Disclaimer: this is not research-level, but I've read some non research-level questions/answers on quasinilpotent operators here, some of them involving renowned users. So I thought I'd give it a try. I've already asked two persons who know functional analysis very well without success. I apologize in advance if a majority thinks this is not suitable for MO. Especially since I have most likely overlooked a trivial argument.
Let $A$ be a quasinilpotent bounded linear operator on a separable Hilbert space $H$, i.e. $\rho(A)=0$.
There is nothing that can be said in general on the spectral radius of $A+A^*$. It could be any nonnegative number, as shown by the first example one can think of: $A$ the $2\times 2$ matrix with $t$ in $(1,2)$ position and $0$ elsewhere.
Well, not entirely true: if $\rho(A+A^*)=0$, it follows easily that $A=0$.
I wonder what happens with the extra assumption that $A+A^*$ is positive instead, i.e. has nonnegative spectrum. In finite dimension, it suffices to consider the trace to conclude that $A=0$. So I believe the same conclusion holds when $A$ is trace-class. But I can't figure out whether:
$$ \rho(A)=0\quad\mbox{and}\quad A+A^*\geq 0\quad\Rightarrow\quad A=0 $$ holds in general, as I can't see in particular how to do the finite-dimensional case without taking the trace.
Thank you.