Is the real part of this inverse matrix positive definite?

Question

Consider a real number $x > 0$, a real symmetric matrix ${\bf A} = {\bf A}^\mathrm{T}$, and the imaginary unit $i$. I want to show that the real part of $$(x{\bf I} + i{\bf A})^{-1}$$ is positive definite.

It's probably a one-liner with the right approach but I just can't find it. The matrix inversion lemma (Woodbury) didn't help me. I'm confident that the statement is true for a physical reason and after checking it numerically for a million random ${\bf A}$ and tiny $x$-values.

A more general form of the statement (also backed up numerically) is: the real part of $$({\bf D} + i{\bf A})^{-1}$$ is positive definite where ${\bf D} = \mathrm{diag}_n(x_n)$ with all $x_n > 0$.

Answer 1

HINT: For the first party: $B\colon =(x I + i A)^{-1}$ is normal, and all its eigenvalues have real part $>0$. The real part of $B$ it will have eigenvalues the real part of the eigenvalues of $B$, so it will be positive definite.

But in general, if $M$ is invertible and the real part of $M$ is positive, then the real part of $M^{-1}$ is positive. To see this, use $$\langle M v, v \rangle + \langle v, M v\rangle > 0$$ and take $w \colon = M^{-1} v$. This takes care of the second part too.