TITLE: Quasinilpotent operators satisfying a polynomial identity
QUESTION [9 upvotes]: Suppose $T$ is a bounded quasinilpotent operator (that is, spectrum is zero) on an infinite dimensional Banach space such that $p(T)=0$ for some polynomial $p$. Does it follow that $T$ is actually nilpotent? In the finite dimensional case, quasinilpotent and nilpotent are the same, but not in the infinite dimensional case.

REPLY [2 votes]: Because $p(T)=0$, then the spectral mapping theorem gives
$$\{0\}=\sigma(p(T))=p(\sigma(T))=p(\{0\}).$$
So $p(\lambda)=q(\lambda)\lambda^n$ for some $n > 0$, with $q(0)\ne 0$. Therefore $q(T)$ is invertible because its spectrum does not include $0$. Hence $q(T)T^n=0$ implies $T^n=0$.