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Nilpotent operator or not given characteristic polynomial?

  1. Mar 4, 2012 #1
    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:

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

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


    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.
     
  2. jcsd
  3. Mar 5, 2012 #2

    morphism

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    Are you familiar with the Cayley-Hamilton theorem?
     
  4. Mar 5, 2012 #3
    Yeah - I hadn't thought about using that for a proof, but it would show that
    [itex] (-1)^{n}T^n = 0 \Rightarrow T^n = 0 [/itex]
    so T is nilpotent.

    So what does this part (b) mean? [itex] T^n = 0 [/itex] 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?
     
  5. Mar 5, 2012 #4

    morphism

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    You're not missing anything: the field F doesn't affect this argument at all.
     
  6. Mar 6, 2012 #5

    mathwonk

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