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If a mxn matrix A, m>=n has a reduced QR-decomposition

  1. Sep 15, 2010 #1
    Hi
    I am arguing with a friend about the following:
    He claims that if a mxn matrix A, m>=n has a reduced QR-decomposition where R has j nonzero diagonal elements, then the rank of A is at least j. I claim that it is exactly j.
    It was some years ago since i read linear algebra so i was hoping someone here could help us out.
     
  2. jcsd
  3. Sep 19, 2010 #2
    Re: QR-decomposition

    Do you really mean what you wrote? Or you mean "exactly j nonzero diagonal elements?" Because, being precise, your condition can be also understood as "where R has at least j nonzero elements, perhaps more".
     
  4. Sep 23, 2010 #3
    Re: QR-decomposition

    Yes, A's rank should be exactly j. In a full decomposition the remaining columns in Q associated with the zero-valued R's should span A's null space. Combined they (all of Q) covers the full m space, which of course makes sense since Q is an orthonormal basis.
     
  5. Sep 23, 2010 #4
    Re: QR-decomposition

    Another way to put it is that: rank(A) = rank(QR) = rank(R), since Q covers the full basis. And the rank of R is j.
     
  6. Sep 23, 2010 #5
    Re: QR-decomposition

    Then, you are right. To say "Q covers the full basis" is a little bit not quite precise. Q is an orthogonal (or unitary) matrix, thus, in particular, invertible. That is all you need.
     
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