positive real matrix-valued function as linear combination of positive-real functions

Question

In my question I am considering $z$ as the complex variable in the $z$-transform $X(z)$ of a discrete-time sequence $x[n]$:

• I have $M$ square complex matrices $\mathbf{R}_m$ of size $N\times N$.
• I have $M$ complex, matrix-valued functions $\mathbf{H}_m(z)$ of complex variable $z$, also of size $N \times N$ .
• Each $\mathbf{H}_m(z)$ is diagonal, with all $N$ diagonal entries containing the same scalar function $H_m(z)$, i.e. $\mathbf{H}_m(z) = H_m(z)\mathbf{I}$.
• All scalar functions $H_m(z)$ are positive-real.

Is there a way to prove that $\sum_{m=1}^M \mathbf{R}_m \mathbf{H}_m(z)$ will be positive-real if all matrices $\mathbf{R}_m$ are Hermitian positive definite?

The only information that I found on discrete-time positive-real functions, is here, but it does not refer to matrix-valued functions. It says: a complex valued function of a complex variable $X(z)$ is said to be positive-real if

1. $z$ real $\implies$ $X(z)$ real
2. $|z| \geq 1 \implies \Re\{X(z)\}\geq0$

Answer 1

So you're looking at $F(z) = \sum_{m=1}^M H_m(z) R_m$. According to Wikipedia, a matrix-valued function is positive-real if

1. Each element of $F(z)$ is analytic in the open right half-plane.
2. Each element of $F(z)$ is real when $z$ is positive and real.
3. The Hermitian part $(F(z) + F(z)^*)/2$ is positive semi-definite when $\text{Re}(z) \ge 0$.

(1) and (3) are true because $H_m$ is assumed positive-real and $$\frac{F(z) + F(z)^*}{2} = \sum_{m=1}^M \text{Re}(H_m(z)) R_m$$ and a sum of positive semidefinite matrices is positive semidefinite. But (2) is not necessarily true, unless you assume $R_m$ have real entries.