# Thogonol Basis Set Homework - Version 040 EXAM03...

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Unformatted text preview: Version 040 – EXAM03 – gilbert – (56780) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The vectors u 1 and u 2 shown in 2 2 u 1 u 2 x 2 x 1 are eigenvectors corresponding to eigenvalues λ 1 = − 3 and λ 2 = − 1 respectively for a 2 × 2 matrix A . Which of the following graphs contains the vector A ( u 1 + u 2 )? 1. 2 2 x 2 x 1 2. − 2 − 2 correct 3. 2 2 x 2 x 1 4. − 2 − 2 5. 2 2 x 2 x 1 6. 2 2 x 2 x 1 Explanation: As the graph of u 1 , u 2 shows, u 1 = bracketleftbigg 2 1 bracketrightbigg u 1 = bracketleftbigg 1 2 bracketrightbigg . Version 040 – EXAM03 – gilbert – (56780) 2 But then A ( u 1 + u 2 ) = A u 1 + A u 2 = λ 1 u 1 + λ 2 u 2 = − 3 u 1 − u 2 = bracketleftbigg − 6 − 3 bracketrightbigg + bracketleftbigg − 1 − 2 bracketrightbigg = bracketleftbigg − 7 − 5 bracketrightbigg . Consequently, A ( u 1 + u 2 ) is contained in the graph − 2 − 2 002 10.0 points If eigenvectors of an n × n matrix A are a basis for R n , then A is diagonalizable. True or False? 1. TRUE correct 2. FALSE Explanation: An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvec-tors. On the other hand, n vectors in R n are a basis for R n if and only if the vectors are linearly independent. So if eigenvectors of A form a basis for R n , they must be linearly independent, in which case A will be diagonalizable. Consequently, the statement is TRUE . 003 10.0 points Use the fact that A = 1 − 4 9 − 7 − 1 2 − 4 1 5 − 6 10 7 ∼ 1 − 1 5 − 2 5 − 6 to determine an orthogonal basis for Col( A ). 1. − 4 2 − 6 , 1 − 1 − 1 2. − 4 2 − 6 , 1 − 1 5 3. 1 − 1 5 , − 4 1 1 correct 4. 1 − 1 5 , 1 − 4 − 1 Explanation: The pivot columns of A provide a basis for Col( A ). But by row reduction, A ∼ 1 − 1 5 − 2 5 − 6 ∼ 1 − 1 5 1 − 5 / 2 3 . Thus the pivot columns of A are a 1 = 1 − 1 5 , a 2 = − 4 2 − 6 . We apply Gram-Schmidt to produce an or-Version 040 – EXAM03 – gilbert – (56780) 3 thogonal basis: set u 1 = a 1 and u 2 = a 2 − parenleftbigg a 2 · u 1 bardbl u 1 bardbl 2 parenrightbigg u 1 = − 4 2 − 6 − ( − 36) 27 1 − 1 5 = − 4 2 − 6 + 4 / 3 − 4 / 3 20 / 3 = − 8 / 3 2 / 3 2 / 3 . Consequently, the set of vectors 1 − 1 5 , − 4 1 1 is an orthogonal basis for Col( A )....
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