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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?

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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.

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  • $\begingroup$ Thank you. I feel stupid now for not thinking about Haar unitaries. $\endgroup$ – quasinilpotent Aug 9 '18 at 11:57

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