Consider the following block matrix: $$ A = \begin{bmatrix} 0 & I\\ -M & -I \end{bmatrix} $$ Suppose matrix $M$ is positive definite and satisfies $M\succeq \alpha I$, where $\alpha>0$ is a constant. When will matrix $A$ be Hurwitz stable, i.e., all of the eigenvalues have negative real parts?
1 Answer
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The eigenvalues of $A$ are $(-1 \pm \sqrt{1-4s})/2$ where $s$ is an eigenvalue of $M$. If $s \ge 1/4$, these have real part $-1/2$, while if $0 < s < 1/4$, they are both real and negative. So it's always true when $M$ is positive definite.
answered Oct 19, 2021 at 23:52
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$\begingroup$ How do you compute the eigenvalues of $A$? $\endgroup$– EvanCommented Oct 20, 2021 at 13:55
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$\begingroup$ Look at $A \pmatrix{u\cr v\cr} - \lambda \pmatrix{u\cr v\cr}$ explicitly. $\endgroup$ Commented Oct 20, 2021 at 14:08