Nilpotent operator or not given characteristic polynomial?

[fishshoe]

Homework Statement

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?

Homework Equations

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

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.

 

[mighty2000]

First, let's define what it means for a linear transformation to be nilpotent. A linear transformation T is nilpotent if there exists a positive integer k such that T^k = 0, i.e. the transformation becomes the zero transformation after being applied k times.

Now, let's consider the given characteristic polynomial p_T(t) = (-1)^n t^n. This polynomial has only one root, namely 0. This means that 0 is the only eigenvalue of T over any field \textbf{F}. This fact is independent of the choice of the field, as the characteristic polynomial only depends on the transformation T and not on the field.

(a) Are all such operators nilpotent?

To answer this question, we need to consider the Jordan canonical form of a linear transformation T. The Jordan canonical form is a block diagonal matrix where each block corresponds to a Jordan block, which is of the form:

J_k(\lambda) = \begin{bmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda & 1 \\ 0 & 0 & \cdots & 0 & \lambda \end{bmatrix}

where \lambda is the eigenvalue and k is the size of the block.

Now, since 0 is the only eigenvalue of T, the Jordan canonical form of T will only have blocks corresponding to J_k(0). This means that T will have the form:

J = \begin{bmatrix} J_{k_1}(0) & 0 & \cdots & 0 \\ 0 & J_{k_2}(0) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & J_{k_m}(0) \end{bmatrix}

where k_1 + k_2 + \cdots + k_m = n, the dimension of the vector space V.

Now, if we apply T to a vector v, we get:

T(v) = Jv = \begin{bmatrix} J_{k_1}(0)v_1 \\ J_{k_