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Math Help - Hard eigenvector/eigenvalue proof

  1. #1
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    Hard eigenvector/eigenvalue proof

    Hard eigenvector/eigenvalue proof-eigenvectors.png

    Struggling with this:

    I started by saying A is invertible if and only if Ax = 0 has a trivial solution.

    So there exists x = 0 where Ax = 0

    0 = lamda. 0 = 0

    I'm not quite sure this proves that lamda is not equal to zero though?!
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  2. #2
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    \displaystyle Ax=\lambda x

    Multiple both sides by A inverse

    \displaystyle A^{-1}Ax=A^{-1}\lambda x\rightarrow x=A^{-1}\lambda x\rightarrow \frac{1}{\lambda}x=A^{-1}x
    Last edited by dwsmith; November 25th 2010 at 05:51 PM.
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  3. #3
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    Quote Originally Posted by chr91 View Post
    Click image for larger version. 

Name:	eigenvectors.png 
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    Struggling with this:

    I started by saying A is invertible if and only if Ax = 0 has a trivial solution.
    You've pretty much got everything backwards here. Ax= 0 for x= 0 for any A. What is true is that A is invertible if and only if there is NO non-trivial solution. That is if x= 0 is the only x such that Ax= 0.

    Saying that \lambda is an eigenvalue of A means there exist non-zero x such that Ax= \lambda x. Use the fact that A is invertible to multiply both sides by A^{-1}.

    So there exists x = 0 where Ax = 0

    0 = lamda. 0 = 0

    I'm not quite sure this proves that lamda is not equal to zero though?!
    Last edited by HallsofIvy; November 26th 2010 at 06:26 AM.
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  4. #4
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