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How can I conclude from that fact, that $K_0$ and $K_1$ are positive definite diagonal matrices, that

$$ A = \left( \begin{matrix} 0 & I \\ -K_0 & -K_1\end{matrix} \right) $$

is Hurwitz? Here, $I$ is the identity matrix and the dimension of all blocks in $A$ is $n \times n$. Hence, the dimension of $A$ is $2n \times 2n$.

Rodrigo de Azevedo
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Let $b_1,\dots,b_n$ and $c_1,\dots,c_n$ denote the diagonal entries of $K_0$ and $K_1$ respectively. As I explain in my post here, there exists a permutation matrix $P$ such that $$ PAP^T = \pmatrix{A_1 \\ & \ddots \\ && A_n}, $$ where $$ A_k = \pmatrix{0 & 1\\-b_k & -c_k}, \quad k = 1,\dots,n. $$ Thus, $A$ is Hurwitz if and only if each $2 \times 2$ $A_k$ is Hurwitz. We quickly conclude that every $A_k$ must be Hurwitz because the positivity of $b_k,c_k$ ensures that $A_k$ has negative trace and positive determinant.

Alternatively, because the block-entries of $A$ commute, we have $$ \det(A - \lambda I) = \det\pmatrix{-\lambda I & I\\ -K_0 & -K_1 - \lambda I} = \det[(- \lambda I)(-K_1 - \lambda I) - (I)(-K_0)]\\ = \det[\lambda^2 - K_1 \lambda + K_0]\\ = \det \pmatrix{\lambda^2 + c_1 \lambda + b_1\\ & \ddots \\ && \lambda^2 + c_n \lambda + b_n}. $$

Ben Grossmann
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  • @Rodrigo Thank you! – Ben Grossmann Jul 16 '20 at 09:13
  • @RodrigodeAzevedo Yes, because $K_1$ is "positive definite diagonal" – Ben Grossmann Jul 16 '20 at 09:14
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    @RodrigodeAzevedo You are right; I was thinking of the direct formula for the coefficients of the characteristic polynomial and forgot that the trace requires a minus sign. This is also reassuring because quadratic polynomials have roots with negative real part iff their coefficients are positive. – Ben Grossmann Jul 16 '20 at 16:22