# How to prove the convergence region for matrix Laurent expansion?

I know how to do it for scalar Laurent series. However, consider

$$\mathbf{F}(z) = \sum_{k=0}^{\infty} C A^k B z^{-k},$$ where $$F, A,B,C$$ are matrix with proper dimension. $$z \in \mathbb{C}$$.

I feel like the convergence region is $$|z| > \rho(A)$$, $$\rho(A)$$ is the spectral radius, but I don't know how to prove it. I cannot find anything online about convergence region for matrix Laurent series.

Let $$A_k:=CA^kB$$ and let $$|| \cdot||$$ denote a multiplicative norm on the space of $$n \times n$$- matrices. Then we have (we assume that $$B \ne 0 \ne C$$)
$$||\frac{1}{z^k}A_k||^{1/k} \le \frac{1}{|z|}||C||^{1/k}||A_k||^{1/k}||B||^{1/k}$$
Since $$\lim_{k \to \infty}\frac{1}{|z|}||C||^{1/k}||A_k||^{1/k}||B||^{1/k} = \frac{1}{|z|}\rho(A)$$, we get
$$\lim \sup ||\frac{1}{z^k}A_k||^{1/k} \le \frac{1}{|z|}\rho(A)$$. This shows that we have convergence, if $$\frac{1}{|z|}\rho(A)<1$$, hence convergence if $$|z| > \rho(A).$$