Crossposted from Mathematics.

Consider the $2\times 2$ matrix \begin{align*} Q = \begin{bmatrix} 1 & 1 \\ 0 & a \end{bmatrix} - \epsilon \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1-\epsilon & 1 \\ -\epsilon & a \end{bmatrix} \end{align*} with $\epsilon>0$ and $a\in \mathbb{R}$ such that $\lvert a\rvert<1$.

One can use the Jury criterion to conclude that there exists a sufficiently small $\epsilon$ such that $Q$ has both the eigenvalues in the open unit disk for all $\lvert a\rvert<1$.

I would like to extend this result to a block-wise case. Consider the $2n \times 2n$ matrix \begin{equation} \tag{1}\label{1} X = \begin{bmatrix} A & I \\ 0 & B \end{bmatrix} - \epsilon \begin{bmatrix} I & 0 \\ C & 0 \end{bmatrix} = \begin{bmatrix} A-\epsilon I & I \\ -\epsilon C & B \end{bmatrix} \end{equation} where $I \in \mathbb{R}^{n\times n}$ is the identity matrix, $A\in\mathbb{R}^{n\times n}$ has the eigenvalues in the closed unit disk (for simplicity, we may consider $A$ with a simple eigenvalue at $1$ and the other $n-1$ inside the open unit disk), $B\in \mathbb{R}^{n\times n}$ has all the eigenvalues in the open unit disk, $C\in \mathbb{R}$ is symmetric and positive semidefinite.

Can we choose $\epsilon>0$ sufficiently small such that $X$ has all the eigenvalues inside the unit disk?

Suppose that $B$ and $C$ commute, i.e., $BC=CB$, then they are jointly diagonalizable via a nonsingular matrix $T\in \mathbb{R}^{n\times n}$. Therefore we can write \begin{align} \tag{2}\label{2} \begin{bmatrix} T & 0 \\ 0 & T \end{bmatrix} \begin{bmatrix} A-\epsilon I & I \\ -\epsilon C & B \end{bmatrix} \begin{bmatrix} T^{-1} & 0 \\ 0 & T^{-1} \end{bmatrix} = \begin{bmatrix} T A T^{-1} -\epsilon I & I \\ -\epsilon \Lambda_C & \Lambda_B \end{bmatrix}. \end{align}

If also $T A T^{-1}$ were diagonal, then we could reduce the problem to $n$ subproblems. I.e., by noticing that the eigenvalues of \begin{align} \tag{3}\label{3} \begin{bmatrix} a & 0 & b & 0\\ 0 & c & 0 & d\\ e & 0 & f & 0\\ 0 & g & 0 & h \end{bmatrix} \end{align} are given by the union of the eigenvalues of \begin{align*} \begin{bmatrix} a & b \\ e & f \end{bmatrix} \text{ and } \begin{bmatrix} c & d \\ g & h \end{bmatrix}, \end{align*} then we can study the eigenvalues of $n$ submatrices in the form of $Q$.

Is it possible to manage the possibility that $T A T^{-1}$ is not diagonal? E.g. does there exist a closed form transformation that reduces the case \eqref{1} or \eqref{2} to the desired case \eqref{3}?