Given is a compact set of matrices$$\mathcal{M} = \{\mathbf{M}(\xi)~|~ 0 < \check{\xi} \leq \xi \leq \hat{\xi}\}, $$ where
$$\mathbf{M}(\xi) = \mathbf{Z}\mathcal{K}(\xi) = \left[\begin{matrix}\mathbf{I}_n & \mathbf{0}\\ \mathbf{F} & \mathbf{G}\end{matrix}\right]\left[\begin{matrix} \Phi(\xi) & \Gamma(\xi) \\ \mathbf{0}& \mathbf{I}_m\end{matrix}\right], \text{with } ~\Phi(\xi) = e^{\mathbf{A}\xi} \text{ and } \Gamma(\xi) = \mathbf{A}^{-1}(\Phi(\xi) - \mathbf{I}_n)\mathbf{B},$$ is a real $(m+n)\times(m+n)$ matrix and $\mathbf{A}\in\mathbb{R}^{n\times n}$ is non-singular.
Let $\hat{\mathbf{M}} \triangleq \mathbf{M}(\hat{\xi}) \text{ and } \check{\mathbf{M}} \triangleq \mathbf{M}(\check{\xi}).$ Then my question is as follows:
Assuming that $\lim_{k\rightarrow \infty}\check{\mathbf{M}}^k = 0$ and $\lim_{k\rightarrow \infty}\hat{\mathbf{M}}^k = 0$, can we prove that the left wise infinite product $$ \prod_{k=1}^\infty{\mathbf{M}_k} = 0 \text{ for any } \mathbf{M}_k \in \mathcal{M} $$ and under what conditions. In particular I don't know how to prove that the joint spectral radius $\rho(\mathcal{M}) < 1$
My intiution is that it is true, but I don't know how to prove it correctly. MATLAB simulations also point towards this trend.