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**Unformatted text preview: **Version 053 – EXAM03 – gilbert – (57245) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the singular value σ 1 for the ma- trix A = 1 − 1 − 2 2 . 1. σ 1 = √ 5 2. σ 1 = 2 √ 2 3. σ 1 = 3 correct 4. σ 1 = √ 6 5. σ 1 = √ 7 Explanation: By definition, σ is a singular value of A when σ = √ λ and λ is an eigenvalue of A T A ; σ 1 is the largest of these singular values. Now A T A = bracketleftbigg 1 − 2 − 1 2 bracketrightbigg 1 − 1 − 2 2 = bracketleftbigg 5 − 4 − 4 5 bracketrightbigg . But then det( A T A − λI ) = bracketleftbigg 5 − λ − 4 − 4 5 − λ bracketrightbigg = λ 2 − 10 λ + 9 = ( λ − 9)( λ − 1) . Consequently, σ 1 = 3 . 002 10.0 points If B = PDP- 1 with P an orthogonal ma- trix and D a diagonal matrix, then B is a symmetric matrix. True or False? 1. FALSE 2. TRUE correct Explanation: Note that for B to be symmetric, B T = B . Also note that all diagonal matrices are by definition symmetric. Consider B T . B T = ( PDP T ) T = ( P T ) T D T P T = PD T P T = PDP T = B. Thus B T = B and B is by definition symmet- ric. Consequently, the statement is TRUE . 003 10.0 points If v 1 , v 2 , . . ., v p span a subspace W of R n and if x is a vector in R n such that x · v j = 0 , j = 1 , . . ., p, then x is in W ⊥ . True or False? 1. TRUE correct 2. FALSE Explanation: By definition W ⊥ consists of all vectors in R n that are orthogonal to any set that spans W . So if Span { v 1 , v 2 , . . ., v p } = W and if x is a vector in R n such that x · v j = 0 for all 1 ≤ j ≤ p , then x is in W ⊥ . Consequently, the statement is TRUE . 004 10.0 points Version 053 – EXAM03 – gilbert – (57245) 2 A vector space V is a subspace of itself. True or False? 1. FALSE 2. TRUE correct Explanation: A set H is a subspace of a vector space V when (i) H contains the zero vector , (ii) the sum u + v of any u , v in H is in H , (iii) the scalar multiple c u of any scalar c and any u in H is in H . Since V has these three properties, V is a subspace of itself. Consequently, the statement is TRUE . 005 10.0 points Find the distance from y = bracketleftbigg 3 − 1 bracketrightbigg to the line L in R 2 through the origin and u = bracketleftbigg − 1 − 2 bracketrightbigg . 1. dist( y , L ) = 14 √ 5 2. dist( y , L ) = 14 3. dist( y , L ) = 7 4. dist( y , L ) = 7 √ 5 correct 5. dist( y , L ) = 21 √ 5 6. dist( y , L ) = 21 Explanation: The line in R 2 through the origin and u is the subspace L = Span { u } , and each y in R 2 has a unique orthogonal decomposition y = proj L y + ( y − proj L y ) where y − proj L y is perpendicular to L . But then dist( y , L ) = bardbl y − proj L y bardbl ....

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