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Math Help - Separating system of equations

  1. #1
    _V_
    _V_ is offline
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    Separating system of equations

    Hi,

    Given a matrix A = \begin{bmatrix} B & 0 & 0 & -XB \\ 0 & B & 0 & -YB \\ 0 & 0 & B & -ZB \end{bmatrix}

    where B is a non-negative m x n matrix and X, Y, Z are m x m diagonal matrices.

    How would I produce the following:

    \begin{bmatrix} B^T{B} & 0 & 0 & -B^{T}XB \\ 0 & B^T{B} & 0 & -B^{T}YB \\ 0 & 0 & B^{T}B & -B^{T}ZB \\ 0 & 0 & 0 & M \end{bmatrix}

    where M = M_x + M_y + M_z
    M_x = B^{T}X^{2}B - (B^{T}XB)(B^{T}B)^{-1}(B^{T}XB)
    M_y = B^{T}Y^{2}B - (B^{T}YB)(B^{T}B)^{-1}(B^{T}YB)
    M_z = B^{T}Z^{2}B - (B^{T}ZB)(B^{T}B)^{-1}(B^{T}ZB)


    Starting from:

    A^{T}A = \begin{bmatrix} B^T{B} & 0 & 0 & -B^{T}XB \\ 0 & B^T{B} & 0 & -B^{T}YB \\ 0 & 0 & B^{T}B & -B^{T}ZB \\ -B^{T}XB & -B^{T}YB & -B^{T}ZB & B^{T}X^{2}B + B^{T}Y^{2}B + B^{T}Z^{2}B \end{bmatrix}

    I'm also wondering if the inverses could be avoided somehow, since B^{T}B may not be invertible.
    Last edited by _V_; December 1st 2011 at 03:55 AM.
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