Prove span(span(s)) = span(s) Let $S$ be a subset of a vector space $V$ . Prove that $\text{Span}(\text{Span}(S)) = \text{Span}(S)$.
What I tried: I know that this proof revolves around the fact that a linear combination of elements = a linear combination of more fundamental elements.
 A: If $v\in $ Span(Span(S)), then $v$ can be written as a linear combination of elements of Span(S). Hence, $v = a_1w_1 +...+ a_nw_n$, where $w_i \in$ Span(S). Now, $w_i \in $ Span(S) $\implies$ $w_i = b_1z_1 +...+b_mz_m$ where $z_m$ is in $S$. Now, just replace each term in the first equation with its expansion and you have written $v$ as a linear combination of elements of $S$. 
A: The span $\text{span}(T)$ of some subset $T$ of a vector space $V$ is the smallest subspace containing $T$.
Thus, for any subspace $U$ of $V$, we have $\text{span}(U) = U$. This holds in particular for $U=\text{span}(S)$, since the span of a set is always a subspace.
A: Let $V$ be a vector space over a field $F$. By definition, $\text{Span}(S) = \left\{x \, | \, x = a_1 s_1 + \dots + a_n s_n, a_i \in F, s_i \in S \right\}$ and $\text{Span}(\text{Span}(S)) = \left\{x \, | x = a_1 s_1 + \dots + a_n s_n, a_i \in F, s_n \in \text{Span}(S) \right\}$. So to show equality, you can just show $\text{Span}(S) \subset \text{Span}(\text{Span}(S))$ and $\text{Span}(\text{Span}(S)) \subset \text{Span}(S)$.
