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