13 2 choose u1a qt 0 i 7211 where i is the r r

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Unformatted text preview: 3. The vector v1(a) is obtained as the least squares solution 9] of (7.2.5b) by solving TT s = l (7.2.12a) for s using forward substitution and setting s v1(a) = QT 0 : (7.2.12b) We readily verify that s s Lv1 (a) = LQT 0 = TT 0] 0 = TT s = l and, using (7.2.12b) and (7.2.11), T v1 (a)U1 (a) = sT 0]QQT 0 = 0: I The orthogonalization task (7.2.7) can also be done by the QR procedure. 1. Determine an orthogonal matrix Q such that QUj = Pj 0 (7.2.13a) where Pj is an r r upper triangular matrix and the zero matrix is l r. (Spatial arguments in (7.2.7) at xj have been omitted for clarity. Although the symbol Q has been used in (7.2.10 - 7.2.12), the matrix in (7.2.13a) is generally not the same.) 2. Select I Uj+1 = QT 0 : 14 (7.2.13b) Using (7.2.13a) Uj = QT Pj : 0 According to (7.2.13b) Thus, QT = Uj+1 Vj+1] Uj = Uj+1 Vj+1] Pj = Uj+1Pj : 0 3. In order to satisfy vjT+1Uj+1 = 0 choose wjT = ;vjT Uj+1: (7.2.14) To verify this, consider (7.2.7b), (7.2.14), and vjT+1Uj+1 = (vjT + wjT UT+1)Uj+1 = vjT Uj+1 + wjT = 0: j...
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