I need to prove that:
if a linear operator $\phi : V \rightarrow V$ on a vector space 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?
Note: A nilpotent operator $\phi$ has been defined as an operator that satisfies $\phi^{n} = 0$ for some $n \geq 1$.
$\phi$ is nilpotent, so $\phi^n = 0$ for some $n$. Now let $v$ be an eigenvector: $\phi v = \lambda v$ for some scalar $\lambda$. Now we get $$ 0 = \phi^n v = \lambda^n v ~\Rightarrow~ \lambda=0 ~. $$
Note that this works in the infinite-dimensional case as well; there is no need to relate $\phi$ to a matrix.
Edit: As suggested in the comments, we can also show that $0$ is always an eigenvalue; in other words, $\phi$ always has at least one eigenvector. Take any $v \neq 0$; we know that $\phi^n v = 0$, so let $k$ be the largest integer such that $\phi^k v \neq 0$. Then $\phi(\phi^k v) = 0$, so $\phi^k v$ is an eigenvector of $\phi$, with eigenvalue $0$.
Suppose that $\phi$ has another eigenvalue $\lambda \ne 0$ so that $\phi(x)=\lambda x $ ($x$ is an eigenvector corresponding to $\lambda$)
Then, $ \phi^n (x)= \phi^{(n-1)}(\phi(x))=\phi^{(n-1)}(\lambda x)=\lambda \phi^{n-1}(x)=\cdots=\lambda ^{n}x\ne 0$.
We have a contradiction, so $\phi$ can't have another eigenvalue except $0$.
