# 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

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Where did "which is not a linear combination of any other $Ax_j$" come from? Note that especially in this area things can be set up from lots of different starting points, and then in one approach $A$ may be proved from $B$ whereas in another approach $B$ is proved from $A$, so it might help if you make more explicit what you're taking as known. – joriki Apr 5 '11 at 9:18
You switched from $v$ to $x$ halfway. – joriki Apr 5 '11 at 9:24