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In given $A,B$ in size of $nXn$ and $AB≠0$ , $BA≠0$.
It's true that rank(AB)=rank(BA)?
Please don't give me full answer to this question, i want just a HINT!!

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Can you find $C$ and $D$ with $CD=0$ and $DC\ne0$? Then take $A$ and $B$ to be the diagonal sums of $C$ and $D$ with the one-by-one matrix $(1)$.

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  • $\begingroup$ Can you explain how did you get to that? $\endgroup$ – user445359 Nov 22 '17 at 20:49
  • $\begingroup$ @StackUser It's a way of dodging your condition that both $AB$ and $BA$ are non-zero. $\endgroup$ – Lord Shark the Unknown Nov 22 '17 at 20:51
  • $\begingroup$ Yes, but how it's achieve that $r(AB)$≠$r(BA)$? $\endgroup$ – user445359 Nov 22 '17 at 20:52
  • $\begingroup$ What is meant by "diagonal sum" is that you take the matrix $C$, for exmaple and add a row and a column with all zeros except $1$ on the diagonal and call this $A$. Do the same to $D$ to get $B$. $\endgroup$ – Somos Nov 22 '17 at 22:17