Nilpotent operator or not given characteristic polynomial?

fishshoe

Hey, I'm working on a proof for a research-related assignment. I posted it under homework, but it's a little abstract and I was hoping someone on this forum might have some advice:

1. The problem statement, all variables and given/known data
Suppose [itex]T:V \rightarrow V [/itex] has characteristic polynomial [itex] p_{T}(t) = (-1)^{n}t^n[/itex].
(a) Are all such operators nilpotent? Prove or give a counterexample.
(b) Does the nature of the ground field [itex]\textbf{F}[/itex] matter in answering this question?

2. Relevant equations
Nilpotent operators have a characteristic polynomial of the form in the problem statement, and [itex]\lambda=0[/itex] is the only eigenvalue over any field [itex]\textbf{F}[/itex].


3. The attempt at a solution
I originally thought that any linear transformation with the given characteristic polynomial would therefore have a block upper or lower triangular form with zeros on the diagonals, and therefore be nilpotent. But I'm confused by part (b), and the more I think about it, I'm not sure how to rule out that another more complex matrix representation of a non-nilpotent transformation might have the same form. And I have no idea how the choice of the field affects it. The very fact that they asked part (b) makes me think it does depend on the field, but I can't figure out why.

morphism

Are you familiar with the Cayley-Hamilton theorem?

fishshoe

Yeah - I hadn't thought about using that for a proof, but it would show that
[itex] (-1)^{n}T^n = 0 \Rightarrow T^n = 0 [/itex]
so T is nilpotent.

So what does this part (b) mean? [itex] T^n = 0 [/itex] means T is nilpotent no matter if the field is complex or reals, right? Is it just a mean-hearted distraction? Am I missing something here?

morphism

You're not missing anything: the field F doesn't affect this argument at all.

mathwonk

so a teacher asking us to know what we are talking about is mean spirited? ouch,