# 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}?

• Please do not cross-post simultaneously. If you do not get a satisfactory answer on MSE in a week then you can consider posting here. May 19, 2021 at 16:43
• May 23, 2021 at 23:05

## 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 $$0 be the maximum modulus of all eigenvalues of $$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.

• Why is $(\lambda I - B)^{-1}C$ symmetric? Jan 18 at 10:19
• @FedericoPoloni Because $(\lambda I - B)^{-1}$ and $C$ are (i) both symmetric (since $B$ and $C$ are symmetric by assumptions) and (ii) they commute. Jan 18 at 14:47
• @FedericoPoloni The purpose of my post was to give a partial answer, by addressing the special case when B is symmetric (and also A as per my latest edit). Hope this is sufficient to revert the downvote. Jan 18 at 15:05
• I am still trying to figure out why exactly, but one of us must be wrong because my answer contains a counterexample with $C$ symmetric positive definite and $B$ symmetric. If you can solve the dilemma let me know. :) Jan 18 at 15:07
• @FedericoPoloni Exactly, I suspected that the lack of symmetry could trigger counterexamples. In fact, non-symmetric matrices can get arbitrarily far from being positive definite even if all their eigenvalues are positive, thereby jeopardising my line of proof. Your nice counterexample is exactly what I was looking for. Jan 18 at 15:16

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 $$\epsilon$$.