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!



Solving equation of matrix valued functions
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](https://en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis)#No_entire_function_dominates_another_entire_function) 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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. $\blacksquare$


