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Suppose $V$ is a complex vector space and $T$ is a linear map on $V$.

Prove or give a counterexample: $\ $ $0$ is the only eigenvalue of $T$ $\iff$ $T$ is nilpotent.

Two things already known :

(1) $\ $ If $T$ is nilpotent, then $0$ is the only eigenvalue of $T$.

(2) $\ $ On finite-dimensional complex vector spaces, if $0$ is the only eigenvalue of $T$, then $T$ is nilpotent.

So we only need to consider one direction. The question is equivalent to:

On infinite-dimensional complex vector spaces, $\,$ $0$ is the only eigenvalue $\,$ $\Longrightarrow$ $\,$$T$ is nilpotent $\ ?$

Any insights are much appreciated.

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Consider the operator $T:(x_i)_{i\in\mathbb{N}}\mapsto(0,0,x_2,x_3,\ldots)$, that is $T\mathbf{x}=R\mathbf{x}-x_1\mathbf{e}_2$, where $R$ is the right-shift operator.

Then $0$ is an eigenvalue since $T\mathbf{e}_1=\mathbf{0}$.
But there are no other eigenvalues (as in the proof that $R$ has no eigenvalues). Moreover, $T$ is not nilpotent since $T^n\mathbf{e}_2=\mathbf{e}_{2+n}\ne\mathbf{0}$

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  • $\begingroup$ This operator is very clever! Thanks a lot :‑) $\endgroup$
    – Sky subO
    Mar 10 at 16:02

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