If matrix $A \in M_n(R)$ is defined such as for each $B \in M_n(R)$ we have $\operatorname{rank}(AB) = \operatorname{rank}(BA)$ find all possible ranks of $A$.
So obviously, $I$ and $0$ are both one of the possibilities for $A$. As it turns out, for each invertible matrix $A$ this equation is true. So we know that $n$ and $0$ are two possibilities of $\operatorname{rank}(A)$.
But what about other ranks? I personally think that this equality is only true if $\operatorname{rank}(A)$ is either $0$ or $n$. But I've got no way to prove it. I tried to struggle with nullspace of $A$ and create a $B$ with it in which $AB = 0$ but $BA \neq 0$ but I could not prove that this matrix always exists.