TITLE: Prove that the only eigenvalue of a nilpotent operator is 0?
QUESTION [25 upvotes]: 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 $n \geq 1$.

REPLY [18 votes]: 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 $0$.