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Math Help - linear operator and eigenvector

  1. #1
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    linear operator and eigenvector

    Hey guys, I am stuck on this question so any help would be really nice. I have read it over and over and can't get it.

    linear operator and eigenvector-text.q.12.png

    Thanks
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  2. #2
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    Re: linear operator and eigenvector

    Quote Originally Posted by shimara View Post
    Click image for larger version. 

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    (a) By hypothesis AX=\lambda X and B^tY=\mu Y . Choose M=XY^t then,

    T(M)=AMB=A(XY^t)B=(AX)(B^tY)^t=

    (\lambda X)(\mu Y)=\lambda \mu (XY^t)=\lambda \mu M

    As M\neq 0 (why?) then, M is an eigenvector of T .

    (b) Try it.
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    Re: linear operator and eigenvector

    Thanks fernando, for part a I don't understand why M \ne 0.

    For part b, aren't the eigenvalues of A and B, \lambda and \mu respectively?
    And these are also the eigenvalues of T?
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  4. #4
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    Re: linear operator and eigenvector

    show that M can only be 0 if either X or Y is 0, since M = XY^t. (that is, the entry mij of M is (xi)(yj)).

    but neither X nor Y can be 0, because they are ____________s.

    λ can't be an eigenvalue of T, unless 1 is an eigenvaule of B^t. similar reasoning applies for μ.

    T takes M and "combines A and B (via M)". you should expect the eigenvalues to "combine" in a similar way.
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    Re: linear operator and eigenvector

    Thanks deveno, X and Y are non-zero because they are eigenvectors.

    Quote Originally Posted by Deveno View Post
    T takes M and "combines A and B (via M)". you should expect the eigenvalues to "combine" in a similar way.
    Sorry deveno, I don't really get this part. I know the eigenvalues will be combined like you say, but I don't understand how to find them.
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  6. #6
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    Re: linear operator and eigenvector

    Fernando showed you that M = XY^t is an eigenvector for T, with eigenvalue λμ. the problem is already telling you the eigenvalues for A and B are given (the λ's and the μ's). so...what are the eigenvalues of T? (you're over-thinking this, the answer isn't that complicated).
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    Re: linear operator and eigenvector

    Quote Originally Posted by Deveno View Post
    Fernando showed you that M = XY^t is an eigenvector for T, with eigenvalue λμ. the problem is already telling you the eigenvalues for A and B are given (the λ's and the μ's). so...what are the eigenvalues of T? (you're over-thinking this, the answer isn't that complicated).

    Yeah, I am definitely over-thinking this, all I can stupidly think of at this time of day is \lambda \mu. Can you just put me out of my misery and tell me the answer?

    And thanks for being patient with my stupidity at the moment.
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  8. #8
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    Re: linear operator and eigenvector

    that's just "one" eigenvalue. might not A and B^t have a bunch of different eigenvalues?
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    Re: linear operator and eigenvector

    Quote Originally Posted by Deveno View Post
    that's just "one" eigenvalue. might not A and B^t have a bunch of different eigenvalues?
    Because A and B are general complex matrices they can have many different eigenvalues, so \lambda \mu is only one of many eigenvalues of T? This confuses me more.
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  10. #10
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    Re: linear operator and eigenvector

    you want to consider all possible products λμ, where λ is an eigenvalue for A, and μ is an eigenvalue for B (B and B^t have the same eigenvalues).
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    Re: linear operator and eigenvector

    Quote Originally Posted by Deveno View Post
    you want to consider all possible products λμ, where λ is an eigenvalue for A, and μ is an eigenvalue for B (B and B^t have the same eigenvalues).
    omg, you were right, it is really simple and I was way over thinking it. Thanks deveno.
    So I would write it like:
    the eigenvalues of T are \lambda_i \mu_j for i,j \in R.
    I will make sure to go over this and fully comprehend all your wonderful help!
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  12. #12
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    Re: linear operator and eigenvector

    i'm not sure what the set "R" is. it would be better to just say: where \lambda_i is an eigenvalue of A, and \mu_j is am eigenvalue for B.
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    Re: linear operator and eigenvector

    oh ok, thanks again deveno.
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