# Condition for 3×3 block matrix to be stable

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.

$$\newcommand{\al}{\alpha}\newcommand\la\lambda\newcommand\R{\mathbb R}$$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 $$$$\la_\pm:=\la_\pm(\al,h,m):=\frac{h+2 m-2 \al \pm\sqrt{4 \alpha ^2+h^2-4 \al m}}{2 (h+m)}.$$$$ Taking now any $$\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 $$(-1,1)$$. $$\quad\Box$$

• The answer is now completed "by hand". Commented Apr 26, 2023 at 21:12
• This is really a nice answer. But this sentence may contain a typo: "So, $\lambda$ is a nonzero eigenvalue of $A$ iff $d(\lambda)=1$" should be "$d(\lambda)=0$". . Commented Apr 27, 2023 at 3:21
• This answer finds all such $M,\alpha$ when $M$ and $H$ can be mutually diagonalizable. But for the scenario where $M$ is not a diagonal matrix after $H$ has been diagonalized, there may be choices of $M$ such that $A$ is stable. Nevertheless, @iosif-pinelis solves 99% of my question. Thank you! Commented Apr 27, 2023 at 3:30
• @Zishuo : Your request was to "design a symmetric positive definite matrix $M\in\mathbb{R}^{n\times n}$ and positive scalar $\alpha$ such that" -- emphasis mine. Isn't this request completely fulfilled now? Commented Apr 27, 2023 at 3:36