Commuting matrices with a transpose conjugate

Question

Two matrices are simultaneously diagonalizable iff they commute. Can anything be said about the simultaneous diagonalizability if two matrices commute but with a dagger? This happens when both A and B are Hermitian.

That is, $AB = (BA)^\dagger$

Answer 1

If $A$ and $B$ are simultaneously diagonalizable, they must commute. So if in addition $AB = (BA)^\dagger$, we have $(BA)^\dagger = BA$, i.e. their product is Hermitian. Now $A$ and $B$ commute with $BA$. Thus the eigenspaces of $BA$ are invariant under both $A$ and $B$. There are two cases:

1. On an eigenspace $V = V_\lambda$ of $BA$ for a nonzero $\lambda$, we have $AB = \lambda I$, so $B|_V = \lambda (A|_V)^{-1}$.
2. On the null space $V = \ker(BA)$ of $BA$, we have $BA = 0$. Thus $AV \subseteq \ker(B)$ and $BV \subseteq \ker(A)$.