# Nilpotent operator or not given characteristic polynomial?

• fishshoe
In summary: J_{k_m}(0)v_m \end{bmatrix}where v_i is a vector of length k_i.Since J_{k_i}(0) is a nilpotent matrix of size k_i, we can see that after applying T k_i times, each block will become the zero matrix. Therefore, T^k = 0, and thus T is nilpotent.In summary, all linear transformations with characteristic polynomial p_T(t) = (-1)^n t^n are nilpotent, as long as the dimension of the vector space is equal to the number of blocks in the Jordan canonical form, and the blocks are all of the form J_k(0
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.

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_

## 1. What is a nilpotent operator?

A nilpotent operator is a linear operator on a vector space that, when composed with itself enough times, becomes the zero operator. This means that there exists some positive integer n such that the operator raised to the n-th power is equal to the zero operator.

## 2. How can I determine if an operator is nilpotent?

To determine if an operator is nilpotent, you can raise the operator to different powers until you reach the zero operator. If this occurs before reaching the dimension of the vector space, then the operator is nilpotent. Additionally, if the operator's characteristic polynomial has only 0 as a root, then the operator is nilpotent.

## 3. What is the characteristic polynomial of a nilpotent operator?

The characteristic polynomial of a nilpotent operator is always equal to xn, where n is the dimension of the vector space on which the operator acts. This is because all of the eigenvalues of a nilpotent operator are equal to 0.

## 4. Can a non-nilpotent operator have the characteristic polynomial xn?

Yes, a non-nilpotent operator can have the characteristic polynomial xn. This is because the characteristic polynomial only tells us the eigenvalues of an operator, and it is possible for an operator to have all eigenvalues equal to 0 without being nilpotent.

## 5. How is the nilpotency of an operator related to its Jordan form?

In the Jordan form of a matrix, the nilpotent blocks are represented by square matrices with 0's along the main diagonal and 1's along the superdiagonal. Therefore, an operator is nilpotent if and only if its Jordan form consists only of nilpotent blocks. This means that the nilpotency of an operator is directly related to its Jordan form.

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