Does either each entry OR the trace of the resolvent of a real matrix map the upper half plane to itself? If so, how to prove it?

51 Views Asked by At

Let $M \in \mathbb{R}^{d \times d}$ be a real matrix. Let $\sigma(M)$ denote the set of eigenvalues of $M$. Let $R_M:\mathbb{C} \setminus \sigma(M) \to \mathbb{C}^{d \times d} $ be the function $R_M(z):= (M-zI)^{-1} \in \mathbb{C}^{d \times d} $ denote the resolvent of the matrix $M$.

Then is it true that:

(1) If the imaginary part of $z$ is positve, i.e. $I(z) > 0$, then must each entry of $I(R_M(z)) > 0$, i.e. be positive as well?

(2) If the above is NOT true, then can we at least have (somewhat weakly):

If $I(z) > 0$, then $I(trace(R_M(z))) > 0$?

It's easy to see in dimension $d=1$, but how do we prove it in higher dimensions? Thanks!