Does $\text{rank}(AB) = \text{rank}(BA) = r$?

Question

$A$ and $B$ are both $n \times n$ matrices. $A$ has rank $r$ and $B$ has rank $n$ (invertible).

My question is: Does $\text{rank}(AB) = \text{rank}(BA) = r$ for any $A$?

Answer 1

Since $B$ is invertible, $rank(AB)=rank(A)=r$ for any such $A$. It follows from the fact that $rank(AB) \le \min\{rank(A),rank(B)\}$

Observe that $Ker(B) \subset Ker(AB)$ and $Range AB \subset RangeA$. Can you use Rank-Nullity Theorem to conclude from here ??

Answer 2

For any matrix $A$, the rank of $A$ is equal to $r$ iff there exists invertible matrices $U, V$ such that $$A=UE^{(r)}V,$$ where $E^{(r)}_{i,i}=1$ for $1\le i\le r$ and zero others. So for any invertible matrix $B$ this criteria works with matrices $U, VB$ or $BU, V$