TITLE: Prove the operators $T+U$ and $U$ have the same eigenvalues where $T$ is nilpotent
QUESTION [7 upvotes]: Let $V$ be an $n$-dimensional vector space on $\mathbb{C}$, and $T$ a nilpotent operator on $V$. 
Let $U$ be in $L(V)$ s.t. $UT = TU$. Prove that the operators $T+U$ and $U$ have the same eigenvalues. 
So far I know that the characteristic polynomial of $T$ is $x^n$, but I'm having a difficult time moving forward from there. Any help is appreciated.

REPLY [2 votes]: First, I think we all agree that this is obvious for $T=0$, so let's assume that $T\neq0$ for the rest of this post.
First, let's look at the commutativity relation between $T$ and $U$. Since $TU=UT$, if we pick an eigenvector $v$ of $U$ with eigenvalue $\lambda$, what can we say about $UT\left(v\right)$? When looking at this, just remember that $UT\left(v\right)=U\left(T\left(v\right)\right)$, and you should see what's happening when you mess around using the commutativity.
Next, for each eigenvalue $\lambda$ of $U$, let's select an eigenvector $v$. Since $T$ is nilpotent, there exists $n$ so that $T^{n-1}\neq0$ and $T^{n}=0$. So, what can you say about $\left(T+U\right)\left(T^{n-1}v\right)$? This should be one direction of the proof for you.
The other direction shouldn't be all that hard. Just show that $T+U$ commutes with $T$ and see where you can get from there using (modified versions of) the arguments in the previous two paragraphs.