Normal complex matrix-valued function Is there any examples of an $n\times n$ normal-invertible matrix of entire complex functions entries i.e., $HH^{*}=H^{*}H$, and $\det(H(z))\neq 0, \forall z\in \mathbb{C}$, where $H^{*}(z)=\left(\overline{H(z)} \right)^{T}$? Or a general characterization of such matrices!
 A: For easy examples, take diagonal matrices whose diagonal entries are entire functions with no zeros (thus exponentials of entire functions).  You can conjugate by a (constant) unitary matrix.  I'm not sure if this characterizes all such matrices.
EDIT: It seems that does characterize them.  Suppose for $z$ in a nonempty open subset $U$ of $\mathbb C$, an eigenvalue $\lambda(z)$ of $H(z)$ has constant multiplicity $m$.  Since it is a root of the non-constant polynomial $\det(H(z) - \lambda I)$ with analytic (in $z$) coefficients, $\lambda(z)$ is analytic in $z$.
For $z_0 \in U$, there are $m$ linearly independent eigenvectors of $H(z_0)$ with eigenvalue $\lambda(z_0)$.  Forming a matrix with these as columns, some $m \times m$ submatrix is nonsingular.  Multiply on the right by the inverse of that submatrix, and you get a matrix $V$ whose columns are eigenvectors and where this particular submatrix is the $m \times m$ identity matrix.  The remaining entries of $V$ are the unique solution of a system of linear equations: in block-matrix form $$ \pmatrix{H_{11}(z_0) & H_{12}(z_0)\cr H_{21}(z_0) & H_{22}(z_0)} \pmatrix{v_1 \cr I_m} = \lambda(z_0) \pmatrix{v_1 \cr I_m} $$
Replacing $z_0$ by $z$, we find that for $z$ in a neighbourhood of $z_0$ the system
$$   \pmatrix{H_{11}(z) & H_{12}(z)\cr H_{21}(z) & H_{22}(z)} \pmatrix{v_1(z) \cr I_m} = \lambda(z) \pmatrix{v_1(z) \cr I_m} $$
has a unique solution $v_1(z)$; this solution is analytic in $z$, and the columns
of $\pmatrix{v_1(z)\cr I_m}$ are $m$ linearly independent eigenvectors of $H(z)$ for eigenvalue $\lambda(z)$.  However, since $H(z)$ is normal, eigenvectors of $H(z)$ for eigenvalue $\lambda(z)$ are also eigenvectors of $H^*(z)$ for eigenvalue $\overline{\lambda(z)}$.  The same proof as above then says that $v_1(z)$ should be conjugate-analytic.  But the only functions (on a connected open set) that are both analytic and conjugate-analytic are constant.  Thus
the eigenvectors can be chosen to be constant.
