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Math Help - Eigenvectors of recursive set of similarity transformations

  1. #1
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    Eigenvectors of recursive set of similarity transformations

    Hi!

    I need some help with the following:
    Consider the sequence
    A_{1} = A \\<br />
A_{2} = P ^{-1}_{1} A_{1} P_{1} \\<br />
A_{3} = P ^{-1}_{2} A_{2} P_{2}  \\<br />
.  \\<br />
. \\<br />
. ... \\ <br />
A_{k+1} = P ^{-1}_{k} A_{k} P_{k} .

    1) Write an expression for the eigenvectors of A_{k+1} using the eigenvectors of A_{k}. Then use this to write an expression for the eigenvectors of A_{k} using the matrices P_{1},..., P_{k-1}, and the eigenvectors of A_{1} = A.

    I am completely lost and have no idea where to start. Any pointing me in the right direction would be greatly appreciated!

    j

    edit: I can't get line breaks to work in the above code...(?)
    Last edited by Jodles; October 27th 2009 at 02:00 AM. Reason: troubles with linebreaks in latex code
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  2. #2
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    Quote Originally Posted by Jodles View Post
    Hi!

    I need some help with the following:
    Consider the sequence
    A_{1} = A \\<br />
A_{2} = P ^{-1}_{1} A_{1} P_{1} \\<br />
A_{3} = P ^{-1}_{2} A_{2} P_{2} \\<br />
. \\<br />
. \\<br />
. ... \\ <br />
A_{k+1} = P ^{-1}_{k} A_{k} P_{k} .

    1) Write an expression for the eigenvectors of A_{k+1} using the eigenvectors of A_{k}. Then use this to write an expression for the eigenvectors of A_{k} using the matrices P_{1},..., P_{k-1}, and the eigenvectors of A_{1} = A.

    I am completely lost and have no idea where to start. Any pointing me in the right direction would be greatly appreciated!

    j

    edit: I can't get line breaks to work in the above code...(?)

    \mbox {If v is an eigenvector of } A_k \mbox{ with eigenvalue }\lambda \,,\,\, then\; A_kv = \lambda v. \mbox { As P is invertible } \exists u\in V\; s.t.\; Pu=v , and so:

    A_{k+1}(u)=P_k^{-1}A_kP_k(u)=P_k^{-1}A_k(v)=P_k^{-1}(\lambda v)=\lambda P_k^{-1}v=\lambda u

    Try to take it from here now.

    Tonio
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  3. #3
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    Tonio, thank you! You saved my day!

    Here's my take on the second part ("Write an expression for the eigenvectors of A_{k} using the matrices P_{1},..., P_{k-1}, and the eigenvectors of A_{1} = A).

    (I'm using v_k and v_{k+1} instead of v and u, respectively.)

    From the first part we now have: v_{k+1} = P_k^{-1} v_k. Obviously v_k would be v_k = P_{k-1}^{-1} v_{k-1} (1) ,
    and further:
    v_{k-1} = P_{k-2}^{-1} v_{k-2}
    v_{k-2} = P_{k-3}^{-1} v_{k-3}
    ... and so on...

    Inserted into (1), we get:
    v_k = P_{k-1}^{-1} P_{k-2}^{-1}P_{k-3}^{-1} v_{k-3}

    Thus a pattern evolves:
    v_k = P_{k-1}^{-1} \cdot P_{k-2}^{-1} \cdot ... \cdot P_{1}^{-1} v_{1} (2)
    where v_1 is the eigenvector of A_1 = A, i.e. Av_1 = \lambda v_1

    Does that look somewhat alright? Something I've missed? Or better ways to write it? (especially (2))...

    J
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  4. #4
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    Quote Originally Posted by Jodles View Post
    Tonio, thank you! You saved my day!

    Here's my take on the second part ("Write an expression for the eigenvectors of A_{k} using the matrices P_{1},..., P_{k-1}, and the eigenvectors of A_{1} = A).

    (I'm using v_k and v_{k+1} instead of v and u, respectively.)

    From the first part we now have: v_{k+1} = P_k^{-1} v_k. Obviously v_k would be v_k = P_{k-1}^{-1} v_{k-1} (1) ,
    and further:
    v_{k-1} = P_{k-2}^{-1} v_{k-2}
    v_{k-2} = P_{k-3}^{-1} v_{k-3}
    ... and so on...

    Inserted into (1), we get:
    v_k = P_{k-1}^{-1} P_{k-2}^{-1}P_{k-3}^{-1} v_{k-3}

    Thus a pattern evolves:
    v_k = P_{k-1}^{-1} \cdot P_{k-2}^{-1} \cdot ... \cdot P_{1}^{-1} v_{1} (2)
    where v_1 is the eigenvector of A_1 = A, i.e. Av_1 = \lambda v_1

    Does that look somewhat alright? Something I've missed? Or better ways to write it? (especially (2))...

    J

    Looks fine to me though all those indexes get me dizzy. I'd do a specific example with k= 2 or 3 and see if we get the correct pattern.

    Tonio
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  5. #5
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    Hehe, alright! Thanks again!

    All the best,
    Joachim
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