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I know how to prove enter image description here

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But what about we have different size of AP matrix?


2 Answers 2


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.

  • $\begingroup$ do u think if i make as what i done befoer but change PA to AP $\endgroup$
    – user146264
    Commented Jun 25, 2014 at 17:51
  • $\begingroup$ 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. $\endgroup$
    – Gaussler
    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 $\mathrm{rank}(AP) = \mathrm{rank}(A).$


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