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Math Help - locally nilpotent operator.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    locally nilpotent operator.

    Let T:V \to V nilpotent linear transformation over some field, let assume that dimV=n.
    Prove that T nilpotent if and only if the characteristic polynomial of T is x^n



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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let T:V \to V nilpotent linear transformation over some field, let assume that dimV=n.
    Prove that T nilpotent if and only if the characteristic polynomial of T is x^n



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    Remember, T is nilpotent if there exists some k \in \mathbb{N} such that T^k = 0. Use this along with the Cayley-Hamilton theorem to prove this fact (the <= side should be trivial).
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Defunkt View Post
    Remember, T is nilpotent if there exists some k \in \mathbb{N} such that T^k = 0. Use this along with the Cayley-Hamilton theorem to prove this fact (the <= side should be trivial).
    No... I thought about it and couldn't do it.
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    If x^n is the char. polynomial of T then by Cayley-Hamilton, T^n=0 therefore T is nilpotent by definition.

    For the other direction, simply prove that T can not have any non-zero eigenvalues (use the fact that if \lambda is an eigenvalue of T then \lambda^k is an eigenvalue of T^k).
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