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.
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.