Nilpotence of a linear endomorphism

Question

Let $F$ be a field, $V$ a finite-dimensional $F$-vector space and $f \in \operatorname{End}_{F}(V)$.

Show: If $\mathbb{F}=\mathbb {C}$ and there is no one-dimensional subspace $U$ of $V$ with $ f(U)=U$, then $f$ nilpotent.

Proof: We have to show that there exists an $n \in \mathbb{N}$ such that $f^n = 0$.

Let $v \in U$ be an eigenvector to the eigenwert $\lambda \in \mathbb{F}$. Since $f(U)=U$ we know that $U$ is invariant and $f^n(v)=\lambda^n v \ \forall v \in U$

Now, how to go from here? Is it true that $f \neq 0$ because of $dim(U) \neq 1$?

Thanks!

Answer 1

An endomorphism of a finite dimensional complex vector space is nilpotent if and only if it has no nonzero eigenvalues. If you accept this general fact, then the claim becomes obvious, since an eigenvector with nonzero eigenvalue spans a subspace $U$ with $f(U)=U$.