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QR Factorization

  1. Jan 23, 2007 #1

    hotvette

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    Can someone tell me what I'm doing wrong? Starting with normal equation for least squares:

    (1) ATAx = ATb

    and introducing a diagonal permutation matrix P such that PA = K:

    (2) A = P-1K

    (3) AT = KT[P-1]T = KTP-1 since P is diagonal, PT = P

    Substituting (2) and (3) into (1):

    (4) (KTP-1)(P-1K)x = KTP-1b

    (5) KTP-1P-1Kx = KTP-1b

    Introducing QR factorzation of K, where Q is orthogonal and R is upper triangular:

    (6) K = QR, KT = RTQT

    Equation (5) becomes:

    (7) RTQTP-1P-1QRx = RTQTP-1b

    Pre multiplying both sides by [RT]-1:

    (8) QTP-1P-1QRx = QTP-1b

    Pre-multiplying both sides by Q (QQT = I since Q is orthogonal):

    (9) P-1P-1QRx = P-1b

    Pre-multiplying both sizes by P:

    (10) P-1QRx = b

    Pre-multiplying both sizes by P:

    (11) QRx = Pb

    Pre-multiplying both sizes by QT:

    (12) Rx = QTPb, where PA = K = QR

    Since R is upper triangular, x can be solved via back substitution. But, something the wrong the above development because a numerical example doesn't work out correct. The standard QR factorization of the least squares normal equation results in:

    (13) R'x = Q'Tb, where A = Q'R'

    When I solve an actual numerical example, I get different results using (12) and (13). Where did I go wrong? :yuck:
     
    Last edited: Jan 23, 2007
  2. jcsd
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