Condition for 3×3 block matrix to be stable

Question

Given a square symmetric matrix $H\in\mathbb{R}^{n\times n}$, design a symmetric positive definite matrix $M\in\mathbb{R}^{n\times n}$ and positive scalar $\alpha$ such that the following ${3n\times 3n}$ matrix is Schur stable (all eigenvalues in open unit disk):

$$ A=\begin{bmatrix} I & 0&-\alpha I\\ I&0&0\\ (M+H)^{-1} & 0 &(M+H)^{-1}(M-2\alpha I) \end{bmatrix} $$ where $I$ is the ${n\times n}$ Identity matrix. This problem originates from the stability analysis of a discrete-time linear system. I tried numerical examples and it is easy to find such $M$ and $\alpha$.

I have no idea how to design such $M$ in general or deduce such requirements on $M$.

One thought is that construct a symmetric positive definite $P$ such that $A^\top P A \prec P$. But the freedom of $P$ is so large that I do not know where to start.

Answer 1

$\newcommand{\al}{\alpha}\newcommand\la\lambda\newcommand\R{\mathbb R}$Such a construction of $M$ and $\al$ is always possible.

Indeed, take any complex $\la$. Rearranging columns and rows of the matrix $A-\la I_{3n}$, we see that $\la$ is an eigenvalue of $A$ iff \begin{equation*} D(\la):=\begin{vmatrix} -\la I&I&0\\ 0&(1-\la)I&-\al I \\ 0&B&C-\la I \end{vmatrix}=0, \end{equation*} where $|\cdot|$ denotes the determinant, \begin{equation*} B:=(M+H)^{-1},\quad C:=B(M-2\al I), \end{equation*} $I:=I_n$.

Note that $D(\la)$ is the determinant of a block-triangular matrix, so that \begin{equation*} D(\la)=(-\la)^n \begin{vmatrix} (1-\la)I&-\al I \\ B&C-\la I \end{vmatrix}. \end{equation*} So, $D(1)\ne0$, since $B=(M+H)^{-1}$ is nonsingular.

So, without loss of generality (wlog), $\la\ne1$, and then, by "The general case", \begin{equation*} \begin{aligned} D(\la)&=(-\la)^n(1-\la)^n\,|C-\la I-B((1-\la)I)^{-1}(-\al I)| \\ & =(-\la)^n(1-\la)^n\,|B(M-2\al I)-\la I+\al(1-\la)^{-1}B| \\ & =(-\la)^n(1-\la)^n\,|B|\,|(M-2\al I)-\la(M+H)+\al(1-\la)^{-1}I| \\ & =(-\la)^n(1-\la)^n\,|B|\,d(\la), \end{aligned} \end{equation*} where \begin{equation*} d(\la):=|(1-\la)M-\la H+\al((1-\la)^{-1}-2)I|. \end{equation*}

So, $\la$ is a nonzero eigenvalue of $A$ iff $d(\la)=0$.

By diagonalization, wlog the matrix $H$ is diagonal, with (say) real $h_1,\dots,h_n$ on its diagonal. Letting now $M$ be diagonal as well, with positive real $m_1,\dots,m_n$ on its diagonal, we see that \begin{equation*} d(\la)=\prod_{i=1}^n f_{\al,h_i}(\la,m_i), \end{equation*} where $f_{\al,h}(\la,m):=(1-\la)m-\la h+\al((1-\la)^{-1}-2)$.

For $\la\ne1$, the equation $f_{\al,h}(\la,m)=0$ for $\la$ is equivalent to a quadratic equation, with roots \begin{equation} \la_\pm:=\la_\pm(\al,h,m):=\frac{h+2 m-2 \al \pm\sqrt{4 \alpha ^2+h^2-4 \al m}}{2 (h+m)}. \end{equation} Taking now any $\al\in(\max(0,-h),\infty)$ and then choosing $m=\frac{4\al^2+h^2}{4\al}$, we get $\la_+=\la_-=\frac h{2\al+h}\in(-1,1)$.

So, for any real $\al>\max(0,-h_1,\dots,-h_n)$ we can find positive real $m_1,\dots,m_n$ such that all the roots $\la$ of the equation $d(\la)=0$ are in the interval $(-1,1)$.

Thus, we will have all the eigenvalues of $A$ in the interval $(-1,1)$. $\quad\Box$