis is 0,"},{"Start":"08:38.920 ","End":"08:41.010","Text":"1, 2, these are the 0,"},{"Start":"08:41.010 ","End":"08:42.885","Text":"1, 2 from here."},{"Start":"08:42.885 ","End":"08:45.540","Text":"1 more example, take 4,"},{"Start":"08:45.540 ","End":"08:47.500","Text":"5, 6 this time."},{"Start":"08:47.500 ","End":"08:50.290","Text":"Same setup we put in 4,"},{"Start":"08:50.290 ","End":"08:51.760","Text":"5, 6 for x, y,"},{"Start":"08:51.760 ","End":"08:58.680","Text":"z. Compute these 3 scalar coefficients."},{"Start":"08:58.680 ","End":"09:03.730","Text":"Make these into a vector and that\u0027s the coordinate vector of 4,"},{"Start":"09:03.730 ","End":"09:06.475","Text":"5, 6 with respect to the basis B."},{"Start":"09:06.475 ","End":"09:13.460","Text":"In general, we can say that our w which is the arbitrary vector x, y,"},{"Start":"09:13.460 ","End":"09:18.425","Text":"z, has this expression here as"},{"Start":"09:18.425 ","End":"09:24.380","Text":"the coordinate vector with respect to the basis."},{"Start":"09:24.380 ","End":"09:27.030","Text":"That\u0027s all for this clip."}],"ID":9993},{"Watched":false,"Name":"Change-of-Basis Matrix part 1","Duration":"3m 47s","ChapterTopicVideoID":9932,"CourseChapterTopicPlaylistID":31680,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9932.jpeg","UploadDate":"2017-08-07T11:38:06.4170000","DurationForVideoObject":"PT3M47S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.945","Text":"Now beginning a new topic in vector spaces,"},{"Start":"00:03.945 ","End":"00:07.075","Text":"the concept of change of basis matrix."},{"Start":"00:07.075 ","End":"00:10.615","Text":"It\u0027s going to be in 5 parts."},{"Start":"00:10.615 ","End":"00:15.050","Text":"It\u0027s directly related to the concept of coordinate vectors"},{"Start":"00:15.050 ","End":"00:17.820","Text":"so I suggest you review this."},{"Start":"00:17.820 ","End":"00:20.253","Text":"What I\u0027m going to do is explain it"},{"Start":"00:20.253 ","End":"00:25.095","Text":"through a series of examples, solve exercises."},{"Start":"00:25.095 ","End":"00:27.005","Text":"Just to keep things simple,"},{"Start":"00:27.005 ","End":"00:32.525","Text":"I\u0027m going to have all the examples from the vector space R^3,"},{"Start":"00:32.525 ","End":"00:35.510","Text":"although you can see how it will be clearly"},{"Start":"00:35.510 ","End":"00:40.010","Text":"generalizable to any dimension or any vector space."},{"Start":"00:40.010 ","End":"00:45.815","Text":"Let\u0027s begin. Here\u0027s the first exercise that we\u0027re going to solve,"},{"Start":"00:45.815 ","End":"00:52.445","Text":"and we\u0027re given 2 bases, B_1 and B_2."},{"Start":"00:52.445 ","End":"00:59.330","Text":"I expect you to just take it on trust that these are bases or verify it yourself."},{"Start":"00:59.330 ","End":"01:01.820","Text":"I don\u0027t want to waste time doing this."},{"Start":"01:01.820 ","End":"01:06.095","Text":"In fact, B_2 will not be used in this particular exercise,"},{"Start":"01:06.095 ","End":"01:07.760","Text":"but these variants,"},{"Start":"01:07.760 ","End":"01:11.465","Text":"this will appear in the following clips, so it\u0027s included."},{"Start":"01:11.465 ","End":"01:16.250","Text":"Let\u0027s take a general vector x,y,z in R^3."},{"Start":"01:16.250 ","End":"01:19.730","Text":"The first part of the exercise is to compute"},{"Start":"01:19.730 ","End":"01:25.980","Text":"the coordinate vector of v relative to the basis B_1."},{"Start":"01:26.050 ","End":"01:30.410","Text":"Essentially we\u0027re just repeating some of the material from"},{"Start":"01:30.410 ","End":"01:35.255","Text":"coordinate vectors, but that\u0027s okay."},{"Start":"01:35.255 ","End":"01:37.465","Text":"B_1 is this,"},{"Start":"01:37.465 ","End":"01:42.250","Text":"we create this augmented matrix by taking the rows"},{"Start":"01:42.250 ","End":"01:46.428","Text":"or the vectors here as columns of this"},{"Start":"01:46.428 ","End":"01:50.265","Text":"augmented matrix and here we put x, y, z."},{"Start":"01:50.265 ","End":"01:55.170","Text":"Now we go through a series of row operations"},{"Start":"01:55.170 ","End":"01:58.265","Text":"until we get the identity matrix here."},{"Start":"01:58.265 ","End":"02:00.260","Text":"I\u0027ll go through this quickly,"},{"Start":"02:00.260 ","End":"02:05.600","Text":"subtract the top row from the second row to get a 0 here."},{"Start":"02:05.600 ","End":"02:09.080","Text":"Here we get y minus x, and then"},{"Start":"02:09.080 ","End":"02:14.520","Text":"we subtract the third row from the second row to get a 0 here."},{"Start":"02:14.740 ","End":"02:19.625","Text":"What we get is we take this x here,"},{"Start":"02:19.625 ","End":"02:24.450","Text":"and this comes from this column here."},{"Start":"02:24.550 ","End":"02:33.430","Text":"This, we put here, and this is this column here."},{"Start":"02:33.670 ","End":"02:37.070","Text":"Then what\u0027s here goes here,"},{"Start":"02:37.070 ","End":"02:41.795","Text":"and it\u0027s the third element of the basis,"},{"Start":"02:41.795 ","End":"02:44.225","Text":"really, that\u0027s what these 3 are."},{"Start":"02:44.225 ","End":"02:52.280","Text":"That\u0027s the long-hand form of the coordinate vector."},{"Start":"02:52.280 ","End":"02:59.634","Text":"You could just write it for short that the coordinate vector of v"},{"Start":"02:59.634 ","End":"03:08.940","Text":"with respect to the basis B_1 is x,"},{"Start":"03:08.940 ","End":"03:13.485","Text":"y minus x minus c and z."},{"Start":"03:13.485 ","End":"03:17.660","Text":"To give you an example of how we use this,"},{"Start":"03:17.660 ","End":"03:20.150","Text":"if we have, let\u0027s say 4, 5,"},{"Start":"03:20.150 ","End":"03:24.785","Text":"6 and we want to write it in terms of the basis B_1,"},{"Start":"03:24.785 ","End":"03:28.440","Text":"we compute these 3 from,"},{"Start":"03:28.440 ","End":"03:32.405","Text":"these would be my x, y, z, so x is 4,"},{"Start":"03:32.405 ","End":"03:38.435","Text":"y minus x minus z is minus 5 and z is 6,"},{"Start":"03:38.435 ","End":"03:44.690","Text":"and this is how we write this vector in terms of the 3 basis vectors."},{"Start":"03:44.690 ","End":"03:47.580","Text":"We\u0027re done for part 1."}],"ID":9994},{"Watched":false,"Name":"Change-of-Basis Matrix part 2","Duration":"3m 29s","ChapterTopicVideoID":9933,"CourseChapterTopicPlaylistID":31680,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9933.jpeg","UploadDate":"2017-08-07T11:38:18.2770000","DurationForVideoObject":"PT3M29S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.655","Text":"We\u0027re continuing with the concept of change of basis matrix."},{"Start":"00:05.655 ","End":"00:08.898","Text":"Again, with an exercise which is in fact"},{"Start":"00:08.898 ","End":"00:11.475","Text":"a continuation of the previous exercise,"},{"Start":"00:11.475 ","End":"00:15.570","Text":"it\u0027s almost identical to the previous one."},{"Start":"00:15.570 ","End":"00:17.055","Text":"In both cases,"},{"Start":"00:17.055 ","End":"00:22.230","Text":"we have B_1 and B_2 are bases of 3,"},{"Start":"00:22.230 ","End":"00:25.830","Text":"but this time we have to compute"},{"Start":"00:25.830 ","End":"00:35.715","Text":"the general vector as a coordinate vector relative to B_2 this time,"},{"Start":"00:35.715 ","End":"00:39.075","Text":"we had B_1 in the previous exercise."},{"Start":"00:39.075 ","End":"00:42.880","Text":"Here it\u0027s B_2, that\u0027s the only difference."},{"Start":"00:42.980 ","End":"00:46.100","Text":"The solution, well, it\u0027s pretty much routine because"},{"Start":"00:46.100 ","End":"00:49.210","Text":"we\u0027re just repeating what we did before,"},{"Start":"00:49.210 ","End":"00:53.040","Text":"only this time with B_2 instead of B_1,"},{"Start":"00:53.040 ","End":"00:59.784","Text":"where we take the vectors in the basis and put them as columns"},{"Start":"00:59.784 ","End":"01:04.735","Text":"in this augmented matrix, and here it\u0027s always the x, y, z."},{"Start":"01:04.735 ","End":"01:10.655","Text":"The idea is to get this matrix,"},{"Start":"01:10.655 ","End":"01:14.240","Text":"the restricted part to be the identity matrix"},{"Start":"01:14.240 ","End":"01:17.225","Text":"through a series of row operations."},{"Start":"01:17.225 ","End":"01:22.661","Text":"The first row operation will be to subtract the second row"},{"Start":"01:22.661 ","End":"01:27.455","Text":"from the third row, and that will give us the 0 here,"},{"Start":"01:27.455 ","End":"01:29.735","Text":"and here we get z minus x."},{"Start":"01:29.735 ","End":"01:33.126","Text":"Next, we\u0027re going to subtract"},{"Start":"01:33.126 ","End":"01:39.305","Text":"the second row from the third row to get 0 here."},{"Start":"01:39.305 ","End":"01:41.870","Text":"After we\u0027ve done that, this is what we get."},{"Start":"01:41.870 ","End":"01:45.785","Text":"Notice we already have the identity matrix here,"},{"Start":"01:45.785 ","End":"01:48.325","Text":"which is what we want."},{"Start":"01:48.325 ","End":"01:52.160","Text":"This in fact tells us how to write x, y,"},{"Start":"01:52.160 ","End":"01:55.190","Text":"z in terms of the basis B_2,"},{"Start":"01:55.190 ","End":"01:58.025","Text":"and let me just go back a little bit."},{"Start":"01:58.025 ","End":"02:03.675","Text":"This is the first element of the basis,"},{"Start":"02:03.675 ","End":"02:11.300","Text":"that\u0027s this, and we take the entry here and put it here."},{"Start":"02:11.300 ","End":"02:14.570","Text":"Then we have this entry which goes here,"},{"Start":"02:14.570 ","End":"02:20.575","Text":"and this is the second element of B_1 as you can see from here."},{"Start":"02:20.575 ","End":"02:25.845","Text":"This is what we put here and the"},{"Start":"02:25.845 ","End":"02:35.505","Text":"third member of the basis."},{"Start":"02:35.505 ","End":"02:37.310","Text":"That\u0027s in longhand."},{"Start":"02:37.310 ","End":"02:40.720","Text":"We can also write this in shorthand form"},{"Start":"02:40.720 ","End":"02:44.510","Text":"like so where we just abbreviate and we just say"},{"Start":"02:44.510 ","End":"02:48.248","Text":"the coordinate vector of V"},{"Start":"02:48.248 ","End":"02:52.910","Text":"with respect to the basis B_2 consists of these 3,"},{"Start":"02:52.910 ","End":"02:54.710","Text":"and that\u0027s the 3 I got from here,"},{"Start":"02:54.710 ","End":"02:56.870","Text":"or if you like, from here."},{"Start":"02:56.870 ","End":"03:00.870","Text":"As an example, I\u0027ll take the same 4,"},{"Start":"03:00.870 ","End":"03:03.340","Text":"5, 6 are used in the previous."},{"Start":"03:03.340 ","End":"03:08.135","Text":"We can write this in terms of the basis B_2."},{"Start":"03:08.135 ","End":"03:13.910","Text":"The coefficients are just computed from here or from here,"},{"Start":"03:13.910 ","End":"03:18.960","Text":"x would be 4, y is 5,"},{"Start":"03:18.960 ","End":"03:23.790","Text":"and z minus x minus y is 6 minus 4 minus 5,"},{"Start":"03:23.790 ","End":"03:26.235","Text":"which is the minus 3."},{"Start":"03:26.235 ","End":"03:29.860","Text":"This completes Part 2."}],"ID":9995},{"Watched":false,"Name":"Change-of-Basis Matrix part 3","Duration":"3m 52s","ChapterTopicVideoID":9934,"CourseChapterTopicPlaylistID":31680,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9934.jpeg","UploadDate":"2017-08-07T11:38:32.8470000","DurationForVideoObject":"PT3M52S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.901","Text":"Continuing with the concept of change of basis matrix"},{"Start":"00:03.901 ","End":"00:10.290","Text":"through a series of exercises, it\u0027s all 1 evolving exercise."},{"Start":"00:10.290 ","End":"00:13.470","Text":"We have the same 2 bases,"},{"Start":"00:13.470 ","End":"00:17.190","Text":"B_1 and B_2 as in the previous exercises."},{"Start":"00:17.190 ","End":"00:23.055","Text":"This time we\u0027re finally coming to the concept of change of basis matrix,"},{"Start":"00:23.055 ","End":"00:26.775","Text":"which I\u0027ll define through the example."},{"Start":"00:26.775 ","End":"00:30.390","Text":"The solution is broken up into 3 steps."},{"Start":"00:30.390 ","End":"00:38.070","Text":"In step 1, we compute the coordinate vector relative to the old basis."},{"Start":"00:38.070 ","End":"00:42.420","Text":"I guess I should say, when we say from B_1 to B_2,"},{"Start":"00:42.420 ","End":"00:48.149","Text":"this 1 is the old, and this 1 is the new."},{"Start":"00:48.149 ","End":"00:51.295","Text":"But we don\u0027t have to do this computation,"},{"Start":"00:51.295 ","End":"00:58.920","Text":"because we\u0027ve already done it in the previous part 1, I believe."},{"Start":"00:58.920 ","End":"01:01.610","Text":"We have the solution,"},{"Start":"01:01.610 ","End":"01:04.160","Text":"I just copied it."},{"Start":"01:04.160 ","End":"01:07.250","Text":"The next step is to write each vector in"},{"Start":"01:07.250 ","End":"01:10.835","Text":"the new basis as the linear combination of the old basis."},{"Start":"01:10.835 ","End":"01:15.000","Text":"What this means is that we\u0027re going to apply this formula here"},{"Start":"01:15.000 ","End":"01:21.645","Text":"to each of the 3 vectors in the new basis."},{"Start":"01:21.645 ","End":"01:29.690","Text":"Let me just emphasize again what\u0027s old and what is new that we have."},{"Start":"01:29.690 ","End":"01:36.412","Text":"The old is B_1, the new is B_2, and we want to write each of these"},{"Start":"01:36.412 ","End":"01:41.455","Text":"in terms of the old from this formula."},{"Start":"01:41.455 ","End":"01:44.515","Text":"Just take this and put it in a box."},{"Start":"01:44.515 ","End":"01:46.570","Text":"We have 3 computations."},{"Start":"01:46.570 ","End":"01:53.720","Text":"The first vector, just plugging in x,y,z is 1, 0, 1, and here,"},{"Start":"01:53.720 ","End":"01:56.820","Text":"this is really the only computation,"},{"Start":"01:56.820 ","End":"02:01.120","Text":"y minus x minus z comes out 0 minus 1 minus 1,"},{"Start":"02:01.120 ","End":"02:04.570","Text":"which is minus 2, and this is what we get."},{"Start":"02:04.570 ","End":"02:07.370","Text":"Then the next 1,"},{"Start":"02:07.370 ","End":"02:13.080","Text":"and I\u0027ll leave you to check that these numbers are what we get."},{"Start":"02:13.080 ","End":"02:19.547","Text":"The third 1 here is plugging in x,y,z is 0, 0,1,"},{"Start":"02:19.547 ","End":"02:22.105","Text":"and this is what we get,"},{"Start":"02:22.105 ","End":"02:25.305","Text":"and that\u0027s step 2."},{"Start":"02:25.305 ","End":"02:28.140","Text":"Now, the last step,"},{"Start":"02:28.140 ","End":"02:32.330","Text":"step 3, is to take these coefficients,"},{"Start":"02:32.330 ","End":"02:38.160","Text":"the 1 \u0027s that are in color, and put them into a matrix,"},{"Start":"02:38.160 ","End":"02:44.010","Text":"but the transpose of that matrix."},{"Start":"02:44.010 ","End":"02:47.395","Text":"The transpose, which means that the rows become columns"},{"Start":"02:47.395 ","End":"02:52.490","Text":"like this 1 minus 2, 1 becomes the first column,"},{"Start":"02:52.490 ","End":"02:57.775","Text":"and this 0, 0, 1 becomes the 2nd column,"},{"Start":"02:57.775 ","End":"03:02.940","Text":"and 0 minus 1,1 becomes the third column."},{"Start":"03:02.940 ","End":"03:09.465","Text":"The notation is we write M,"},{"Start":"03:09.465 ","End":"03:14.085","Text":"matrix, the new 1 on the top,"},{"Start":"03:14.085 ","End":"03:20.050","Text":"which is B2, and the old bases at the bottom here."},{"Start":"03:20.330 ","End":"03:23.970","Text":"I just have to point out,"},{"Start":"03:23.970 ","End":"03:26.310","Text":"and that\u0027s the asterisk here,"},{"Start":"03:26.310 ","End":"03:29.390","Text":"that this is not universally agreed on."},{"Start":"03:29.390 ","End":"03:37.670","Text":"Some people write the inverse of this matrix as the change of basis matrix,"},{"Start":"03:37.670 ","End":"03:41.735","Text":"so if you see that, don\u0027t be alarmed."},{"Start":"03:41.735 ","End":"03:45.305","Text":"Many things in mathematics are not universally agreed on,"},{"Start":"03:45.305 ","End":"03:47.375","Text":"some defined things other ways."},{"Start":"03:47.375 ","End":"03:52.710","Text":"But other than that, we\u0027re done with this 3rd clip."}],"ID":9996},{"Watched":false,"Name":"Change-of-Basis Matrix part 4","Duration":"2m 45s","ChapterTopicVideoID":9935,"CourseChapterTopicPlaylistID":31680,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9935.jpeg","UploadDate":"2017-08-07T11:38:42.7830000","DurationForVideoObject":"PT2M45S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.075","Text":"Continuing with change of basis matrix and this ongoing exercise,"},{"Start":"00:06.075 ","End":"00:10.665","Text":"this time we have the reverse of the previous exercise."},{"Start":"00:10.665 ","End":"00:16.725","Text":"There we had to compute the change of basis matrix from B_1 to B_2,"},{"Start":"00:16.725 ","End":"00:21.195","Text":"and this time we\u0027re going to do it from B_2 to B_1,"},{"Start":"00:21.195 ","End":"00:25.275","Text":"and it\u0027s going to be the same 3 steps."},{"Start":"00:25.275 ","End":"00:35.430","Text":"The first step is to compute the coordinate vector relative to the old basis."},{"Start":"00:35.430 ","End":"00:39.615","Text":"At this time, this one is going to be considered the"},{"Start":"00:39.615 ","End":"00:45.595","Text":"old and this one is the new."},{"Start":"00:45.595 ","End":"00:47.660","Text":"We already did that."},{"Start":"00:47.660 ","End":"00:51.460","Text":"I believe it was in step 2."},{"Start":"00:51.460 ","End":"00:55.640","Text":"I meant to say Part 2 of the 5-part series."},{"Start":"00:55.640 ","End":"00:57.305","Text":"Yeah, that\u0027s what we computed."},{"Start":"00:57.305 ","End":"01:00.650","Text":"Step 2 is still to write each vector"},{"Start":"01:00.650 ","End":"01:03.890","Text":"in the new basis as a linear combination of the old basis,"},{"Start":"01:03.890 ","End":"01:07.130","Text":"except that the roles of old and new have switched."},{"Start":"01:07.130 ","End":"01:10.250","Text":"Old is B_2 and new is B_1."},{"Start":"01:10.250 ","End":"01:19.955","Text":"What it means is that we take this formula and I put it in a nice box."},{"Start":"01:19.955 ","End":"01:23.245","Text":"We have to apply it 3 times,"},{"Start":"01:23.245 ","End":"01:33.530","Text":"once to each of the members of the new basis, beginning with 1,1,0."},{"Start":"01:35.810 ","End":"01:38.010","Text":"These are x, y, and z,"},{"Start":"01:38.010 ","End":"01:41.320","Text":"we plugin for x, y and z minus x minus y."},{"Start":"01:41.320 ","End":"01:42.849","Text":"This one, for example,"},{"Start":"01:42.849 ","End":"01:49.890","Text":"would be 0 minus 1 minus 1,"},{"Start":"01:49.890 ","End":"01:52.185","Text":"which gives us minus 2."},{"Start":"01:52.185 ","End":"01:54.960","Text":"The next one is this,"},{"Start":"01:54.960 ","End":"01:58.940","Text":"and I\u0027ll leave you to verify the computation."},{"Start":"01:58.940 ","End":"02:02.310","Text":"Similarly with the third one."},{"Start":"02:02.310 ","End":"02:06.755","Text":"We\u0027re going to do what we did before with the coefficients,"},{"Start":"02:06.755 ","End":"02:10.475","Text":"the ones colored in blue."},{"Start":"02:10.475 ","End":"02:16.305","Text":"We just make a matrix from these and then transpose it."},{"Start":"02:16.305 ","End":"02:23.135","Text":"For example, these 3 coefficients become the first column,"},{"Start":"02:23.135 ","End":"02:29.600","Text":"and these 3 coefficients become the second column."},{"Start":"02:29.600 ","End":"02:33.455","Text":"This, this, and this go in the third column."},{"Start":"02:33.455 ","End":"02:36.970","Text":"Rows become columns, that is transpose."},{"Start":"02:36.970 ","End":"02:45.129","Text":"This concludes step 3 and part 2 of the series of clips."}],"ID":9998},{"Watched":false,"Name":"Change-of-Basis Matrix part 5","Duration":"4m 15s","ChapterTopicVideoID":9930,"CourseChapterTopicPlaylistID":31680,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/9930.jpeg","UploadDate":"2017-08-07T11:37:33.7370000","DurationForVideoObject":"PT4M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:04.605","Text":"This is the last in the 5-part series."},{"Start":"00:04.605 ","End":"00:10.425","Text":"This exercise puts together all the concepts in the previous parts."},{"Start":"00:10.425 ","End":"00:15.480","Text":"Once again, we have B_1 and B_2, the same as in the previous parts,"},{"Start":"00:15.480 ","End":"00:19.560","Text":"and a general vector v which is x, y, z."},{"Start":"00:19.560 ","End":"00:23.835","Text":"Here, we have to show 3 things."},{"Start":"00:23.835 ","End":"00:29.515","Text":"The first part shows us that if we take the right change of basis matrix,"},{"Start":"00:29.515 ","End":"00:34.250","Text":"we can convert the coordinate vector relative to 1 base"},{"Start":"00:34.250 ","End":"00:38.020","Text":"into the coordinate vector relative to the other base."},{"Start":"00:38.020 ","End":"00:40.355","Text":"It\u0027s a little bit confusing."},{"Start":"00:40.355 ","End":"00:45.200","Text":"We call this the change of basis from B_2 to B_1,"},{"Start":"00:45.200 ","End":"00:49.685","Text":"even though really it should be the other way,"},{"Start":"00:49.685 ","End":"00:56.915","Text":"because we multiply by the B_1 coordinate vector and get the B_2 coordinate vector."},{"Start":"00:56.915 ","End":"01:05.265","Text":"We just write it in symbols though that\u0027s not even standard, the names,"},{"Start":"01:05.265 ","End":"01:11.055","Text":"like this is from B_1 to B_2, from B_2 to B_1, although we called it B_2 to B_1."},{"Start":"01:11.055 ","End":"01:15.390","Text":"Anyway, let\u0027s just stick to the way it was written."},{"Start":"01:15.430 ","End":"01:18.335","Text":"The third part, either way you look at it,"},{"Start":"01:18.335 ","End":"01:21.395","Text":"it says that one is the inverse of the other."},{"Start":"01:21.395 ","End":"01:28.260","Text":"Let\u0027s start with part 1, or better still,"},{"Start":"01:28.260 ","End":"01:37.025","Text":"let\u0027s review the values of all these matrices and vectors."},{"Start":"01:37.025 ","End":"01:42.990","Text":"Here they all are just copied from the previous parts."},{"Start":"01:43.820 ","End":"01:49.740","Text":"Number 1 says that we have to multiply,"},{"Start":"01:49.740 ","End":"01:52.995","Text":"let\u0027s see, B_1 on the top, B_2 at the bottom,"},{"Start":"01:52.995 ","End":"01:59.030","Text":"this one, and if we multiply it by this vector,"},{"Start":"01:59.030 ","End":"02:01.595","Text":"then we should get this vector;"},{"Start":"02:01.595 ","End":"02:03.755","Text":"this times this is this."},{"Start":"02:03.755 ","End":"02:06.205","Text":"Let\u0027s see."},{"Start":"02:06.205 ","End":"02:08.925","Text":"I want this times this to be this,"},{"Start":"02:08.925 ","End":"02:10.335","Text":"that\u0027s what it says here."},{"Start":"02:10.335 ","End":"02:12.110","Text":"I should have really explained also"},{"Start":"02:12.110 ","End":"02:18.560","Text":"that when we take a vector and represent it as a matrix,"},{"Start":"02:18.560 ","End":"02:22.785","Text":"we make it a column matrix,"},{"Start":"02:22.785 ","End":"02:26.910","Text":"so that this x, y minus x minus z, z is this"},{"Start":"02:26.910 ","End":"02:28.980","Text":"x, y minus x minus z, z."},{"Start":"02:28.980 ","End":"02:32.505","Text":"I\u0027ll leave you to verify it,"},{"Start":"02:32.505 ","End":"02:35.790","Text":"perhaps I\u0027ll do one of them,"},{"Start":"02:35.790 ","End":"02:36.870","Text":"maybe the last one."},{"Start":"02:36.870 ","End":"02:43.925","Text":"If I take minus 2 times x and minus 1 times this and 0 times this,"},{"Start":"02:43.925 ","End":"02:47.135","Text":"all I have to do is minus 1 times this reverses it,"},{"Start":"02:47.135 ","End":"02:50.380","Text":"so it\u0027s x plus z minus y,"},{"Start":"02:50.380 ","End":"02:53.535","Text":"and then I subtract 2x."},{"Start":"02:53.535 ","End":"02:55.340","Text":"Anyway, you get this,"},{"Start":"02:55.340 ","End":"02:57.950","Text":"I suggest just verify this."},{"Start":"02:57.950 ","End":"03:09.165","Text":"The second part says that we want the B_2 on the top, B_1 at the bottom."},{"Start":"03:09.165 ","End":"03:14.090","Text":"This one times this one has to be this one."},{"Start":"03:14.090 ","End":"03:15.980","Text":"Let\u0027s check that."},{"Start":"03:15.980 ","End":"03:19.490","Text":"These are the right ones."},{"Start":"03:19.490 ","End":"03:25.545","Text":"What we have is we wanted this times this to equal this,"},{"Start":"03:25.545 ","End":"03:27.710","Text":"and if you do the computation,"},{"Start":"03:27.710 ","End":"03:29.675","Text":"you\u0027ll see that it\u0027s true."},{"Start":"03:29.675 ","End":"03:36.025","Text":"The last part is to show that these 2 are inverses of each other."},{"Start":"03:36.025 ","End":"03:38.310","Text":"To show that they\u0027re inverses,"},{"Start":"03:38.310 ","End":"03:43.400","Text":"the easiest thing is to multiply them together and see that we get the identity matrix."},{"Start":"03:43.400 ","End":"03:51.300","Text":"Again, I\u0027ll leave it up to you to check because you know how to multiply matrices."},{"Start":"03:51.440 ","End":"03:58.280","Text":"The main thing here that we\u0027ve shown is that we can get a coordinate vector"},{"Start":"03:58.280 ","End":"04:05.170","Text":"with respect to one basis in terms of a coordinate vector relative to another basis"},{"Start":"04:05.170 ","End":"04:09.890","Text":"by multiplication by the appropriate change of basis matrix"},{"Start":"04:09.890 ","End":"04:12.335","Text":"or change of coordinate matrix."},{"Start":"04:12.335 ","End":"04:15.960","Text":"We are finally 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1.1

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