Nilpotent operator

Remarks

Example: on $X = \ell^2 \oplus \ell^2$ let $T(x, y) = (0, x)$. Then $T^2(x, y) = T(0, x) = (0, 0)$, so $T$ is nilpotent of index $2$, although its range is infinite-dimensional.

Non-example: the Volterra operator $(Vf)(x) = \int_0^x f(t) \, dt$ on $L^2[0, 1]$ has spectrum $\set{0}$, but $(V^n 1)(x) = x^n / n!$ is not the zero function for any $n$, so $V$ is not nilpotent.

If $T$ is nilpotent and $X \ne \set{0}$, then $\sigma(T) = \set{0}$: for $\lambda \ne 0$ the series $\lambda^{-1} \sum_{k=0}^{n-1} (T/\lambda)^k$ is a finite sum that inverts $\lambda I - T$.