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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online