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Given $n\times n$ matrices with entire functions entries (holomorphic on all of the complex plane $\mathbb{C}$)

$A(z)=[a_{ij}(z)],B(z)=[b_{ij}(z)]$, i.e.,

$a_{ij}(z),b_{ij}(z)$ are entire functions for all $i,j=1,\dots,n$.

Can we find a form of a third $n\times n$ matrix $C(z)=[c_{ij}(z)]$ (in terms of $A$ and $B$, and under some conditions on $A$ and $B$ as needed) such that

$$A(z)A^{*}(v)+B(z)B^{*}(v)=C(z)C^{*}(v)$$ for all $z,v\in\mathbb{C}$?

Edit: where $A^{*}(v)=\left(\overline{A(v)} \right)^{T}$ (the Hermitian transpose of $A(v)$).

I tried adding and subtracting $A(z)B^{*}(v)$ to the LHS but it doesn't work!

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  • $\begingroup$ I don't understand. Is it true that $A^*(v) = (A(\bar v))^T$ and not $A^*(v) = (\bar A(\bar v))^T$? $\endgroup$ Jun 24, 2020 at 9:39
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    $\begingroup$ @DieterKadelka I think it is meant to say $A^*(v)=A(v)^*$, simply the hermitian transpose of A. So if $A(v)=(a_{ij}(v))$ then $A^*(v)=(\bar a_{ji}(v))$ $\endgroup$
    – Ivan Meir
    Jun 24, 2020 at 10:22
  • $\begingroup$ @ivan, that's true, thank you! $\endgroup$
    – MathGuest
    Jun 24, 2020 at 10:54

1 Answer 1

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First take the case $n=1$. Let $A(z)=(a(z))$, $B(z)=(b(z))$, $a(z),b(z)$ entire functions. Then $A(z)A^*(z)+B(z)B^*(z)=|a(z)|^2+|b(z)|^2$. Therefore we need $C(z)=(c(z))$ with $c(z)$ an entire function s.t $A(z)A^*(z)+B(z)B^*(z)=C(z)C^*(z)=|C(z)|^2=|a(z)|^2+|b(z)|^2$ for all $z\in \mathbb {C}$.

Therefore $|C(z)|^2\geq |a(z)|^2$ and $|C(z)|^2\geq |b(z)|^2$ for all $z\in \mathbb {C}$.

Liouville's theorem says that no entire function dominates another unless they are related by a constant multiple. Hence we must have $a(z)=\lambda b(z)$for some $\lambda \in \mathbb {C}$ and then $c(z)=\sqrt{1+|\lambda|^2}b(z)$.

Thus for $n=1$ we require that $A$ and $B$ are scalar multiples of one another and this also satisfies the more general equation with $z$ and $w$.

I believe in general you may need $A(z)=T(z)M$ and $B(z)=T(z)N$ where $M, N \in \mathbb{C}_{n\times n}$, constant matrices and $T(z)$ is any $n \times n$ matrix with entire entries.

This works because then

$$A(z)A^*(w)+B(z)B^*(w)=T(z)(MM^*+NN^*)T^*(w)$$$$=(T(z)Q)(T(w)Q)^*=C(z)C(w)^*$$

where $C(z)=T(z)Q$ a matrix with entire entries and $QQ^*=MM^*+NN^*$. $Q$ exists because $MM^*+NN^*$ is hermitian and has a Cholesky Decomposition.

To prove this we need the following lemma:

Lemma: For any $n \times n$ matrix $A(z)$ with entire entries there exists a matrix $A'(z)$ also with entire entries such that $A'(z)A(z)=A(z)A'(z)=D(z)I$ where $D(z)$ is entire.

Proof: $A'$ is just the Adjugate Matrix of $A$ and $D(z)=\det(A).$ $\blacksquare$

Theorem:

If $A(z)$, $B(z)$ and $C(z)$, all $n \times n$ matrices with entire entries, satisfy $$A(z)A^*(z)+B(z)B^*(z)=C(z)C^*(z) \tag{1} $$ for all $z \in \mathbb {C}$ then $A=C(z)M$, $B=C(z)N$ for constant $n \times n$ matrices $N$ and $M$ with $MM^*+NN^*=I$.

Proof:

By the lemma there exists an entire matrix $C'(z)$ s.t $C'(z)C(z)=C(z)C'(z)=\det(C(z))$.

Multiply equation (1) on the left by $C'$ and the right by $C'^*$. This gives

$$a(z)a^*(z)+b(z)b^*(z)=|\det(C(z))|^2 I $$

where $a(z)=C'(z)A(z)$, $b(z)=C'(z)B(z)$, both matrices with entire entries.

If $a(z)=(a_{ij}(z))$, $b(z)=(b_{ij}(z))$ then the main diagonal of the LHS, $L(z)=(l_{ij})$ is given by $$l_{ii}=\sum_{j=1}^n(|a_{ij}|^2+|b_{ij}|^2)=|\det(C(z))|^2. \tag{2}$$

Hence for all the entire functions $a_{ij}$ and $b_{ij}$ we have $$|a_{ij}|\leq |\det(C(z))|.$$ and $$|b_{ij}|\leq |\det(C(z))|$$ where $\det(C(z))$ is also an entire function.

Again by Liouville's Theorem we must have $|a_{ij}|=\lambda_{ij} \det(C(z))$ and $|b_{ij}|=\mu_{ij} \det(C(z))$ for some $\lambda_{ij}, \mu_{ij} \in \mathbb {C}.$

This clearly implies that $a=\det(C(z)) M$ and $b=\det(C(z)) N$ for constant $n \times n$ matrices $M=(\lambda_{ij})$ and $N=(\mu_{ij})$.

Hence by the definitions of $a$ and $b$,

$C'(z)A(z)=\det(C(z)) M$ and $C'(z)B(z)=\det(C(z)) N$.

Note that if $\det(C(z))=0$ then by equation (2) $A(z)=0$, $B(z)=0$ and hence $C(z)=0$. Since then $A(z)=C(z)I$ and $B(z)=C(z)I$ we may take $M=N=I$.

We can therefore assume henceforth that $\det(C(z))\neq 0$.

Multiplying both equations on the left by $C(z)$ gives $\det(C(z))A(z)=\det(C(z)) C(z) M$ and $\det(C(z))B(z)=\det(C(z)) C(z) N.$

Hence we have, as $\det(C(z))\neq 0$,

$A(z)=C(z) M$ and $B(z)= C(z) N.$

Substituting back into equation (1) gives

$$A(z)A^*(z)+B(z)B^*(z)=C(z)(MM^*+NN^*)C^*(z)$$$$=C(z)C(z)^*$$

for all $z$, which implies $MM^*+NN^*=I$ since $C(z)$ is invertible. $\blacksquare$

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  • $\begingroup$ the equation says that $A(z) A^{*}(w)+... $, but you used $A(z) A^{*} (z) ... $ in the first line! $\endgroup$
    – MathGuest
    Jun 24, 2020 at 12:17
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    $\begingroup$ I was taking the special case $z=w$ which obviously must hold if the equation holds for all $z$ and $w$. $\endgroup$
    – Ivan Meir
    Jun 24, 2020 at 12:38
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    $\begingroup$ If you take $C(z) = $ any 2x2 matrix-valued function, $A=C(z)\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}$, $B=C(z)\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$, the equality holds but $B$ is not a matrix multiple of $A$ nor viceversa. $\endgroup$ Jun 24, 2020 at 16:07
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    $\begingroup$ @FedericoPoloni Thank you Federico that is a great example! I think it works because $AB^*=BA^*=0$ and so $(AA^*+BB^*)=(A+B)(A+B)^*$. I had originally thought that case was covered by my condition but obviously not! Food for thought :-) $\endgroup$
    – Ivan Meir
    Jun 24, 2020 at 18:31
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    $\begingroup$ I've updated the conditions on $A$ and $B$ in my solution. $\endgroup$
    – Ivan Meir
    Jun 24, 2020 at 19:58

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