Linear Algebra

### Theorem 5.4: Basis Extension Theorem #### Statement: Let \( V \) be a finite-dimensional vector space and \( \mathbf{v}_1, \ldots, \mathbf{v}_m \) be linearly independent vectors in \( V \). Then there exist \( n = \dim(V) - m \) vectors \( \mathbf{u}_1, \ldots, \mathbf{u}_n \in V \) such that \( \{\mathbf{v}_1, \ldots, \mathbf{v}_m, \mathbf{u}_1, \ldots, \mathbf{u}_n\} \) is a basis for \( V \). #### Proof: 1. **Initial Setup**: Given that \( \mathbf{v}_1, \ldots, \mathbf{v}_m \) are linearly independent vectors in \( V \), we start by considering the span of these vectors, denoted as \( \text{span}(\mathbf{v}_1, \ldots, \mathbf{v}_m) \). 2. **Dimension Analysis**: Since \( \mathbf{v}_1, \ldots, \mathbf{v}_m \) are linearly independent, the dimension of their span is \( m \), i.e., \( \dim(\text{span}(\mathbf{v}_1, \ldots, \mathbf{v}_m)) = m \). 3. **Vector Space Dimension**: The dimension of the entire vector space \( V \) is given as \( \dim(V) \). By the properties of dimensions in vector spaces, we know that \( m \leq \dim(V) \). 4. **Finding Additional Vectors**: To extend \( \{\mathbf{v}_1, \ldots, \mathbf{v}_m\} \) to a basis of \( V \), we need to find additional vectors. The number of additional vectors required is \( n = \dim(V) - m \). 5. **Extension Process**: - Begin by selecting any vector \( \mathbf{u}_1 \in V \) not in the span of \( \{\mathbf{v}_1, \ldots, \mathbf{v}_m\} \). If \( \mathbf{u}_1 \) is linearly independent with respect to \( \{\mathbf{v}_1, \ldots, \mathbf{v}_m\} \), add it to the set. - Continue this process, selecting each subsequent vector \( \mathbf{u}_i \) such that it is linearly independent of all previously selected vectors \( \{\mathbf{v}_1, \ldots, \mathbf{v}_m, \mathbf{u}_1, \ldots, \mathbf{u}_{i-1}\} \). - This process is repeated until \( n \) additional vectors are found. 6. **Verification of Basis**: - The set \( \{\mathbf{v}_1, \ldots, \mathbf{v}_m, \mathbf{u}_1, \ldots, \mathbf{u}_n\} \) now contains \( \dim(V) \) vectors. - By construction, these vectors are linearly independent. - Therefore, by definition of a basis, this set forms a basis for \( V \). ### Theorem 5.5: #### Statement: Let \( V \) be a finite-dimensional vector space and \( \mathcal{U} \) be a subspace. Then \( \dim(\mathcal{U}) \leq \dim(V) \). #### Proof: 1. **Subspace Property**: By definition, a subspace \( \mathcal{U} \) of \( V \) must satisfy all the axioms of a vector space and every element of \( \mathcal{U} \) must also be an element of \( V \). 2. **Basis of Subspace**: Let \( \{\mathbf{u}_1, \ldots, \mathbf{u}_k\} \) be a basis for \( \mathcal{U} \). By definition, these vectors are linearly independent. 3. **Linear Independence in \( V \)**: Since every vector in \( \mathcal{U} \) is also in \( V \), the vectors \( \{\mathbf{u}_1, \ldots, \mathbf{u}_k\} \) are also linearly independent in \( V \). 4. **Dimension Comparison**: - The dimension of \( \mathcal{U} \) is \( k \), which is the number of vectors in its basis. - Since these \( k \) vectors are linearly independent in \( V \) and \( V \) has a finite dimension, it follows from the properties of vector spaces that \( k \leq \dim(V) \). 5. **Conclusion**: Therefore, \( \dim(\mathcal{U}) \leq \dim(V) \), confirming the theorem's statement. These proofs provide a fundamental understanding of the structure and dimensionality of vector spaces and their subspaces, crucial for advanced studies in linear algebra and its applications in various fields of science and engineering.

Math

Theorem 5.4: Basis Extension Theorem Let \( V \) be a finite-dimensional vector space and \( \mathbf{v}_1, \ldots, \mathbf{v}_m \) be linearly independent vectors in \( V \). Then there exist \( n = \dim(V) - m \) vectors \( \mathbf{u}_1, \ldots, \mathbf{u}_n \in V \) such that \( \{\mathbf{v}_1, \ldots, \mathbf{v}_m, \mathbf{u}_1, \ldots, \mathbf{u}_n\} \) is a basis for \( V \). Theorem 5.5: Let \( V \) be a finite-dimensional vector space and \( \mathcal{U} \) be a subspace. Then \( \dim(\mathcal{U}) \leq \dim(V) \).

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