Eigenvalues in unit disk for a 2×2 block matrix 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}?
 A: Part 1 - Determinant of $X$
As a partial result, it is possible to show that $\lvert\det X\rvert < 1$ for sufficiently small $\epsilon$. In fact, because $B$ and $C$ commute, then thanks to a known property of block matrices, we have that
\begin{equation}
\det\begin{bmatrix}
A-\epsilon I & I\\
-\epsilon C & B
\end{bmatrix}
= \det ((A-\epsilon I)B+\epsilon C).
\end{equation}
From OP's assumptions on the eigenvalues of $A$ and $B$ we have that $\lvert\det A\rvert < 1$ and $\lvert\det B\rvert < 1$, which implies that
\begin{equation}
\lvert\det(AB)\rvert = \lvert\det A\rvert\lvert\det B\rvert < 1.
\end{equation}
Hence, by continuity there exists sufficiently small $\epsilon>0$ such that
\begin{equation}
\lvert\det ((A-\epsilon I)B+\epsilon C)\rvert < 1.
\end{equation}
Part 2 - All eigenvalues except one
Let $0<m<1$ be the maximum modulus of all eigenvalues of $A$ (excluding the eigenvalue $1$) and $B$. The eigenvalues of $A$ and $B$ fulfil
\begin{equation}
\det((A -\lambda I)(B-\lambda I)) = 0
\end{equation}
and, except the eigenvalue $1$ of $A$, their distance from the boundary of the unit disk is at least $1-m$. Then by continuity, for sufficiently small $\epsilon > 0$, all least all solutions except one of the perturbed equation
\begin{equation}
\det((A-\epsilon I -\lambda I)(B-\lambda I) + \epsilon C) = 0
\end{equation}
are in the unit disk, but the solutions of such equation are the eigenvalues of $X$.
Part 3 - The last eigenvalue when $A$ and $B$ are symmetric
Let $v_A$ be the eigenvector corresponding to the single eigenvalue $1$ of $A$. Then $\begin{bmatrix}v_A\\ 0\end{bmatrix}$ is the eigenvector corresponding to the (single) eigenvalue $1$ of the matrix
\begin{equation}
\begin{bmatrix}
A & I\\
0 & B
\end{bmatrix}.
\end{equation}
Hence, for sufficiently small $\epsilon > 0$, the perturbed block matrix $X$ has a (still single, hence real) eigenvalue $\lambda$ that tends to $1$ as $\epsilon \rightarrow 0$. Let $\begin{bmatrix}v\\ w\end{bmatrix}$ be the corresponding eigenvector. From the bottom blocks of X we have
\begin{equation}
w = -\epsilon(\lambda I - B)^{-1}Cv.
\end{equation}
Because $B$ and $C$ commute, so do $(\lambda I - B)^{-1}$ and $C$, which implies that the spectrum of $(\lambda I - B)^{-1}C$ is, up to some ordering, the product of the spectra (see @user91684's answer to Eigenvalues of the product of two diagonalizable commuting matrices.) of $(\lambda I - B)^{-1}$ and $C$. Because (i) $\lambda$ is arbitrarily close to 1, (ii) the eigenvalues of $B$ are in $\mathopen]-1+m,1-m\mathclose[$ and (iii) $C$ is positive semidefinite, then the eigenvalues $\lambda_1,\dotsc,\lambda_n$ of $-\epsilon(\lambda I - B)^{-1}C$ are nonpositive and let $v_1,\dotsc, v_n$ be the respective eigenvectors (which can be chosen orthonormal since $-\epsilon(\lambda I - B)^{-1}C$ is symmetric). Then we can write
\begin{equation}
v = \sum_{i=1}^n \xi_i v_i;\\
w = \sum_{i=1}^n \lambda_i\xi_i v_i.
\end{equation}
It follows that
\begin{equation}
\tag{A0}\label{A0}
\langle w,v\rangle = \sum_{i=1}^n \lambda_i\lVert\xi_i v_i\rVert^2 \leq 0.
\end{equation}
From the top blocks of $X$ we have
\begin{equation}
\tag{A1}\label{A1}
(A-\epsilon I)v + w = \lambda v.
\end{equation}
By testing \eqref{A1} against $v$, using \eqref{A0} and the symmetry of $A$ we have
\begin{align}
\tag{A2}\label{A2}
\lambda\lVert v\rVert^2 = \langle w,v\rangle + \langle (A - \epsilon I)v,v\rangle &\leq \langle (A - \epsilon I)v,v\rangle \leq \rho(A - \epsilon I)\lVert v\rVert^2\\
&\leq (1-\epsilon) \lVert v\rVert^2.
\end{align}
Because $v_A \neq 0$, for sufficiently small $\epsilon$, by continuity, also $\lVert v\rVert \neq 0$, so we cancel $\lVert v\rVert$ on both sides of \eqref{A2}, which implies that
\begin{equation}
\lambda \leq 1-\epsilon
\end{equation}
when $\epsilon$ is sufficiently small.
A: Clearly all eigenvalues apart from the eigenvalue $\lambda(\epsilon)$ with $\lambda(0) = 1$ stay in the open unit disk for $\epsilon$ sufficiently small. To see what happens to the last eigenvalue, use eigenvalue first-order perturbation theory, for instance Theorem 1 in Greenbaum, Li, and Overton - First-order Perturbation Theory for Eigenvalues and Eigenvectors: if $v$ and $z^*$ are the right and left eigenvalue of $A$ associated to $\lambda(0)=1$, then the right and left eigenvectors of the $2\times 2$ matrix $X$ for $\epsilon=0$ are $\begin{bmatrix}v\\0\end{bmatrix}$ and $\begin{bmatrix}z^* & z^*(I-B)^{-1}\end{bmatrix}$ respectively, and plugging these expressions into the theorem one gets
$$
\frac{d\lambda(\epsilon)}{d\epsilon} = \frac{1}{z^*v} \begin{bmatrix}z^* & z^*(I-B)^{-1}\end{bmatrix}\begin{bmatrix}-I & 0\\-C & 0\end{bmatrix} \begin{bmatrix}v\\0\end{bmatrix} = -1 - \frac{1}{z^*v} z^*(I-B)^{-1}Cv.
$$
If this derivative is negative, then for a sufficiently small $\epsilon > 0$ we have that $\lambda(\epsilon)$ is in the unit disk.
However, it is possible to choose $z,v,B$ such that $z^*(I-B)^{-1}v < 0$, and $C = \alpha I$; then for sufficiently large $\alpha > 0$ the derivative is larger than $0$ and I don't think that the result holds: $X(\epsilon)$ does not have eigenvalues in the unit disk for a sufficiently small $\epsilon > 0$.
For instance, $A = \begin{bmatrix}1 & 8 \\ 0 & 0.9\end{bmatrix}$, $B = \begin{bmatrix}1/4 & -1/4\\ -1/4 & 1/4\end{bmatrix}$, $C=I$ numerically gives me matrices $X$ with eigenvalues larger than $1$ for small $\epsilon$.
