Nilpotent operator or not given characteristic polynomial?

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Discussion Overview

The discussion revolves around the properties of linear operators with a specific characteristic polynomial, particularly whether such operators are nilpotent and how the nature of the ground field affects this property. The scope includes theoretical exploration and mathematical reasoning related to linear algebra.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that any linear transformation with the characteristic polynomial \( p_{T}(t) = (-1)^{n}t^n \) should be nilpotent, suggesting that it would have a block upper or lower triangular form with zeros on the diagonals.
  • Another participant introduces the Cayley-Hamilton theorem, indicating that it implies \( T^n = 0 \), which would suggest that \( T \) is nilpotent regardless of the field.
  • There is a question raised about the relevance of the ground field \( \textbf{F} \) in determining whether the operator is nilpotent, with some participants expressing uncertainty about its impact.
  • A later reply asserts that the nature of the field does not affect the argument regarding nilpotency.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the ground field to the nilpotency of the operator. While some argue that the field does not matter, others question its significance, indicating that the discussion remains unresolved regarding the field's influence.

Contextual Notes

There is uncertainty about the implications of the characteristic polynomial and the potential existence of non-nilpotent transformations that could share the same polynomial form. Additionally, the dependence on the definitions of nilpotency and the characteristics of the field is not fully explored.

fishshoe
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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.
 
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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,
 

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