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Math Help - Eigenvalue problem

  1. #1
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    Eigenvalue problem



    This is my attempt at a solution:

    A^2(x) = A(A(x))
    A^2(x) = lambda^2(x)

    Therefore x is an eigenvector of A^2 with eigenvalue lambda^2. The statement for general N>0 follows in the same way.

    Am I missing anything?
    Last edited by temaire; April 4th 2010 at 05:23 PM.
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  2. #2
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    Quote Originally Posted by temaire View Post


    This is my attempt at a solution:

    A^2(x) = A(A(x))
    A^2(x) = lambda^2(x)

    Therefore x is an eigenvector of A^2 with eigenvalue lambda^2. The statement for general N>0 follows in the same way.

    Am I missing anything?

    Not really: now just round up the proof doing some induction.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Not really: now just round up the proof doing some induction.

    Tonio
    I haven't learned what induction is. Or maybe its a word for something I already know. What do you mean by it?
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  4. #4
    Newbie vincent's Avatar
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    A proof by induction works like this: first show your statement holds for n=1 or 0 or the first n for which it is true, then assume it to be true for n=m (this is your induction hypothesis) and then show it to be true for n=m+1 using your hypothesis. if \lambda is an eigenvalue of  A, Ax=\lambda x and A^2x=A\lambda x =\lambda Ax=\lambda^2 x so \lambda^n is an eigenvalue of A^n for n=2. suppose it is true for n=m,then A^{m+1}x=AA^mx=A\lambda^mx=\lambda^mAx=\lambda^{m+  1}
    it is essentially what you said but a a bit more rigorous
    induction is a really useful tool, you should practice using it
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