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Apr 6, 2020 at 19:26 vote accept Guest
Apr 3, 2020 at 23:59 comment added KhashF @NathanielJohnston I thought the same first. But notice that the $A^\#(z_0)$ is the Hermitian adjoint of $A(\bar{z_0})$ not $A(z_0)$.
Apr 3, 2020 at 23:47 comment added Nathaniel Johnston Yes, this is true, and it has nothing to do with entire functions. It's just the fact that if $A$ and $B$ are normal matrices (your conditions (1) and (2)) that commute (condition (3)) then they are simultaneously orthonormally diagonalizable, so each of $A$, $B$, $A^{\#}$, and $B^{\#}$ commute with each other.
Apr 3, 2020 at 23:46 answer added KhashF timeline score: 4
Apr 3, 2020 at 23:32 history edited Guest CC BY-SA 4.0
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Apr 3, 2020 at 23:16 history asked Guest CC BY-SA 4.0