3
$\begingroup$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

$\newcommand{\al}{\alpha}\newcommand\la\lambda\newcommand\R{\mathbb R}$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 $(-1,1)$. $\quad\Box$

$\endgroup$
7
  • $\begingroup$ The answer is now completed "by hand". $\endgroup$ Commented Apr 26, 2023 at 21:12
  • $\begingroup$ 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 "$d(\lambda)=0$". . $\endgroup$
    – Zishuo
    Commented Apr 27, 2023 at 3:21
  • $\begingroup$ @Zishuo : Thank you for your comment. This typo is now fixed. Do you have other concerns about this answer? $\endgroup$ Commented Apr 27, 2023 at 3:27
  • $\begingroup$ 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! $\endgroup$
    – Zishuo
    Commented Apr 27, 2023 at 3:30
  • $\begingroup$ @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? $\endgroup$ Commented Apr 27, 2023 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.