Prove that the only nilpotent operator is 0?

Question

I need to prove that if $\phi:V \to V$ is nilpotent, then its only eigenvalue is $0$.

I know how to prove that this for a nilpotent matrix, but I'm not sure in the case of an operator. How would I be able to relate $\phi$ to a matrix?

The $f$ you defined is not nilpotent (it is of order $2$). – Clément Guérin Mar 23 '15 at 15:02 — Clément Guérin, Mar 23 '15 at 15:02
Thanks, just realized i wasn't looking at the right thing – Max Mar 23 '15 at 15:08 — Max, Mar 23 '15 at 15:08

score 1 · Answer 1 · answered Mar 23 '15 at 15:03

1

$\phi$ is nilpotent iff here is $n$ such that $\phi^{n} = 0$.

Thus $\phi$ is a root of the polynomial $x^{n}$.

It follows that its minimal polynomial is a divisor of $x^{n}$, hence of the form $x^{k}$, for some $k > 0$.

Thus $0$ is the only eigenvalue.

answered Mar 23 '15 at 15:03

Andreas Caranti

1 Answers1