1

Let $G$ be a complex normal matrix (that is, $GG^\dagger = G^\dagger G$) having all eigenvalues with strictly negative real part (i.e. is Hurwitz stable), and let $H$ be a skew-Hermitian matrix with the same dimensions as $G$ that does not necessarily commute with $G$.

I'm seeking to prove that $G + H$ is Hurwitz stable.

I'm able to come up with heuristic arguments based on physical intuition of the system I'm looking at, and I've verified the condition holds in trials where $G$ and $H$ are randomly generated, but a rigorous proof has eluded me.

Any assistance would be much appreciated.

COTO
  • 585

1 Answers1

2

It suffices to note that the matrix $(G + H) + (G + H)^\dagger$ is negative definite, as is explained here (the proof is only given for the real case but can be extended to the complex case).

To see that this is true for our matrix, note that $$ (G + H) + (G + H)^\dagger = G + G^\dagger + H - H = G + G^\dagger $$ Because $G$ is normal, the spectral theorem implies that the eigenvalues of $G + G^\dagger$ will be $\lambda + \bar \lambda$ for all eigenvalues $\lambda$ of $G$. That is, the eigenvalues of $G + G^\dagger$ are $2 \operatorname{Re}\{\lambda\}$ (for each eigenvalue $\lambda$ of $G$). Since these eigenvalues are negative and since $G + G^\dagger$ is Hermitian, we conclude that $G + G^\dagger$ is negative definite, as was desired.

Ben Grossmann
  • 225,327