91Ó°ÊÓ

Linear independence is a critical concept when working with vectors and vector spaces. To determine if a set of vectors is linearly independent, we need to check if the only solution to the linear combination equals the zero vector. Mathematically, if we have vectors \(\textbf{v}_1, \textbf{v}_2, ..., \textbf{v}_n \), they are linearly independent if the equation \[ c_1 \textbf{v}_1 + c_2 \textbf{v}_2 + \ ... \ + c_n \textbf{v}_n = \textbf{0} \] only has the trivial solution \(\textbf{c} = (0, 0, ..., 0)\). This process involves setting up a system of linear equations. In our example, we started by checking the vectors in \[W = \text{span}(\begin{bmatrix} 1 \ 1 \ 1 \ \end{bmatrix}, \begin{bmatrix} -1 \ 2 \ -1 \ \end{bmatrix})\]. The equation was \[\begin{bmatrix} c_1 - c_2 \ c_1 + 2c_2 \ c_1 - c_2 \ \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \ \end{bmatrix}\], where we found \(c_1 = 0 \text{and} c_2 = 0\). This confirmed their linear independence. Identifying linearly independent vectors is essential as they ensure that we are capturing all necessary dimensions of the vector space without redundancy.

A basis of a vector space is a set of vectors that spans the entire vector space and is linearly independent. This means any vector in the space can be written as a unique linear combination of the basis vectors. For example, in \[V = \mathbb{R}^{3}\], we extended our subspace basis \(\begin{bmatrix} 1 \ 1 \ 1 \ \end{bmatrix}, \begin{bmatrix} -1 \ 2 \ -1 \ \end{bmatrix}\) to a full basis by finding an additional linearly independent vector. We chose \[ \begin{bmatrix} 1 \ 0 \ 0 \ \end{bmatrix} \], and verified its independence. By combining these vectors, we formed a basis for \(\mathbb{R}^{3}\). A proper basis has these properties: \

• Linearly Independent
• Spanning the vector space

\ . The span of a set of vectors that includes all the dimensions of a space ensures that any vector in the space can be represented through the basis. This becomes useful in various fields such as in solving systems of linear equations and performing vector transformations.

A subspace is a subset of a vector space that is also a vector space itself. To qualify as a subspace, a set must be closed under vector addition and scalar multiplication, and contain the zero vector. In our exercise, \[ W = \text{span} \left( \begin{bmatrix} 1 \ 1 \ 1 \ \end{bmatrix}, \begin{bmatrix} -1 \ 2 \ -1 \ \end{bmatrix} \right)\] is a subspace of \(\mathbb{R}^{3}\). This means any linear combination of the basis vectors of \ W\ belongs to \ W\. When extending a basis, ensuring the new vector falls outside the subspace and is linearly independent from it is crucial to cover the full vector space. By adding the vector \[\begin{bmatrix} 1 \ 0 \ 0 \ \end{bmatrix}\], which is not in the span of \ W\, we effectively increase the dimensionality of our spanning set to encompass all of \(\mathbb{R}^{3}\). Understanding subspaces helps in simplifying complex vector spaces into manageable pieces and transforming problems into smaller contexts.