Title says it all really.
$A = PD_AP^{-1}$, $B = PD_{B}P^{-1}$. We know $A = A^*$. Can we say for sure that $B = B^*$?
2026-03-27 09:37:45.1774604265
$A,B$ are simultaneously diagonalizable. $A = A^*$. Is $B = B^*$?
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No. Suppose that $A=\operatorname{Id}$ and that $B$ is diagonalizable, but $B\neq B^*$. Then $A$ and $B$ are obviously simultaneously diagonalizable, $A=A^*$, and $B\neq B^*$.