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A short one on matrix multiplication

  1. Jul 23, 2007 #1

    Päällikkö

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    Let A and B be real invertible n x n matrices so that B = (AT)-1.
    Show that Bm = (I - B1A1T)(AmT)+, where
    B1 = [b1, ..., bm],
    Bm = [bm+1, ..., bn],
    A1 = [a1, ..., am],
    Am = [am+1, ..., an].


    Any pointers on how one would go about proving the above? I'm fresh out of ideas.

    I'm not sure if the above statement is even actually true. Using random matrices in matlab, I have not found any counterexamples, though.
     
  2. jcsd
  3. Jul 23, 2007 #2

    Dick

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    What does the '+' symbol exponent on a column vector mean?
     
  4. Jul 23, 2007 #3

    Päällikkö

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    I'm not sure if that's a hint or a question: the pseudoinverse.
     
  5. Jul 23, 2007 #4

    Dick

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    It's a question - just trying to understand the problem.
     
  6. Jul 23, 2007 #5

    Hurkyl

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    So, as block matrices,

    A = [ A1 | Am ]
    B = [ B1 | Bm ]

    Have you tried just multiplying things out in block form?
     
  7. Jul 23, 2007 #6

    Päällikkö

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    How do you get to a situation where you get to use the matrices A and B?

    Do you mean something like
    (I - B1A1T)(AmT)+ = (I - BF(AF)T)((AG)T)+,
    where F = [I O]T (where I is m x m and F is n x m) and G = [O I]T (where I is (n-m) x (n-m) and G is n x (n-m)).

    I doubt you meant that, could you elaborate? Anyways, the above I've tried to no avail.

    EDIT: or do you mean starting from the equation B = [B1 Bm] = (AT)-1 = ([A1 Am]T)-1 and trying to solve for Bm? That I haven't actually tried but it seems a bit like a long shot. I'll see if I can get it going.

    EDIT: No, can't get it going.
     
    Last edited: Jul 23, 2007
  8. Jul 23, 2007 #7

    Dick

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    Ok, try getting it going this way. Look at matrices as blocks, as Hurkyl suggests. Your premise shows B.A^T=1. The partitioning splits this into B1.A1^T+Bm.Am^T=1. So Bm.Am^T=1-B1.A1^T. Now you just have to show that (Am^T)^+ is a good enough inverse to Am^T that you can multiply both sides by it and get what you want. I'm still scratching my head over how to write this down formally, but maybe you can figure it out.
     
  9. Jul 23, 2007 #8

    Dick

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    Yeah, and the last step is easy as well. Am^T has linearly independent rows. Writing out (Am^T)^+ explicitly (as in the wikipedia entry on pseudoinverses) and then multiplying Am^T by it gives an n-mxn-m unit matrix.
     
  10. Jul 24, 2007 #9

    Päällikkö

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    Ah indeed, partitioning it in that manner makes it all clear. Thank you very much.
     
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