If $AB^∗$ and $B^∗A$ are both normal, show that $B A^∗A = A A^∗B$

Question

This might be an easy question, but I haven't been able to prove it.

Prove that if $A B^*$ and $B^* A$ are normal matrices, then $B A^* A = A A^* B$

Any help is appreciated.

Here, $A^*$ is the conjugate transpose of $A$

Answer 1

This follows form the Fuglede-Putnam-Rosenblum Theorem (which istn't trivial) that states: If $X$ and $Y$ are normal matrices and $Z$ is any matrix with $XZ=ZY$, then $X^\ast Z = ZY^\ast$. In case of the question set $X=AB^\ast$, $Y=B^\ast A$ and $Z=A$. Then $XZ = AB^\ast A = ZY$, so $BA^\ast A=X^\ast Z =ZY^\ast= A A^\ast B$.