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Math Help - Eigenvalue/Eigenvector proof.

  1. #1
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    Eigenvalue/Eigenvector proof.

    Suppose that the 2x2 matrix A has only one eigenvalue λ with eigenvector v, and that w is a non zero vector which is not an eigenvector..show that:

    a) v and w are linearly independent
    b) the matrix with respect to the basis {v, w} is
    (λ c
    0 λ) for some c =not to 0
    c) for a suitable choice of w, c = 1


    I am stuck.
    I know how to show that eigenvectors are linearly independent, but how do I show that these two vectors are linearly independent to eachother?
    as for b and c i dont know where to start! Please help!
    Last edited by nlews; January 31st 2010 at 02:37 PM.
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  2. #2
    Super Member girdav's Avatar
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    If E_{\lambda} is the eigenspace corresponding to the eigenvalue \lambda then we can write <br />
E = E_{\lambda}\oplus F.
    What can you say if v and w are linearly dependent?
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  3. #3
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    I'm not sure what that sign means =/
    but if they are dependent then the coefficients in this equation αv + βw = 0 will not be equal to zero. Im not sure how that helps. Sorry!
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  4. #4
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    Quote Originally Posted by nlews View Post
    Suppose that the 2x2 matrix A has only one eigenvalue λ with eigenvector v, and that w is a non zero vector which is not an eigenvector..show that:

    a) v and w are linearly independent
    b) the matrix with respect to the basis {v, w} is
    (λ c
    0 λ) for some c =not to 0
    c) for a suitable choice of w, c = 1


    I am stuck.
    I know how to show that the eigenvalues are linearly independent,
    Unfortunately, I suspect you do NOT know that- it doesn't even makes sense. eigenvalues are numbers and "independent" is only defined for vectors.

    but how do I show that these two vectors are linearly independent to eachother?
    Use the definition of "independent". Suppose av+ bw= 0. Applying A to both sides, a(Av)+ b(Aw)= 3av+ b(Aw)= 0. If a were 0, then either Aw= 0 which means that w is an eigenvector with eigenvalue 0, which by hypothesis is not true, or a= 0. If b were 0, that says 3av= 0 so that v= 0, not an eigenvector. If both are non-zero, we would have w= -(a/b)v so that Aw= -(a/b)Av= -(3a/b)v, again contradicting the fact that w is not an eigenvector.

    as for b and c i dont know where to start! Please help!
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  5. #5
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    sorry yes I obviously meant eigenvectors not eigenvalues. I just meant that I knew the standard proof that eigenvectors are linearly independent if they belong to distinct eigenvalues, and I couldn't see how knowing that would help me here.
    However thank you for your help, that makes sense to me! I had sort of a similar thing but the logic wasn't quite there because I hadn't applied A so thank you! part a done!
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