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.



Condition for 3×3 block matrix to be stable
$\newcommand{\al}{\alpha}\newcommand\la\lambda\newcommand\R{\mathbb R}$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"](https://www.statlect.com/matrix-algebra/determinant-of-block-matrix),
\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 $(-1,1)$. $\quad\Box$


