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Linear Algebra Problem

  1. Feb 2, 2008 #1

    hotvette

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    I have a book that goes though a detailed development of the following:

    [tex]\left[\begin{matrix}J^TJ & J^TV\\ VJ & D \end{matrix}\right]\left[\begin{matrix}p_1 \\ p_2\end{matrix}\right] = \left[\begin{matrix}J^Tr\\ Vr + Ds \end{matrix}\right][/tex]

    where V and D are diagonal matrices and everything is known except [itex]p_1[/itex] and [itex]p_2[/itex].

    Then it says "since the lower right submatrix D is diagonal, it is easy to eliminate [itex]p_2[/itex] from this system and obtain a smaller n x n system to be solved for [itex]p_1[/itex] alone." The implication is that it's so easy, explanation isn't needed. However, I don't see it.

    Can someone explain how to eliminate [itex]p_2[/itex] given D is diagonal?

    I've managed to come up with:

    [tex][VJ(J^TJ)^{-1}J^TV - D]p_2 = VJ(J^TJ)^{-1}J^Tr - Vr - Ds[/tex]

    but this seems far more complicated than what is implied and I don't see how D being diagonal simplifies anything.
     
    Last edited: Feb 2, 2008
  2. jcsd
  3. Feb 6, 2008 #2
    If [itex]D[/itex] is diagonal with non zero eigenvalues, then it has an inverse [itex]D^{-1}[/itex]. Thus from

    [tex]V\,J\,p_1+D\,p_2=V\,r+D\,s[/tex]

    you can solve for [itex]p_2[/itex], i.e.

    [tex]p_2=D^{-1}\,(V\,r+D\,s)-D^{-1}\,V\,J\,p_1[/tex]

    and use this to eliminate [itex]p_2[/itex] from the other equation.
     
  4. Feb 6, 2008 #3

    hotvette

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    Thanks. I can't believe I didn't see it. So simple.
     
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