Matrices and linear independence I wish to prove the following, but I'm not sure if the steps of my proof are correct. Here it goes:
Suppose A is an $m \times m$ matrix with $m$ pivot columns and that $v_1, \ldots, v_p$ is a
linearly independent set of vectors in $\mathbb{R}^m$.
Is $Av_1, \ldots, Av_p$ a linearly independent set of vectors?
Here's an attempt at the proof:
As all the $v_p's$ are linearly independent, $v_i = v_j$ $\implies$ $i=j$ for some $1 \leq i,j \leq p$.
As $A$ has full rank, then $A$ has a unique solution for every right hand side $b$ of the linear equation $Av = b$. That is, each $Av_i$, where $1 \leq i \leq p$ is a unique column vector, say $\bar{v}_p$ which is not a linear combination of any other $Av_j$.
So $Av_1, Av_2, \ldots, Av_p$ are linearly independent.
Any flaws of mistakes in the logic?
Ben
 A: Ok here's an attempt (new attempt) via contradiction.
Suppose that $A$ being a $m \times m$ matrix having exactly $m$ pivot columns and $v_1, v_2 \cdots v_p$ being linearly independent vectors implies that $Av_1, Av_2 \cdots Av_p$ are linearly dependent.
So if $Av_1 \cdots Av_p$ are linearly dependent, there exists constants $c_1 \cdots c_p$, not all zero such that the linear combination $\sum_{i=1}^{p} c_i A v_i = 0$. Now $A$ has exactly $m$ pivot columns so there is unique to solution to every R.H.S. $b$ and so the matrix $A$ is invertible. If it is invertible, I may multiply both sides of the equation by $A^{-1}$, so that I now have 
$\sum_{i=1}^{p} c_i v_i = 0$. But then this means that the $v_p's$ are all linearly dependent, which contradicts the assumption that $v_1, v_2 \cdots v_p$ are all linearly independent.
So indeed $Av_1, Av_2 \cdots Av_p$ are all linearly independent. $\hspace{2in}$ $\blacksquare$
Any objections to this proof against the logic?
Thanks,
Ben
A: Yes. You fail to demonstrate that $Ax_j$ is independent of $Ax_i$ for $i \ne j$.
Let's look at it:

As $A$ has full rank, then $A$ has a unique solution for every right hand side $b$ of the linear equation $Ax = b$.

This is true but does not imply, 

That is, each $Ax_i$, where $1 \leq i \leq p$ is a unique column vector, say $\bar{x}_p$ which is not a linear combination of any other $Ax_j$.

the first part of which is just the definition of a function and the last part of which is not implied by anything you've thus far said. What being able to uniquely solve $Ax = b$ gets you is not that $Ax$ is unique for each $x$ (which is what you're asserting) but that $x$ is unique for each $b$. As an example consider the matrix, $A$ with $1$ in the top left corner and $0$ elsewhere. What is the vector, $v$ for which $Av \ne Av$? You'll see that all matrices have the property that you're claiming. 
Your claim is true of course. If you know about the matrix inverse, then that makes it super easy. Just apply $A^{-1}$ to $\sum c_iAx_i = 0$ and see what happens.
Otherwise, you can just solve $Ax = \sum c_iAx_i$ and $Ax = 0$ and compare the results.
