# Nilpotent operator or not given characteristic polynomial?

1. Mar 4, 2012

### fishshoe

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

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

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.