# Prove rank(AP) = rank(A) if P is an invertible n × n matrix and A is any m × n matrix?

I know how to prove

But what about we have different size of AP matrix?

This follows from the fact that $\operatorname{rank}(X) = \operatorname{rank}(X^T)$ for all matrices $X$. Hence $$\operatorname{rank}(AP) = \operatorname{rank}((AP)^T)=\operatorname{rank}(P^TA^T)=\operatorname{rank}(A^T)=\operatorname{rank}(A),$$ since $P^T$ is invertible, and because of what you already know.
• Well, it wouldn't work in the form stated; for one, you would have to multiply by $x$ from the left instead. But technically, yes, you could even just go through the proof with $P^T A^T$ instead of $PA$ and then apply the above calculation. Commented Jun 25, 2014 at 19:32
We can use the result $\mathrm{rank}(AB)\leq\mathrm{min}\{\mathrm{rank}(A),\,\mathrm{rank}(B)\}$ as follows:
$\mathrm{rank}(A) = \mathrm{rank}(A\underbrace{PP^{-1}}_{=I}) \leq \mathrm{rank}(AP) \leq \mathrm{rank}(A).$
Thus $\mathrm{rank}(AP) = \mathrm{rank}(A).$