# Commuting matrices with a transpose conjugate

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

• Title: Careful, the matrices $A$,$B$ need not commute. – Dietrich Burde Oct 8 '19 at 11:13
• Note that this occurs for arbitrary Hermitian $A,B$, but in this case the matrices $A,B$ need not be simultaneously diagonalizable. – Omnomnomnom Oct 8 '19 at 11:47
• Two diagonalizable matrices that commute are simultaneously diagonalizable. – Robert Israel Oct 8 '19 at 12:04

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 $$BV \subseteq \ker(A)$$.