# Prove that $\mathrm{span}(S) = S$ for a subspace $S$.

Prove that if $S$ is a subspace of a vector space $V$, then $\mathrm{span}(S) = S$.

What I tried: I considered using the properties of vector spaces or maybe using an example where $S \subseteq \mathbb{R}^2$, but those strategies didn't amount to much. Any thoughts?

• What is your definition of span? Jan 12 '16 at 17:29
• just a normal spanning set i.e. a set of elements where you can take a linear combination of them and get S Jan 12 '16 at 17:32

$$\mathrm{Span}(S)$$ is the set of finite linear combinations of elements in $$S$$. In particular, every element in $$S$$ is in $$\mathrm{Span}(S)$$. Thus $$S\subseteq\mathrm{Span}(S)$$. Now, set $$u\in\mathrm{Span}(S)$$. Then $$u=a_1s_1+...+a_ns_n$$ for some scalars $$a_i$$ and $$s_i\in S$$. As $$S$$ is a subspace, it is closed under plus and product by scalar. Then $$u\in S$$. This proves $$\mathrm{Span}(S)\subseteq S$$.
Note that the span of a set $S\subset V$ is the intersection of all subspaces of $V$ that contain $S$, i.e. $$\operatorname{Span}(S) = \bigcap_{S\ \subset\ T\ \leqslant V}T.$$ The intersection of subspaces is again a subspace, so $$\bigcap_{S\ \subset\ T\ \leqslant V}T$$ is a subspace containing $S$, which implies that $$\operatorname{Span}(S)\subset \bigcap_{S\ \subset\ T\ \leqslant V}T.$$ But $\operatorname{Span}(S)$ itself is a subspace containing $S$, therefore $$\operatorname{Span}(S)\supset \bigcap_{S\ \subset\ T\ \leqslant V}T.$$
It follows immediately that if $S\leqslant V$, then $\operatorname{Span}(S)=S$.