Linear Span is Linear Subspace/Proof 1

Proof
First, suppose that $S = \emptyset$.

By definition of linear combination of empty set, it follows that $\map \span \emptyset = \set {\bszero}$, 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 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:

It follows that:

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