Explanation Given: A = [ 1 2 1 3 1 3 1 6 ] , b = [ 5 0 2 1 ] . Approach: Let A be an ( m × n ) matrix, and suppose that the column vectors of A are linearly independent. Let S be an orthogonal matrix such that S A is upper trapezoidal. Given a vector b in R m , let S A and S b be denoted as, S A = [ R O ] and S b = [ c d ] . Then, 1. R is non singular. 2. There is a unique least-squares solution of A x = b , x * . 3. The vector x * satisfies R x * = c . Let u be a nonzero vector in R n and let I be the ( n × n ) identity matrix. The ( n × n ) matrix S given by, S = I − 2 u T u u u T is called a Householder matrix. Calculation: Consider a Householder matrix Q 1 such that the last three component in Q 1 A 1 is zero. Here, A 1 = [ 1 , 1 , 1 , 1 ] T . Now, Q 1 = I − b u u T and b = 2 u T u , u i = 0 , i = 1 , ... , k − 1 and u k = v k − s , Here, s = ± v k 2 + v k + 1 2 + ... + v n 2 , and u i = v i for i = k + 1 , ... , n For k = 1 s = ± v 1 2 + v 2 2 + v 3 2 + v 4 2 = ± 1 2 + 1 2 + 1 2 + 1 2 = ± 2 Choose s = − 2 so that s v 1 ≤ 0 Therefore, u 1 = [ 3 , 1 , 1 , 1 ] T that is, u 1 = [ 3 1 1 1 ] So, u 1 T u 1 = [ 3 1 1 1 ] [ 3 1 1 1 ] = 9 + 1 + 1 + 1 = 12 u 1 T A 1 = [ 3 1 1 1 ] [ 1 1 1 1 ] = 3 + 1 + 1 + 1 = 6 u 1 T A 2 = [ 3 1 1 1 ] [ 2 3 3 6 ] = 6 + 3 + 3 + 6 = 18 b = 2 12 = 1 6 Q 1 A 1 = ( I − b u 1 u 1 T ) A 1 = A 1 − u 1 T A 1 6 u 1 = A 1 − 6 6 u 1 = A 1 − u 1 Q 1 A 2 = ( I − b u 1 u 1 T ) A 2 = A 2 − u 1 T A 2 6 u 1 = A 2 − ( 18 ) 6 u 1 = A 2 − 3 u 1 Now, Q 1 A = [ Q 1 A 1 , Q 1 A 2 ] Q 1 A = [ ( A 1 − u 1 ) ( A 2 − 3 u 1 ) ] = [ [ 1 1 1 1 ] − [ 3 1 1 1 ] [ 2 3 3 6 ] − 3 [ 3 1 1 1 ] ] = [ − 2 − 7 0 0 0 0 0 3 ] The matrix [ − 2 − 7 0 0 0 0 0 3 ] is not in trapezoid form. So, consider a Householder matrix Q 2 such that the last two component in Q 2 ( Q 1 A 2 ) is zero. Here, A 2 = [ − 7 , 0 , 0 , 3 ] T . For k = 2 s = ± v 2 2 + v 3 2 + v 4 2 = ± 0 2 + 0 2 + 3 2 = ± 3 Choose s = − 3 so that s v 2 ≤ 0 , Therefore, u 2 = [ 0 , 3 , 0 , 3 ] T that is, u 2 = [ 0 3 0 3 ] So, u 2 T u 2 = [ 0 3 0 3 ] [ 0 3 0 3 ] = 9 + 9 = 18

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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The least-squares solution to A x = b .

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To find: The least-squares solution to Ax=b.

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Chapter 7 Solutions

To find:

The least-squares solution to Ax=b.