📂Linear Algebra

Expansion and Contraction of the Basis

Theorem¹

Let $S$ be a finite subset of the finite-dimensional vector space $V$.

(a) If $S$ generates $V$ but is not a basis of $V$, then elements of $S$ can be appropriately removed to reduce it to a basis of $V$.

(b) If $S$ is linearly independent but not a basis of $V$, then elements can be suitably added to $S$ to extend it to a basis of $V$.

Corollary

Let $W \le V$ be a subspace of the vector space $V$ of dimension $n$. Let $\gamma = \left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{k} \right\}$ be a basis of $W$. Then, suitable elements can be added to $\gamma$ to extend it to a basis $\beta = \left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{k}, \mathbf{v}_{k+1}, \dots, \mathbf{v}_{n} \right\}$ of $V$.

Proof

(a)

$\span(S) = V$, but if $S$ is not a basis of $V$, it means that $S$ is linearly dependent. Therefore, some vector $\mathbf{v}_{1}$ in $S$ can be expressed as a linear combination of the other vectors. By the addition/subtraction theorem, $S \setminus \left\{ \mathbf{v}_{1} \right\}$ also generates $V$. If $S \setminus {\mathbf{v}_{1}}$ is linearly independent, the proof is complete. If not linearly independent, the same logic can consider a generating set $S \setminus \left\{ \mathbf{v}_{1}, \mathbf{v}_{2} \right\}$ for $V$. Repeating this process yields a set that is a basis of $V$ by removing suitable elements from $S$.

(b)

Assume $\dim(V) = n$. If $S$ is linearly independent but not a basis of $V$, it means that $S$ does not generate $V$. Then, by the addition/subtraction theorem, adding some vector $\mathbf{v}_{1} \notin \span(S)$ to $S$ produces $S \cup \left\{ \mathbf{v}_{1} \right\}$, which remains linearly independent. Repeating this process allows for adding suitable vectors to $S$ to obtain a linearly independent set with a number of elements equal to $n$. Since a set in a $n$-dimensional vector space that is linearly independent and consists of $n$ elements is a basis, the proof is complete.

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