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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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 Spring '14
 JosephE.Flaherty

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