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.)
[Math] If $\operatorname{rank} (AB) = \operatorname{rank} (BA)$ for any $B$, then is $A$ invertible
inverselinear algebramatricesmatrix-rank
Best Answer
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 $BA \neq 0$.