2
$\begingroup$

Let $H$ be a real, invertible and positive semi-definite matrix, in the sense that its symmetric part $S$ is positive semi-definite. Consider the matrix $$ G = (I+\alpha H_d)H $$ for some $\alpha > 0$, where $H_d$ is the diagonal part of $H$ (obtained by setting all off-diagonal entries of $H$ to $0$). Prove or disprove that $G$ is stable for small enough $\alpha$, in the sense that its eigenvalues have non-negative real part.

I can neither prove this nor find a counter-example. It is definitely true for $2 \times 2$ matrices, and I think also for $3 \times 3$. In general, there are two special cases worth mentioning:

  • If $H$ has no pure imaginary eigenvalues then its eigenvalues have positive real part, and so does $G$ for small enough $\alpha$. This holds in particular if $H$ is symmetric.
  • If $H$ is antisymmetric then $H_d = 0$ so $G = H$, which has pure imaginary eigenvalues, with zero real part as required.

Note that $H$ is stable and $(I+\alpha H_d)$ is positive definite. Ideally I would have liked to use Sylvester's Law of Inertia to conclude, as suggested here, but neither $H$ nor $G$ are symmetric. I also tried some other sufficient conditions for stability like diagonal dominance, but this does not hold in general.

$\endgroup$
  • $\begingroup$ Cross-posted to MO $\endgroup$ – Robert Israel Jul 1 '18 at 17:28
  • $\begingroup$ I'm quite new here - is that not okay? Since it's both a research and general question, I thought people from both communities might find it interesting. $\endgroup$ – Nao Jul 1 '18 at 17:30
  • 2
    $\begingroup$ It's OK if you tell both communities that you've cross-posted, including a link to the posting on the other site. See e.g. this post on meta $\endgroup$ – Robert Israel Jul 1 '18 at 21:48
  • $\begingroup$ @RobertIsrael ...and if you do it only after not receiving an answer for a few days on the first website. $\endgroup$ – Federico Poloni Jul 2 '18 at 7:05
2
$\begingroup$

Note that $G$ is similar to the matrix $$ (I + \alpha H_d)^{-1/2}G(I + \alpha H_d)^{1/2} = (I + \alpha H_d)^{1/2}H(I + \alpha H_d)^{1/2} $$ Because $H$ is positive semidefinite, any matrix of the form $MHM^T$ is positive semidefinite, and we can take $M = (I + \alpha H_d)^{1/2}$ in particular. Because $G$ is similar to a positive semidefinite matrix, it must be stable.


I see no justification that the matrix $G'$ should generally be positive stable. In fact, if $H$ fails to be invertible, then $G$ cannot be positive stable since it also fails to be invertible.

If $G$ is positive stable, then we can certainly choose $\alpha$ sufficiently small to make $G'$ positive stable since we have $\lim_{\alpha \to 0^+}G'(\alpha) = G$.

The trick we applied before doesn't work unless we happen to know that $H$ is symmetric. If $H$ isn't symmetric, then the matrix $(I + \alpha(H_d - H))$ also fails to be symmetric.

$\endgroup$
  • $\begingroup$ Thanks for your perfect reply. I edited my question to ask one further thing before I accept your answer! $\endgroup$ – Nao Jul 1 '18 at 17:06
  • $\begingroup$ @Nao I've added a bit addressing your edit. In the future, however: if you're going to add something to your original question, it would be best to create a separate post with that followup question (you may want to include a link to the original question). $\endgroup$ – Omnomnomnom Jul 1 '18 at 17:53
  • $\begingroup$ Thanks for your comment, and I will avoid adding an edit next time. I will post a separate question now to see whether someone can find a counter-example where $G'$ is not positive stable... I have tried hard to find one. Note that $H$ is assumed invertible, by the way. $\endgroup$ – Nao Jul 1 '18 at 17:56
  • $\begingroup$ @Nao I missed that bit $\endgroup$ – Omnomnomnom Jul 1 '18 at 17:57
  • $\begingroup$ @Nao have you tried generating random matrices $H$? $\endgroup$ – Omnomnomnom Jul 1 '18 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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