If $\operatorname{rank} (AB) = \operatorname{rank} (BA)$ for any $B$, then is $A$ invertible?

Question

Let $A$ and $B$ be two (nonzero) real square matrices and suppose that $\operatorname{rank} (AB) = \operatorname{rank} (BA)$ for any $B$. Can one prove that $A$ is invertible? (The converse is true and a simple linear algebra question but I'm stuck on this one.)

Answer 1

If $A$ is not invertible, there is a nonzero vector $u$ in the null space of $A$. Also, since you assume $A \neq 0$, there is a nonzero vector $v$ in the column space of $A$. Use these observations to construct a rank one matrix $B$ such that $AB = 0$ and $BA \neq 0$.

Answer 2

Let's construct the $B$ matrix along the lines given by Mike F. In the image of $A$ we have a nonzero vector $v$ which has a nonzero antecedent $w$ so that $v=A w$. In components we define $B_{ij}=u_i w_j$. This matrix is nonzero. Left multiplication by $A$ gives 0 because $u$ is in the kernel of $A$. If we do the right-multiplication by $A$ we find a matrix which is $u_i v_j$ which is nonzero. Strictly speaking there is a transposition when dealing with the right stuff. Great ! thanks.