homework_12_solutions

homework_12_solutions - MA 35100 HOMEWORK ASSIGNMENT#12...

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MA 35100 HOMEWORK ASSIGNMENT #12 SOLUTIONS Problem 1. pg. 229; prob. 8 Consider a linear transformation L ( ~x ) = A ~x from R n to R m , with ker( L ) = { ~ 0 } . The pseudoin- verse L + of L is the transformation from R m to R n given by L + ( ~ y ) = ( the least-squares solution of L ( ~x ) = ~ y ) . a. Show that the transformation L + is linear. Find the matrix A + of L + , in terms of the matrix A of L . b. If L is invertible, what is the relationship between L + ad L - 1 ? c. What is L + ( L ( ~x ) ) , for ~x in R n ? d. What is L ( L + ( ~ y ) ) , for ~ y in R m ? e. Find L + for the linear transformation L ( ~x ) = 1 0 0 1 0 0 ~x. Solution: (a) Consider the linear system A ~x = ~ y . According to Theorem 5.4.6 on page 224 of the text, the least-squares solution is ~x * = ( A T A ) - 1 A T ~ y . That means we have the transformation L + ( ~ y ) = ~x * = A + ~ y in terms of A + = ( A T A ) - 1 A T . Using Definition 2.1.1 on page 44, we see that L + is indeed a linear transformation. (b) Assume now that L is invertible, so that the matrix A is invertible. Theorem 5.3.9 on page 216 states rank( A T ) = rank( A ), and Theorem 3.1.7 on page 109 states that an n × n matrix is invertible if and only if its rank is n , so A T is also invertible. We may now use Theorem 2.4.7 on page 83: A + = ( A T A ) - 1 A T = A - 1 ( A T ) - 1 A T = A - 1 . Hence A + = A - 1 , so that L + = L - 1 if L is invertible. (c) For all ~x R n we have L + ( L ( ~x ) ) = ( A T A ) - 1 A T ( A ~x ) = ( A T A ) - 1 ( A T A ) ~x = ~x. Hence L + ( L ( ~x ) ) = ~x . (d) For all ~ y R m we have L ( L + ( ~ y ) ) = A ( A T A ) - 1 A T ~ y = A ( A T A ) - 1 A T ~ y. We consider Theorem 5.4.7 on page 225. Write A = ~v 1 ~v 2 · · · ~v n ; its column vectors are linearly independent by Theorem 3.2.8 on page 119 because ker( A ) = { ~ 0 } . According to Theorem 3.1.3 on page 105 they form a basis for the subspace V = im( A ) = span ( ~v 1 ,~v 2 , . . . ,~v n ) of R m . Hence we conclude from Theorem 5.4.7 that A ( A T A ) - 1 A T is the matrix of the orthogonal projection onto V . This means L ( L + ( ~ y ) ) = proj V ~ y is the projection of ~ y onto V = im( A ). 1
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(e) The matrix A + of L + is A + = ( A T A ) - 1 A T = 1 0 0 0 1 0 1 0 0 1 0 0 - 1 1 0 0 0 1 0 = 1 0 0 1 - 1 1 0 0 0 1 0 = A T . Hence the pseudoinverse of L is L + ( ~ y ) = 1 0 0 0 1 0 ~ y . Problem 2. pg. 230; prob. 15 Consider an m × n matrix A with ker( A ) = { ~ 0 } . Show that there exists an n × m matrix B such that B A = I n . Hint : A T A is invertible. Solution: If A is an m × n matrix then A T is an n × m matrix, so that A T A is an n × n square matrix. According to Theorem 5.4.2 on page 221 of the text, since ker( A ) = { ~ 0 } we may conclude that A T A is invertible. Consider the n × m matrix B = ( A T A ) - 1 A T , as motivated by Exercise 5.4.8 on page 229 (i.e., Problem 1 above). Then we have B A = ( A T A ) - 1 ( A T A ) = I n . Hence B = ( A T A ) - 1 A T is the desired matrix. Problem 3. pg. 232; prob. 37 The accompanying table lists several commercial airplanes, the year they were introduced, and the number of displays in the cockpit. Plane Year t Displays d Douglas DC-3 ’35 35 Lockheed Constellation ’46 46 Boeing 707 ’59 77 Concorde ’69 133 a. Fit a linear function of the form log( d ) = c 0 + c 1 t to the data points ( t i , log( d i )), using least squares.
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