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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, $AB = (BA)^\dagger$

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