**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.



On the self-adjoint part of a quasinilpotent operator
I think the Volterra operator is a counterexample. Consider $L^2[0,1]$ and define
$$
(Vf)(x) = \int\_0^x f(t) dt.
$$
A calculation shows that 
$$
(V^\*f)(x) = \int\_x^1 f(t) dt.
$$
So $V+V^\*$ is positive, since it is the orthogonal projection onto the constant functions. But $V$ is well-known to be quasinilpotent, since it has no eigenvalues but can be shown to be Hilbert-Schmidt (hence compact).


