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Thread: locally nilpotent operator.

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

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



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



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    Remember, T is nilpotent if there exists some $\displaystyle k \in \mathbb{N}$ such that $\displaystyle T^k = 0$. 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 $\displaystyle k \in \mathbb{N}$ such that $\displaystyle T^k = 0$. 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 $\displaystyle x^n$ is the char. polynomial of T then by Cayley-Hamilton, $\displaystyle T^n=0$ 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 $\displaystyle \lambda$ is an eigenvalue of T then $\displaystyle \lambda^k$ is an eigenvalue of $\displaystyle T^k$).
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