Linear Span is Linear Subspace

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Theorem

Let $V$ be a vector space over a division ring $K$.

Let $S \subseteq V$ be a subset of $V$.


Then the linear span $\map \span S$ is a subspace of $V$.


Proof 1

First, suppose that $S = \O$.

By definition of linear combination of empty set, it follows that $\map \span \O = \set \bszero$, where $\bszero$ denotes the zero vector of $V$.

From Zero Subspace is Subspace, it follows that the trivial vector space $\set \bszero$ is a subspace of $V$.


Suppose instead that $S$ is non-empty.

All $v \in \map \span S$ are of the form:

$v = \ds \sum_{k \mathop = 1}^n \lambda_k v_k$

where $n \in \N$, $\lambda_1, \ldots, \lambda_n \in K$, and $v_1, \ldots, v_n \in S$.

We use the Two-Step Vector Subspace Test to prove that $\map \span S$ is a subspace of $V$.

For any $\lambda \in K$, it follows from Vector Space Axiom $\text V 6$: Distributivity over Vector Addition that:

$\ds \lambda v = \lambda \paren {\sum_{k \mathop = 1}^n \lambda_k v_k} = \sum_{k \mathop = 1}^n \paren {\lambda \lambda_k} v_k$

which shows that $\lambda v \in \map \span S$, which is the first condition of the Two-Step Vector Subspace Test.


For any $w \in \map \span S$, let $w$ be of the form:

$w = \ds \sum_{l \mathop = 1}^m \mu_l w_k$

where $m \in \N$, $\mu_1, \ldots, \mu_m \in K$, and $w_1, \ldots, w_m \in S$.

Define the three sets:

\(\ds N\) \(:=\) \(\ds \set {1 \le k \le n: v_k \notin \set {w_1, \ldots, w_m} }\)
\(\ds M\) \(:=\) \(\ds \set {1 \le l \le m: w_l \notin \set {v_1, \ldots, v_n} }\)
\(\ds \OO\) \(:=\) \(\ds \set {\tuple {k,l}: 1 \le k \le n, 1 \le l \le m, v_k = w_l}\)


It follows that:

\(\ds v + w\) \(=\) \(\ds \sum_{k \mathop = 1}^n \lambda_k v_k + \sum_{l \mathop = 1}^m \mu_l w_k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in N} \lambda_k v_k + \sum_{l \mathop \in M} \mu_l w_l + \sum_{\tuple {k,l} \mathop \in \OO} \paren {\lambda_k + \mu_l} v_k\) Vector Space Axiom $\text V 5$: Distributivity over Scalar Addition

which shows that $v + w \in \map \span S$, which is the second condition of the Two-Step Vector Subspace Test.

$\blacksquare$


Proof 2

This is a special case of Generated Submodule is Linear Combinations.

As such, the statement follows immediately from that theorem.

$\blacksquare$


Sources