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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?

user146264
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2 Answers2

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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.

Gaussler
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  • do u think if i make as what i done befoer but change PA to AP – user146264 Jun 25 '14 at 17:51
  • 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. – Gaussler Jun 25 '14 at 19:32
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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 $\mathrm{rank}(AP) = \mathrm{rank}(A).$

user3135338
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