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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.

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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 $|z| > \rho(A).$

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