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Let $\bf A$ be an $m \times n$ matrix. If $\bf P$ and $\bf Q$ are invertible $m \times m$ and $n \times n$ matrices, respectively

prove $\operatorname{rank}(\mathbf{PA}) = \operatorname{rank}(\bf{A})$

"I know how to prove $\operatorname{rank}(\mathbf{AQ})= \operatorname{rank}(\bf{A})$, where I start with $R(L_{\mathbf{AQ}})=L_{\bf{A}} L_{\bf{Q}}(F^n)=\dots$" but it seems like I can not prove $\operatorname{rank}(\mathbf{PA})=\operatorname{rank}(\bf{A})$ by this approach"

any help from you guys would be great. Thanks

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For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Nov 6 '12 at 22:41
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. – Julian Kuelshammer Nov 6 '12 at 22:43
Thanks for the editing Julian!! – mathwannabe Nov 6 '12 at 22:45
I didn't, Shaktal did. – Julian Kuelshammer Nov 6 '12 at 22:57
hint $\text{rank}(M) = \text{rank}(M^T)$, so if you know how to solve $\text{rank(AQ)}=\text{rank}(A)$ and $(AQ)^T=Q^TA^T$ you're all set. – karakfa Nov 6 '12 at 23:00

Hint: By the rank-nullity theorem, you can prove that two matrices (of the same sizes) have equal rank by instead proving that their null spaces have equal dimension.

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so since {Rank}(PA) =Rank((PA)^t)=Rank(A^t P^t) R(A^t P^t)=R(L_A^tP^t)=R(L_A^t L_P^t) = L_A^t L_P^t(F^m) = L_A^t(F^m)= R(L_A^t) so Rank(PA)=Rank(A^t P^t)= Rank(A^t)= Rank(A) am i correct? – mathwannabe Nov 7 '12 at 1:29
Yes, I believe that works. However, you can save yourself much headache by instead proving the following: $PAx = 0$ if and only if $Ax = 0$. Then apply rank-nullity. – Christopher A. Wong Nov 7 '12 at 1:55
Alright Thanks alot Christopher!!! – mathwannabe Nov 7 '12 at 3:10

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