conditions for which matrix has all eigenvalues inside unit circle or on the unit circle $\mathbf{M}=\begin{bmatrix}
\mathbf{y}&0&0&\cdots&0&0&\mathbf{x}\\
\mathbf{I}&0&0&\cdots&0&0&0\\
0&\mathbf{I}&0&0&0&0&0\\
0&0&\mathbf{I}&0&0&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&0&\cdots&\mathbf{I}&0&0\\
0&0&0&0&0&\mathbf{I}&0\\
\end{bmatrix}$
Given that $\mathbf{y}=\operatorname{diag}\left(y_i\right)$ and $\mathbf{x}=\operatorname{diag}\left(x_i\right) $ where each $i=1,2,\dots, N$, $y_i$ are $N\times N$ matrices whose eigenvalues are strictly less inside unit circle. $I$ are $N\times N$. For each $i$, $x_i$ are also $N\times N$.
Could anyone help me to find out under what condition(s) this matrix $\mathbf{M}$  has the property that modulus of its all eigenvalues strictly less than $1$ given that each block matrix in $\mathbf{y}$ are given that they all have eigenvalues less than $1$, for all $i$?
Thank you.
 A: We could deduce the conditions, given an approximate solution for eigenvector correspondent to the maximum eigenvalue, so, we could reformulate our objective into a simpler one.
When the maximum eigenvalue is a strictly less than 1 ?

Assuming than $M$ is a $(n, n)$ block matrix.
If $M$ is a diagonalizable matrix, then we could use the power iteration method.
\begin{equation} \label{eq1}
\mathbf{v_{k + 1}} = \mathbf{\frac{Mv_{K}}{||Mv_{K}||}}, 
\end{equation}
where $v_{k+i}$, is the best approximation for eigenvector correspondent to the maximum eigenvalue.
\begin{equation} \label{eq2}
\mathbf{v_{0}} = \begin{bmatrix}a_{0}\\a_{1}\\\vdots\\a_{n}\end{bmatrix}
\end{equation}
$a_{0},\ ... \ , a_{i}, \quad$  are vectors
Assuming we have started with $\mathbf{v_{0}} \succ 0$, then
\begin{equation} \label{eq3}
\mathbf{Mv_{0}} = \begin{bmatrix}a_{0} * x + a_{n} * y\\a_{1}\\\vdots\\a_{n}\end{bmatrix}
\end{equation}

We are able to make a variety of assumptions on $x$ and $y$, to guarantee that the recursive relation is decreasing.
