how to prove that invertible matrix and vectors span the same space? Given $M$ is an invertible matrix, and {$\vec{v_1}...\vec{v_k}$} spans $R^n$,
then {A$\vec{v_1}...A\vec{v_k}$} also spans $R^n$
What does matrix invertibility have to do with span?
 A: Here is a proof:
Let $A$ be an $n\times n$ invertible matrix. If $\{v_1,...,v_k\}$ span $\mathbb R^n$, then we can write any element of $\mathbb R^n$ as a linear combination of $\{v_1,...,v_k\}$.  In particular, for each $i$, $1\leq i\leq n$, we can write $$A^{-1}v_i=a^i_1v_1+...+a^i_kv_k,$$ where $a^i_j$ are constants. Multiplying by $A$ on both sides of the above equation, we get that $$v_i=a^i_1(Av_1)+...+a^i_k(Av_k).$$ Since we can write any $v_i$ as a linear combination of $\{Av_1,...,Av_k\}$, and since $\{v_1,...,v_k\}$ is a spanning set, we can write any vector in $\mathbb R^n$ as a linear combination of $\{Av_1,...,Av_k\}$, so this is also a spanning set.
The reason that $A$ being invertible is necessary is that if $A$ is not invertible, there is a nonzero vector $v_1$ such $Av_1=0$. Choose $\{v_2,...,v_n\}$ to be vectors in $\mathbb R^n$ such that none of them is a multiple of $v_1$ and so that $\{v_1,...,v_n\}$ spans $\mathbb R^n$. Then $\{Av_1,...,Av_n\}$ has at most $n-1$ nonzero elements, so it cannot span $\mathbb R^n$.
A: From scratch:
If $\vec v\in \mathbb R^n$, then so is $A^{-1}\vec v$, and so
$A^{-1}\vec v=\sum_{i=1}^{k}c_i\vec v_i$ for some $c_i\in \mathbb R$. Upon multiplying by $A$ you get 
$AA^{-1}\vec v=\vec v=A\sum_{i=1}^{k}c_i\vec v_i=\sum_{i=1}^{k}c_iA\vec v_i$.
