TITLE: Prove $\operatorname{span}(S)=\operatorname{span}(S')$
QUESTION [0 upvotes]: Theorem. If $S=\{\mathbf v_1,\mathbf v_2,\dots,\mathbf v_r\}$ and $S'=\{\mathbf w_1,\mathbf w_2,\dots,\mathbf w_k\}$ are nonempty sets of vectors in a vector space $V$, then
  $$
\operatorname{span}\{\mathbf v_1,\mathbf v_2,\dots,\mathbf v_r\} =
\operatorname{span}\{\mathbf w_1,\mathbf w_2,\dots,\mathbf w_k\}
$$
  if and only if each vector in $S$ is a linear combination of those in $S'$, and each vector in $S'$ is a linear combination of those in $S$.

I am not really sure how to prove this theorem.  I know that if $S$ is a linear combination of $S'$, then $S=span(S')$.
If $S'$ is a linear combination of $S$, then $S'=span(S)$.
Where do I go from here?

REPLY [1 votes]: A vector is in the span of $S'$ so that $v=a_1w_1+a_2w_2+...+a_kw_k$
Each vector in $S'$ is a linear combination of those in $S$
$w_1=b_{11}v_1+b_{12} v_2+...+b_{1k} v_k$
$w_2=b_{21}v_1+b_{22} v_2+...+b_{2k} v_k$
...
$w_k=b_{k1}v_1+b_{k2} v_2+...+b_{kk} v_k$
So $v$ can be represented as 
$v = a_1(b_{11}v_1+b_{12} v_2+...+b_{1k} v_k)+a_2(b_{21}v_1+b_{22} v_2+...+b_{2k} v_k)+...$
This shows that every vector in $S'$ is in $S$, and a similar process shows that every element in $S$ is in $S'$ which means $\text{span}(S)=\text{span}(S')$

REPLY [1 votes]: Suppose $span(S')=span(S)$.  Then in particular, since the elements of $S$ are in $span(S)=spans(S')$, it follows that the elements of $S$ are in $span(S')$, and thus the elements of $S$ can be expressed as the linear combinations of elements of $S'$ by definition of being in $span(S')$.  Similarly, the elements of $S'$ can be expressed as linear combinations of elements in $S$.  Thus, we have our first direction:
Now the second direction, showing that if each element of $S$ can be written as a linear combination of elements in $S'$, and vice-a-versa, then $span(S)=span(S')$.  Linear combinations of linear combinations of the elements of $S$ are just linear combinations of $S$: $$\sum_{i}{a_i\left(\sum_{j}{b_{ij}s_j}\right)}=\sum_{i}{\sum_{j}{a_ib_{ij}s_j}}=\sum_{j}{\left(\sum_i{a_ib_{ij}}\right)s_j}$$ so that since each element of $S$ lies in $span(S')$, we see that $span(S)\subset span(S')$.  But the above analogously shows that $span(S')\subset span(S)$ since $S'$ satisfies the same condition as $S$, from which we see that $span(S)=span(S')$. (Since $A\subset B$ and $B\subset A$ implies that $A=B$)
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