EXAM03-solutions - Version 123 EXAM03 gilbert(57245 This...

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Version 123 – EXAM03 – gilbert – (57245) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind the distance From y = 6 9 3 to the plane in R 3 spanned by u 1 = 2 2 1 , u 2 = 1 2 2 . 1. dist = 11 2. dist = 8 3. dist = 7 4. dist = 10 5. dist = 9 correct Explanation: The plane in R 3 spanned by u 1 , u 2 is the subspace W = Span { u 1 , u 2 } , and each y in R 3 has a unique orthogonal decomposition y = proj W y + ( y proj W y ) where y proj W y is in the orthogonal com- plement W . But then dist( y , W ) = b y proj W y b . Now u 1 , u 2 are non-zero othogonal vectors, so Form a basis For W such that proj W y = p y · u 1 b u 1 b 2 P u 1 + p y · u 2 b u 2 b 2 P u 2 . But when y = 6 9 3 , u 1 = 2 2 1 , u 2 = 1 2 2 , we see that p y · u 1 b u 1 b 2 P u 1 = 2 2 1 , while p y · u 2 b u 2 b 2 P u 2 = 2 1 2 2 . Consequently, proj W y = 2 2 1 2 1 2 2 = 0 6 3 , and so y proj W = 6 9 3 0 6 3 = 3 2 1 2 . Thus dist( y , W ) = 9 . 002 10.0 points ±ind the term σ 2 u 2 v T 2 in the spectral de- composition oF A = b 2 2 1 2 B . 1. σ 2 u 2 v T 2 = 1 5 b 4 2 8 4 B correct 2. σ 2 u 2 v T 2 = 1 5 b 2 4 4 8 B 3. σ 2 u 2 v T 2 = 1 5 b 8 4 4 2 B 4. σ 2 u 2 v T 2 = 1 5 b 4 8 2 4 B Explanation:
Version 123 – EXAM03 – gilbert – (57245) 2 The spectral decomposition of A deter- mined by the Singular Value Decomposition A = [ u 1 u 2 ] b σ 1 0 0 σ 2 Bb v T 1 v T 2 B is given by A = σ 1 u 1 v T 1 + σ 2 u 2 v T 2 . But u 1 = 1 σ 1 A v 1 , u 2 = 1 σ 2 A v 2 , so this spectral decomposition can be written simply as A = A v 1 v T 1 + A v 2 v T 2 , where v 1 , v 2 are normalized eigenfunctions of A T A corresponding to eigenvalues λ 1 λ 2 . For the given matrix A = b 2 2 1 2 B , we see that A T A = b 2 1 2 2 Bb 2 2 1 2 B = b 5 2 2 8 B . Now the eigenvalues of A T A are λ 1 = 9 , λ 2 = 4, and the corresponding normalized eigen- functions are v 1 = 1 5 b 1 2 B , v 2 = 1 5 b 2 1 B , so v 1 v T 1 = 1 5 b 1 2 2 4 B , v 2 v T 2 = 1 5 b 4 2 2 1 B . Thus A v 1 v T 1 = 1 5 b 2 2 1 2 Bb 1 2 2 4 B = 1 5 b 6 12 3 6 B , while A v 2 v T 2 = 1 5 b 2 2 1 2 Bb 4 2 2 1 B = 1 5 b 4 2 8 4 B , Consequently, σ 2 u 2 v T 2 = 1 5 b 4 2 8 4 B , and the full spectral decomposition of A is A = 1 5 b 6 12 3 6 B + 1 5 b 4 2 8 4 B , as can be checked by direct calculation. 003 10.0 points Find the solution of the di±erential equa- tion d u dt = A u ( t ) , u (0) = b 6 2 B , when A is the matrix A = b 2 5 1 4 B .

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