Prove that $\operatorname{span}( \operatorname{span}(S)) \subseteq \operatorname{span}(S)$ I need to prove that $\operatorname{span}\left( \operatorname{span}(S) \right) \subseteq \operatorname{span}(S)$. I know that similar thing has been already asked but no proof was given. I gave it a shot and I'd like to know if it's right and if I can simplify it somehow.
Let $S \in L$ be a non-empty set of vectors and $v \in \operatorname{span}\left( \operatorname{span}(S) \right)$.
If $v \in \operatorname{span}\left( \operatorname{span}(S) \right)$, then $v$ can be written as a linear combination of vectors in $\operatorname{span}(S)$
$$\overrightarrow v = {a_1}\overrightarrow {w_1}  + a_2 \overrightarrow {{w_2}} + \cdots + a_n\overrightarrow {w_n}, $$
where for each $\overrightarrow {w_i}$, where $i \in \left\{ 1,2,\ldots,n \right\}$, $\overrightarrow {w_i} \in \operatorname{span}(S)$ applies.
Since $\overrightarrow {w_i} \in \operatorname{span}(S)$, it can be written as a linear combination of  vectors in $\operatorname{span}( \operatorname{span}(S))$ $$\overrightarrow {w_i} = {b_1}\overrightarrow {z_1}  + b_2\overrightarrow {z_2} + \cdots + b_m\overrightarrow {z_m}, $$
where for each $\overrightarrow {z_j} $, where $j \in \left\{ 1,2,\ldots,m \right\}$, $\overrightarrow {z_j}  \in S$ applies.
Therefore $$\overrightarrow v = a_1\left( b_1 \overrightarrow {z_1} + b_2\overrightarrow {z_2} + \cdots + b_m\overrightarrow {z_m} \right) + a_2\left( b_1\overrightarrow {z_1} + b_2\overrightarrow {z_2} + \cdots + b_m\overrightarrow {z_m} \right) + \cdots + a_n \left( b_1\overrightarrow {z_1}  + b_2 \overrightarrow {z_2} + \cdots + b_m\overrightarrow {z_m} \right),$$
$$\overrightarrow v  = {a_1}{b_1}\overrightarrow {{z_1}}  + {a_2}{b_2}\overrightarrow {{z_2}}  +  \cdot  \cdot  \cdot  + {a_1}{b_m}\overrightarrow {{z_m}}  + {a_2}{b_1}\overrightarrow {{z_1}}  + {a_2}{b_2}\overrightarrow {{z_2}}  +  \cdot  \cdot  \cdot  + {a_2}{b_m}\overrightarrow {{z_m}}  +  \cdot  \cdot  \cdot  + {a_n}{b_1}\overrightarrow {{z_1}}  + {a_n}{b_2}\overrightarrow {{z_2}}  +  \cdot  \cdot  \cdot  + {a_n}{b_m}\overrightarrow {{z_m}}, $$$$\overrightarrow v  = \left( {{a_1}{b_1} + {a_2}{b_1} +  \cdot  \cdot  \cdot  + {a_n}{b_m}} \right)\overrightarrow {{z_1}}  + \left( {{a_2}{b_1} + {a_2}{b_2} +  \cdot  \cdot  \cdot  + {a_n}{b_m}} \right)\overrightarrow {{z_2}}  +  \cdot  \cdot  \cdot  + (a_n b_1 + a_n b_2 + \cdots + a_n b_m) \overrightarrow {z_m}. $$
We have shown that $\overrightarrow v \in \operatorname{span}( \operatorname{span}(S))$ can be written as a linear combination of vectors in $S$, in other words $\operatorname{span}(\operatorname{span}(S)) \subseteq \operatorname{span}(S).$
 A: $\newcommand\span{\operatorname{span}}$
$\newcommand\vec{\overrightarrow}$
I think the basic idea of your proof is certainly correct:
"If $\vec v \in \span(\span S)$, it is a linear combination of vectors in $\span S$, but since vectors in $\span S$ are linear combinations of vectors in $S$, $\vec v$ is a linear combination of linear combinations of vectors in $S$. So it must be a linear combination of vectors in $S$."
However I think your notation around $\vec{w_i}$ being a linear combination is a little unclear, because your $b\,$s make no mention of $i$! This makes your final expression for $v$ confusing, because the $b_1$ in "$a_1 b_1$" is not the same thing as the one in "$a_2 b_1$".
A nice way to address this is to instead let $\vec{w_i} = b_{ij} \vec{z_j}$, using the summation convention (if you're not familiar with it, it just means that on the RHS we're summing over all values of $j$). This makes a clear distinction between each of the coefficients, and also quite nicely makes explicit the fact that really we're talking about a linear map from some finite subset of $S$, ie we're talking about a matrix with entries $b_{ij}$. It's probably also worth pointing out that you're choosing $\{\vec{z_i}\}$ to be the union of the set of all vectors you're taking linear combinations of (which is a union of a finite number of finite sets, so is finite).
In general, if you're familiar with the summation convention, you could put it to work in your proof to make all the long lines and ellipses a little easier to stomach.
It's worth noting that if you can also prove $\span S \subseteq \span(\span S)$, you've shown that $\span S = \span(\span S)$!
A less important thing to point out is just to make sure that you're expanding all the brackets correctly! :) It happens to the best of us, but (in your notation) each of the $b$ terms in the final coefficient of $\vec{z_m}$ should be $b_m$.
Addendum:
By writing $b_{ij}$, all we're saying is that we have two indices that we're using to identify each coefficient. If you've ever done any programming, you could think of this as saying something like b[i][j]. This is why I talk about a matrix, as this sort of naturally gives an interpretation of all these coefficients as living in a rectangular grid.
The reason we really need to do this is because the coefficients in $\vec{z_1}$, for example, might not be the same across different $\vec{w_i}$.
For example, if $\vec{w_1} = \vec{z_1} + 3\vec{z_2}$, but
$\vec{w_2} = 2\vec{z_1} + \vec{z_2}$, then in your original notation, we would have $b_1 = 1$, but also a "different" $b_1 = 2$ (which is very hard to follow!). So with the double-indexed notation, we just say
$b_{11} = 1$, $b_{12} = 3$, $b_{21} = 2$, $b_{22} = 1$, resolving any ambiguity.
The part about the summation convention is simply that
$b_{ij}\vec{z_j}$ is a shorthand for $\sum_j b_{ij}\vec{z_j}$.
A: Here is a shot with double indexes as suggested.
Let $S$ be a subset of a linear space $L$ and $v$ be in $span\left( {span\left( S \right)} \right)$. 
That means that we can write
$$\overrightarrow v  = \sum\limits_{i = 1}^n {{a_i} \cdot \overrightarrow {{v_i}} } $$
where each $\overrightarrow {{v_i}} $ lies in $span\left( S \right)$ and a set $\left\{ {i = \left\{ {1,2,...n} \right\},{a_i} \ne 0} \right\}$ is finite.
For every $i$ we can write
$$\overrightarrow {{v_i}}  = \sum\limits_{j = 1}^m {{b_{ij}} \cdot \overrightarrow {{v_{ij}}} } $$
where each ${\overrightarrow {{v_{ij}}} }$ lies in $S$ and a set $\left\{ {j = \left\{ {1,2,...m} \right\},{b_{ij}} \ne 0} \right\}$ is finite.
By substituting for $\overrightarrow {{v_i}} $ in first equation we get
$$\overrightarrow v  = \sum\limits_{i = 1}^n {{a_i} \cdot \overrightarrow {{v_i}}  = } \sum\limits_{i = 1}^n {{a_i} \cdot } \sum\limits_{j = 1}^m {{b_{ij}} \cdot \overrightarrow {{v_{ij}}} }  = \sum\limits_{i = 1}^n {} \sum\limits_{j = 1}^m {\left( {{a_i} \cdot {b_{ij}}} \right) \cdot \overrightarrow {{v_{ij}}} ,} $$
where the product ${{a_i} \cdot {b_{ij}}}$ is zero only for finitely many pairs of $\left( {i,j} \right)$. That means that $\overrightarrow v $ also lies in $span\left( M \right)$ and overall
$$\left( {\overrightarrow v  \in span\left( {span\left( S \right)} \right) \wedge \overrightarrow v  \in span\left( S \right)} \right) \Rightarrow span\left( {span\left( S \right)} \right) \subseteq span\left( S \right).$$
