# Nilpotent operator or not given characteristic polynomial?

• fishshoe
In summary, the homework statement says that all linear operators with a characteristic polynomial of the form in the problem statement are nilpotent. However, the nature of the ground field doesn't affect this argument.
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:

## 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.

Are you familiar with the Cayley-Hamilton theorem?

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

So what does this part (b) mean? $T^n = 0$ 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?

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

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

## 1. What is a nilpotent operator?

A nilpotent operator is a linear transformation on a vector space that has the property that there exists a positive integer n such that the nth power of the operator is the zero operator.

## 2. How can you determine if an operator is nilpotent or not?

To determine if an operator is nilpotent, you can calculate its characteristic polynomial and check if it has a root of multiplicity greater than or equal to the dimension of the vector space it is acting on. If this is true, then the operator is nilpotent.

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

The characteristic polynomial of a nilpotent operator is a polynomial of degree n, where n is the dimension of the vector space it is acting on, and its roots are all 0.

## 4. Can a non-zero operator have a characteristic polynomial with all roots equal to 0?

Yes, it is possible for a non-zero operator to have a characteristic polynomial with all roots equal to 0. This is because the roots of the characteristic polynomial do not necessarily correspond to the eigenvalues of the operator.

## 5. What is the relationship between the nilpotency of an operator and its characteristic polynomial?

The nilpotency of an operator is closely related to its characteristic polynomial. A nilpotent operator has a characteristic polynomial with all roots equal to 0, while an operator with a characteristic polynomial with all roots equal to 0 is not necessarily nilpotent.

• Linear and Abstract Algebra
Replies
8
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
3
Views
1K
• Linear and Abstract Algebra
Replies
5
Views
2K
• Linear and Abstract Algebra
Replies
5
Views
1K
• Linear and Abstract Algebra
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Linear and Abstract Algebra
Replies
3
Views
2K
• Linear and Abstract Algebra
Replies
1
Views
2K
• Linear and Abstract Algebra
Replies
10
Views
1K