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Math Help - finding orthogonal matrix?

  1. #1
    kpc
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    finding orthogonal matrix?

    I would like to solve the following problem in the 'correct' way. I currently use optimization to find the answer, but I would like to get rid of it.

    Given a real-valued symmetric Matrix M, containing only elements along all borders and the diagonal.
    Also given a symmetric prototype matrix P, containing ones and zeroes. The 1/0 indicate where values are allowed in the result matrix.

    I need to find an orthogonal matrix Q, so that when applying
    M_2 = Q M_1 Q^-1
    the matrix M_2 does not contain any values at the zero positions of P.
    The other way is possible, M_2 may contain a zero, where P is one.
    M_2(i,j) * (1 - P(i,j)) = 0

    P was chosen in such a way, that the transformation is possible.
    Q does not have to be unique. Multiple solutions might be possible.

    For some forms of P, it is possible to derive Q using a series of Givens rotations, annihilating elements one by one. But I could not find a solution for the general case. Neither could i find a strategy in which order to apply the Givens rotations.

    How to derive the Q matrix?

    We could start simple:
    P is chosen in such a way that Q is unique (I think).
    This way M_2 cannot be zero, where P is one.
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  2. #2
    kpc
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    Since nobody even left a hint, maybe the problem is not clear enough.
    So to clarify, here is an example.

    M_1 = \begin{bmatrix}<br />
0 & 0.364 & -0.654 & 0.668 & -0.343 & 0.015 \\ <br />
0.364 & 1.314 & 0 & 0 & 0 & 0.364 \\<br />
-0.654 & 0 & 0.783 & 0 & 0 &0.654 \\<br />
0.668 & 0 & 0 & -0.804 & 0 & 0.668 \\<br />
-0.343 & 0 & 0 & 0 & -1.297 & 0.343 \\<br />
0.015 & 0.364 & 0.654 & 0.668 & 0.343 & 0<br />
 \end{bmatrix}

    P = \begin{bmatrix}<br />
0 & 1 & 0 & 0 & 0 & 1 \\<br />
1 & 1 & 1 & 0 & 1 & 1 \\<br />
0 & 1 & 1 & 1 & 1 & 0 \\<br />
0 & 0 & 1 & 1 & 1 & 0 \\<br />
0 & 1 & 1 & 1 & 1 & 1 \\<br />
1 & 1 & 0 & 0 & 1 & 0<br />
\end{bmatrix}

    M_2 = \begin{bmatrix}<br />
0 & 1.060 & 0 & 0 & 0 & 0.015 \\<br />
1.060 & -0.002 & 0.874 & 0 & -0.326 & 0.032 \\<br />
0 & 0.874 & 0.048 & 0.836 & 0.034 & 0 \\<br />
0 & 0 & 0.836 & -0.067 & 0.872 & 0 \\<br />
0 & -0.326 & 0.034 & 0.872 & 0.017 & 1.060 \\<br />
0.015 & 0.032 & 0 & 0 & 1.060 & 0<br />
\end{bmatrix}

    For this specific P-matrix, Q can be build using a series of rotations:
    \begin{matrix}<br />
\textrm{Pivot} & \textrm{Zeroed element} & \textrm{rotation} \\<br />
\hline<br />
4,5 & 1,5 & \theta=\tan^{-1}(-M_{1,5}/M_{1,4}) \\<br />
3,4 & 1,4 & \theta=\tan^{-1}(-M_{1,4}/M_{1,3}) \\<br />
2,3 & 1,3 & \theta=\tan^{-1}(-M_{1,3}/M_{1,2}) \\<br />
3,4 & 3,6 & \theta=\tan^{-1}(M_{3,6}/M_{4,6}) \\<br />
4,5 & 4,6 & \theta=\tan^{-1}(M_{4,6}/M_{5,6}) \\<br />
3,4 & 2,4 & \theta=\tan^{-1}(-M_{2,4}/M_{2,3}) \\<br />
\end{matrix}

    The question is about finding a general procedure to determine the/a Q-matrix, valid for any P-matrix (provided the transformation is possible)
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  3. #3
    Super Member Random Variable's Avatar
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    I don't quite understand your question, but have you considered using Householder reflections?
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  4. #4
    kpc
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    Could you state what is unclear?
    Let me try again.
    I want to find a Q matrix, so that Q M Q^-1 is transformed into the shape described in the P matrix.

    Only for certain shapes of P, it is easy to build up Q with a series of transformations, one step at the time. Eg. like using the Householder transformations for wiping the lowerleft triangle in QR decomposition.
    However for a general prescribed solutions pattern (P) it seems to me like solving a rubiks cube. As soon as I put the next zero somewhere, my previous zeroes start to fill up again.
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