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Math Help - Eigenvectors

  1. #1
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    Eigenvectors

    Let x and y be eigenvectors of a matrix A such that is also an eigenvector of A. Is an eigenvector of A? Prove or disprove.

    I know that and and but I'm not sure where I can go from there without being given that x and y correspond to the same eigenvalue... I started with: but I don't know how to proceed. Please help!!


    OP edit after getting help: Thank you--I didn't think of the 2 cases, it becomes easy!
    Last edited by mr fantastic; December 12th 2009 at 04:43 PM. Reason: Restored original question.
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  2. #2
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    Quote Originally Posted by casanova View Post
    Let x and y be eigenvectors of a matrix A such that is also an eigenvector of A. Is an eigenvector of A? Prove or disprove.

    I know that and and but I'm not sure where I can go from there without being given that x and y correspond to the same eigenvalue... I started with: but I don't know how to proceed. Please help!!

    \underline{\mbox{First case}} : both eigenvectors are linearly dependent \Longrightarrow y = \alpha x\,,\,\alpha\in\mathbb{F}= the definition field, but then:

    i) they both have the same eigenvalue \lambda;

    ii) x\pm y=(1\pm \alpha)x\Longrightarrow A(x\pm y)=A((\alpha\pm 1)x)=(\alpha\pm 1)Ax=(\alpha\pm a)\lambda x

    \underline{\mbox{Second case}}: both eigenvectors are lin. independent, with eigenvalues \lambda_1\,,\,\lambda_2 but then:

     \lambda_1 x+\lambda_2 y=Ax+Ay=A(x+y)=\lambda_2(x+y)=\lambda_3x+\lambda_3  y\Longrightarrow (\lambda_1-\lambda_3)x+(\lambda_2-\lambda_3)y=0 \Longrightarrow \lambda_1=\lambda_3=\lambda_2 , by lin. independence...

    Now you just end the argument to show that with the given info, anyway x-y is an eigenvector of A.

    Tonio
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