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
The eigenvalues of $A$ are $(1 \pm \sqrt{14s})/2$ 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.


$\begingroup$ Look at $A \pmatrix{u\cr v\cr}  \lambda \pmatrix{u\cr v\cr}$ explicitly. $\endgroup$ Oct 20, 2021 at 14:08