Motivation: I am working on a research problem and have been stuck for a while. I hope someone can help, as it requires only linear algebra. :)

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 always stable and $(I+\alpha H_d)$ is always 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.

  • 1
    $\begingroup$ Cross-posted to MSE $\endgroup$ Jul 1, 2018 at 17:29
  • $\begingroup$ Should I delete this post since it is duplicated in MSE, and more detail is there? $\endgroup$
    – smalldog
    Jul 1, 2018 at 21:16
  • $\begingroup$ Now that it's been answered, please don't delete. $\endgroup$ Jul 2, 2018 at 7:43

1 Answer 1


It's true for all $\alpha \ge 0$. $G$ has the same eigenvalues as $(1+\alpha H_d)^{1/2} H (1+\alpha H_d)^{1/2}$, which is positive semidefinite.


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