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If $A(z) :=[A_{ij}(z)] $ and $B(z) :=[B_{ij}(z)] $ are two invertible $n\times n$ matrices of entire complex valued functions entries $A_{ij}(z)$, and $B_{ij}(z) $ with

(1). $AA^{\#}=A^{\#}A$ where $A^{\#}(z)=\left(\overline{A(\bar{z})} \right)^{T}$.

(2). $BB^{\#}=B^{\#}B$

(3).$ AB=BA$

Is it true that $$B(A A^{\#}) =(AA^{\#}) B$$.

Thanks in advance.

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  • $\begingroup$ 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. $\endgroup$
    – Nathaniel Johnston
    Commented Apr 3, 2020 at 23:47
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    $\begingroup$ @NathanielJohnston I thought the same first. But notice that the $A^\#(z_0)$ is the Hermitian adjoint of $A(\bar{z_0})$ not $A(z_0)$. $\endgroup$
    – KhashF
    Commented Apr 3, 2020 at 23:59

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Let $z_0\in\Bbb{R}$ be arbitrary. The matrices $A_0:=A(z_0)$ and $B_0:=B(z_0)$ are normal; in view of $A_0^*=A^\#(\bar{z_0})=A^\#(z_0)$ and $B_0^*=B^\#(\bar{z_0})=B^\#(z_0)$ they commute with their Hermitian adjoints. They also commute with each other. So $A_0$ and $B_0$ could be simultaneously diagonalized by a unitary matrix. That matrix also diagonalizes $A_0^*$ and $B_0^*$. So the four matrices $A_0=A(z_0),B_0=B(z_0),A_0^*=A^\#(z_0),B_0^*=B^\#(z_0)$ could be simultaneously diagonalized which implies that they all commute with each other. So $B(AA^\#)=(AA^\#)B$ at any point of the real line. Because the entries are entire functions, they coincide throughout the whole complex plane.

Added: As mentioned by @Guest and @MarkSapir, $A^\#(z)$ (respectively $B^\#(z)$) is the Hermitian conjugate of $A(\bar{z})$ (resp. $B(\bar{z})$) rather than that of $A(z)$ (resp. $B(z)$). Indeed, the Hermitian conjugates $A^*(z):=(\overline{A(z)})^{\rm{T}}$ and $B^*(z):=(\overline{B(z)})^{\rm{T}}$ do not vary holomorphically with $z$; they are anti-holomorphic. On the contrary, $A^\#(z)$ and $B^\#(z)$ are holomorphic because in their definitions the complex conjugation is applied both in the domain and the range. To circumvent this difficulty, one notices that $A^\#(z)$ (respectively $B^\#(z)$) is indeed the Hermitian conjugate of $A(z)$ (resp. $B(z)$) if $z$ is real. So by a linear algebra argument, one can establish $B(z)\left(A(z)A^\#(z)\right)=\left(A(z)A^\#(z)\right)B(z)$ for $z\in\Bbb{R}$. This persists on the rest of $\Bbb{C}$ because the entries of these matrices are entire functions (holomorphic on the whole $\Bbb{C}$).

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    $\begingroup$ $A^{*}$ is not the same as $A^{\#} $. $A^{\#} (z) =A^{*} (\bar{ z}) $. $\endgroup$
    – Guest
    Commented Apr 3, 2020 at 23:51
  • $\begingroup$ @Guest You are right, I edit my answer shortly. $\endgroup$
    – KhashF
    Commented Apr 3, 2020 at 23:54
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    $\begingroup$ @MarkSapir I think "entire" means analytic on $\Bbb{C}$: en.wikipedia.org/wiki/Entire_function $\endgroup$
    – KhashF
    Commented Apr 4, 2020 at 1:44
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    $\begingroup$ Sorry, my mistake is that I assumed that all functions involved are analytic. But the function $f(z)=\bar z$ is not analytic. Is it also a problem with the answer? $\endgroup$
    – user6976
    Commented Apr 4, 2020 at 17:35
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    $\begingroup$ @MarkSapir $A^\#(z)=(\overline{A(\bar{z})})^{\rm{T}}$ is holomorphic since you are conjugating twice (Schwarz Reflection); but $(A(z))^*=(\overline{A(z)})^{\rm{T}}$ is an anti-holomorphic function of $z$. $\endgroup$
    – KhashF
    Commented Apr 4, 2020 at 17:52

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