Question on whether, "An entire function, nowhere zero, has an entire logarithm," holds for matrix-valued entire functions as well

Question

It is known that an entire function that is nowhere zero must be the exponential of another entire function.

Does this hold for matrix-valued functions as well? That is, given a matrix-valued entire function, none of whose eigenvalues is zero anywhere (save at complex infinity, trivially), is it true that it must be the exponential of another matrix-valued entire function?

I need (not in mathematical sense) this to hold, because (in particular) it would imply that a suitable pointwise branch of "$\ln(e^{K}e^{L})$", $K$ and $L$ being real skew-symmetric matrices, would exist such that it is real analytic w.r.t. the (upper) elements of $K$ and $L$. (I tried to prove this weaker statement using the BCH-D formula with no apparent success)

Answer 1

Counterexample: Consider the entire function $$ A(z) = \pmatrix{e^z & 0\cr z & 1\cr}$$ An entire logarithm of $A(z)$ must have eigenvalues $z + 2\pi i n$ and $2 \pi i m$ for some (constant) integers $n$ and $m$, with eigenvectors $\pmatrix{(e^z-1)/z\cr 1\cr}$ and $\pmatrix{0\cr 1\cr}$ respectively (for $e^z \ne 1$). Such a matrix must be $$ \pmatrix{z + 2 \pi i n & 0\cr \frac{z^2 + 2 \pi i (n-m) z}{e^z - 1} & 2 \pi i m\cr}$$ But the $(2,1)$ matrix element has a pole at $z = 2\pi i j$ with $j \ne 0, m-n$, so this doesn't work.

Answer 2

This is my comment above slightly expanded. Let's focus on $2\times 2$ matrices for convenience and let $A(z)$ be entire with $\det A=1$ (divide through by a holomorphic square root of $\det A$ if this doesn't hold initially, and recall that $\sqrt{\det A(z)}\not= 0$ does have an entire scalar logarithm).

If $A(z)=e^{B(z)}$, then $e^{\textrm{tr}\: B}=1$, so $\textrm{tr}\: B(z)\equiv 2\pi in$, and by replacing $B$ with $B-i\pi n$ (and perhaps $A$ with $-A$), we can then also assume that $\textrm{tr}\: B=0$.

The eigenvalues of $A$ are $\lambda(z)=T(z)/2 \pm (1/2)\sqrt{T^2(z)-4}$, $T(z)=\textrm{tr}\: A(z)$. These are typically not entire, but only have Puiseux series expansions about the points $z_0$ with $T(z_0)=\pm 2$. As a consequence, $$ \det B(z)=\log\lambda\cdot\log 1/\lambda =-\log^2\lambda(z) $$ will usually be prevented from being holomorphic near such points $z_0$, and hence there is no entire matrix logarithm.

Concrete examples where this happens are now easy to find. For example $$ A(z) = \begin{pmatrix} 1 & z\\ 1 & 1+z\end{pmatrix} $$ works (and the problematic points are $z_0=0, -3$), or of course the example given by Robert, which in normalized form is $$ A(z) = \begin{pmatrix} e^{z/2} & 0\\ ze^{-z/2} & e^{-z/2} \end{pmatrix} , $$ and here $\textrm{tr}\: A=\pm 2$ when $e^{z/2}=\pm 1$, consistent with Robert's calculation.