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).

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Julien May 4 '13 at 0:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.