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Math Help - linear operator nilpotent

  1. #1
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    linear operator nilpotent

    Hi all, another question from my text is confusing me.

    Question: A linear operator L is nilpotent if some positive power L^k=0. Prove that L is nilpotent iff there is a basis of V such that the matrix of L is upper triangular, with all diagonal entries zero.

    Any ideas? Thanks in advance.

    EDIT: If L is nilpotent, then its eigenvalues would all be zero. If we then write L is upper triangular form then L^k will have \lambda_i^k on the diagonal, thus L^k=0 \implies \lambda_i=0.
    Am I correct here?

    June
    Last edited by Juneu436; September 20th 2011 at 05:10 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: linear operator nilpotent

    One way: the minimal polynomial of L has the form \mu (x)=x^s . Now, apply a well known result about canonical forms.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Re: linear operator nilpotent

    Quote Originally Posted by Juneu436 View Post
    Hi all, another question from my text is confusing me.

    Question: A linear operator L is nilpotent if some positive power L^k=0. Prove that L is nilpotent iff there is a basis of V such that the matrix of L is upper triangular, with all diagonal entries zero.

    Any ideas? Thanks in advance.

    EDIT: If L is nilpotent, then its eigenvalues would all be zero. If we then write L is upper triangular form then L^k will have \lambda_i^k on the diagonal, thus L^k=0 \implies \lambda_i=0.
    Am I correct here?

    June
    Perhaps a more direct way is--well, the direct way. Namely, since L can be upper triangularized, say L=UTU^{-1} and so L^k=UT^kU^{-1}=0 if and only if T^k=0, but from there I bet you can conclude since the diagonal entries of T^k are \lambda_1^k,\cdots,\lambda_n^k where \lambda_1,\cdots,\lambda_n are the eigenvalues, with multiplicity, of L


    EDIT: And, this is precisely what you said. Missed that. So, correct!
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