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Matrix with Orthonormal columns

  1. Jul 25, 2009 #1

    hotvette

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    If Q is an m x n (m > n) matrix with orthonormal columns, we know that [itex]Q^TQ = I[/itex] of dimension n x n. I have a question about [itex]QQ^T[/itex]. It is a symmetric m x m matrix but also appears to be singular. Why would it be singular? Probably something basic I've long since forgotten.
     
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  3. Jul 25, 2009 #2

    HallsofIvy

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    Why do you say it "appears to be singular"? You can take Q= I as a matrix with orthonormal matrix such that [itex]QQ^T[/itex] is not singular.
     
  4. Jul 25, 2009 #3

    hotvette

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    Thanks for your reply. I'm dealing specifically with non-square matrices with more rows than columns where I'm trying to factor [itex]QQ^T[/itex] using QR or SVD. The result is always nonsense so I tried an experiment by generating random non-square matrices with orthonormal columns and in all cases [itex]QQ^T[/itex] had condition numbers ~1E15 and the factorization routines broke down. I figured there must be some rule or axiom I'd forgotten.

    This came up when trying to simplify the matrix equation [itex]A^TDAx = A^Tb[/itex] where A is non-square with more rows than columns and D is diagonal. If A is factored into QR the simplification gets stalled because I end up with [itex]QQ^T[/itex] on both sides and can't go further.

    The broader question is whether [itex]A^TDAx = A^Tb[/itex] can be solved for x without having to carry out the matrix multiplications [itex]A^TDA[/itex], where A is non-square with more rows than columns and D is diagonal.
     
    Last edited: Jul 25, 2009
  5. Jul 26, 2009 #4

    hotvette

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    I managed to answer the broader question after realizing I made a goof in the equation. It really is ATDAx = ATDb and can be readily solved by letting B = D1/2A. The problem is then BTBx = BTD1/2b which is easy to solve.

    On the other question, I'm still curious whether a generalization can be made about the nature of QQT in the case where Q is rectangular (m > n) and has orthonormal columns. I've scoured my linear algebra texts and can't find any statements about this.
     
  6. Jul 26, 2009 #5
    [itex]QQ^T[/itex] has rank n, so if n < m then it is singular.
     
  7. Jul 29, 2009 #6

    hotvette

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    Thanks. Actually, I realized it doesn't matter that the columns are orthonormal. I found the relation I was looking for:

    rank(A) = rank(ATA) = rank(AAT)

    It makes intuitive sense since you aren't adding any information by forming the matrix products.

    http://en.wikipedia.org/wiki/Rank_(linear_algebra [Broken])
     
    Last edited by a moderator: May 4, 2017
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