# On the self-adjoint part of a quasinilpotent operator

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.

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).
• How misleading finite dimension...And I did not even try the most famous quasinilpotent operator. Nicely done, thank you. I still wonder if we can arrange for $A+A^*$ to be positive invertible. But I'll try to do that alone. – Julien May 4 '13 at 0:39