If the trace of all positive powers of a $n \times n$ complex matrix is $0$, then the matrix must be nilpotent. <https://math.stackexchange.com/questions/159167/traces-of-all-positive-powers-of-a-matrix-are-zero-implies-it-is-nilpotent>


Is a similar conclusion known to be valid in the setting of finite von Neumann algebras? Let $\tau$ be a faithful normal tracial state on a finite von Neumann algebra $\mathscr{R}$. Let $A \in \mathscr{R}$ be such that $\tau(A^n) = 0$ for $n \in \mathbb{N}$. Does it imply that $A$ is quasinilpotent? Any well-known counter-examples?



Quasinilpotent operator in finite von Neumann algebra
The answer is negative. Consider the algebra of essentially bounded functions on the unit circle with the trace induced by the Lebesgue measure. Then the identity function $z \mapsto z$ is a counterexample. Indeed, it is a unitary element, so its spectrum is contained in the unit circle; it does not even contain zero.


