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Math Help - nilotent proof

  1. #1
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    nilotent proof

    show that T is nilpotent if and only if 0 is the only eigenvalue of T.

    I am having difficulty finding a way to solve this problem. I first solved it by assuming that T was diagonalizable because T \in L(v) over the Complex. but that is not true because it is not necessarily normal.

    HELP!
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  2. #2
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    Quote Originally Posted by mathisthebestpuzzle View Post
    show that T is nilpotent if and only if 0 is the only eigenvalue of T.

    I am having difficulty finding a way to solve this problem. I first solved it by assuming that T was diagonalizable because T \in L(v) over the Complex. but that is not true because it is not necessarily normal.

    HELP!
    let T be nilpotent, say T^m=0, and \lambda be an eigenvalue of T. so Tv=\lambda v, for some 0 \neq v \in V.

    now we have T^2v= \lambda Tv = \lambda^2 v, \ ... \ , T^m v = \lambda^m v. thus \lambda^m v = 0. but v \neq 0. thus \lambda^m = 0, i.e.

    \lambda= 0. conversely, suppose all eigenvalues of T are 0, and \dim V = n. then by Cayley-Hamilton

    theorem we have T^n=0. \ \ \ \square
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  3. #3
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    hey thank you so much i appreciate that alot.
    i'm stuck on a few more in this section...help is pretty rare in this section i would love more insight.
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