# Prove or give a counterexample: Over C,$\,$ $0$ is the only eigenvalue $\iff$ $T$ is nilpotent.

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.

(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.

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 $$T^n\mathbf{e}_2=\mathbf{e}_{2+n}\ne\mathbf{0}$$