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Thread: 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 $\displaystyle T \in L(v)$ 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 $\displaystyle T \in L(v)$ over the Complex. but that is not true because it is not necessarily normal.

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

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

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

    theorem we have $\displaystyle T^n=0. \ \ \ \square$
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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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