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Thread: 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
    $\displaystyle A_{1} = A \\
    A_{2} = P ^{-1}_{1} A_{1} P_{1} \\
    A_{3} = P ^{-1}_{2} A_{2} P_{2} \\
    . \\
    . \\
    . ... \\
    A_{k+1} = P ^{-1}_{k} A_{k} P_{k} .$

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

    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; Oct 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
    $\displaystyle A_{1} = A \\
    A_{2} = P ^{-1}_{1} A_{1} P_{1} \\
    A_{3} = P ^{-1}_{2} A_{2} P_{2} \\
    . \\
    . \\
    . ... \\
    A_{k+1} = P ^{-1}_{k} A_{k} P_{k} .$

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

    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...(?)

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

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

    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 $\displaystyle A_{k}$ using the matrices $\displaystyle P_{1},..., P_{k-1}$, and the eigenvectors of $\displaystyle A_{1} = A$).

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

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

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

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

    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 $\displaystyle A_{k}$ using the matrices $\displaystyle P_{1},..., P_{k-1}$, and the eigenvectors of $\displaystyle A_{1} = A$).

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

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

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

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

    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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