Is the $\operatorname{span}(\operatorname{span}(S))=\operatorname{span}(S)$? $$\operatorname{span}(\operatorname{span}(S))=\operatorname{span}(S)$$
Does that statement hold true?  I think it does since it is just the span of the span, but how would one prove this?
 A: $\operatorname{Span}(s)=V$ is a vector space, and a vector space is by definition closed under the operation of taking linear combinations of its elements. Since the span of $V$ consists of linear combinations of elements of $V$, we see that $\operatorname{Span}(\operatorname{Span}(s))=\operatorname{Span}(V) = \operatorname{Span}(s)$.
A: $\newcommand{\span}{\operatorname{span}}$
Hint: consider a generating family of $\span(S)$ and write everything
explicitely.
full details:
You have that $\span (S)\subset \span(\span (S))$ immediately.
Let $\{e_i, 1\le i \le n\}$ be a generating family of $\span (S)$.
Let $k\ge 1$, $a_1,\dots, a_m\in \span (S)$ with
$$
a_l = \sum_{k=1}^n a_{lk} e_k
$$
and let $x = \sum_{l=1}^m A_l a_l\in \span(\span(S))$.
Then
$$
x = \sum_{l=1}^m A_l a_l = \sum_{l=1}^m A_l \sum_{k=1}^n a_{lk} e_k
= \sum_{k=1}^n\left[\sum_{l=1}^m A_la_{lk}\right] e_k\in \span(S).
$$
Conclusion:
$$
\span (S)\subset \span(\span (S))\\
\span(\span (S))\subset \span (S)\\
\implies \span (S)= \span(\span (S))
$$
A: Let ${\bf x}$ be a vector in ${\rm span}({\rm span}(S))$.  By definition,
$${\bf x}=\lambda_1{\bf s}_1+\cdots+\lambda_n{\bf s}_n$$
for some scalars $\lambda_1,\ldots,\lambda_n$ and some vectors ${\bf s}_1,\ldots,{\bf s}_n$ in ${\rm span}(S)$.  But we can also, by definition of span, write
$${\bf s}_1=\mu_{1,1}{\bf s}_{1,1}+\cdots+\mu_{1,m}{\bf s}_{1,m}$$
and so on, where $\mu_{k,l}$ are scalars and ${\bf s}_{k,l}$ are vectors in $S$.  Substituting these into the previous equation shows that ${\bf x}$ is a linear combination of vector in $S$, so ${\bf x}\in{\rm span}(S)$.
Thus ${\rm span}({\rm span}(S))$ is a subset of ${\rm span}(S)$.
Conversely, if $\bf x$ is in ${\rm span}(S)$ then it is already a linear combination of vectors in ${\rm span}(S)$, and therefore is in ${\rm span}({\rm span}(S))$.
