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}?


2 Answers 2


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.

  • $\begingroup$ Why is $(\lambda I - B)^{-1}C$ symmetric? $\endgroup$ Jan 18 at 10:19
  • $\begingroup$ @FedericoPoloni Because $(\lambda I - B)^{-1}$ and $C$ are (i) both symmetric (since $B$ and $C$ are symmetric by assumptions) and (ii) they commute. $\endgroup$
    – MathMax
    Jan 18 at 14:47
  • $\begingroup$ @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. $\endgroup$
    – MathMax
    Jan 18 at 15:05
  • $\begingroup$ 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. :) $\endgroup$ Jan 18 at 15:07
  • 1
    $\begingroup$ @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. $\endgroup$
    – MathMax
    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 $\epsilon$.


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