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Math Help - Problem of diagonal matrix..

  1. #1
    kin
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    Problem of diagonal matrix..



    i found the diagonal matrix =

    <br />
D = \left(\begin{array}{ccc}0&0&0\\0&1&0\\0&0&1\end{ar  ray}\right)<br />
    but i dont know how to find X^2= A......
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  2. #2
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    Quote Originally Posted by kin View Post


    i found the diagonal matrix =

    <br />
D = \left(\begin{array}{ccc}0&0&0\\0&1&0\\0&0&1\end{ar  ray}\right)<br />
    but i dont know how to find X^2= A......
    Dear kin,

    Can you tell me what you obtained for matrix P? Then I shall try to help you.
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  3. #3
    kin
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    Quote Originally Posted by Sudharaka View Post
    Dear kin,

    Can you tell me what you obtained for matrix P? Then I shall try to help you.

    i find
    <br /> <br />
P= \left(\begin{array}{ccc}-2/3&2&-1\\1/3&1&0\\1&0&1\end{array}\right)<br />

    <br /> <br />
P^-1= \left(\begin{array}{ccc}-3&6&-3\\1&-1&1\\3&-6&4\end{array}\right)<br />
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  4. #4
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    For any invertible matrix P, define U by U= P^{-1}XP. Then X= PUP^{-1} so X^2= (PUP^{-1})(PUP^{-1})= PU^2P^{-1}. Now if also X^2= A, A= PU^2P^{-1} and P^{-1}AP= U^2.

    If P is such that P^{-1}AP= D, a diagonal matrix, the D= U^2 and so U is also a diagonal matrix having entries equal to the square roots of the diagonal elements of D. That is, with D= \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}, U is also \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} and, as before, X= PUP^{-1}.
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  5. #5
    kin
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    Quote Originally Posted by HallsofIvy View Post
    For any invertible matrix P, define U by U= P^{-1}XP. Then X= PUP^{-1} so X^2= (PUP^{-1})(PUP^{-1})= PU^2P^{-1}. Now if also X^2= A, A= PU^2P^{-1} and P^{-1}AP= U^2.

    If P is such that P^{-1}AP= D, a diagonal matrix, the D= U^2 and so U is also a diagonal matrix having entries equal to the square roots of the diagonal elements of D. That is, with D= \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}, U is also \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} and, as before, X= PUP^{-1}.
    actually X= PUP^{-1} = A= PDP^{-1}...!!!!!!!!!!!!!

    by the definition of X, X should not = +A/ - A......!!!!
    Last edited by kin; April 22nd 2010 at 05:50 PM.
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