# If $AB≠0$ , $BA≠0$ , it's true that rank(AB)=rank(BA)? [closed]

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!!

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 $(1)$.
• @StackUser It's a way of dodging your condition that both $AB$ and $BA$ are non-zero. – Lord Shark the Unknown Nov 22 '17 at 20:51
• Yes, but how it's achieve that $r(AB)$≠$r(BA)$? – user445359 Nov 22 '17 at 20:52
• 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 $B$. – Somos Nov 22 '17 at 22:17