Summer 2002 done

Summer 2002 done - (15 pts)1. Let T : R2 R2 be a linear...

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(15 pts) 1. Let T : R 2 -→ R 2 be a linear transformation with the property that T ± 3 1 = ± 2 - 4 , T ± 1 1 = ± 0 2 . Let ~ e 1 , ~ e 2 denote the standard basis for R 2 . (1) Calculate T ( ~ e 1 ) and T ( ~ e 2 ). (2) What is T ( ~x ) for an arbitrary vector ~x = ± x 1 x 2 in R 2 ? (3) Find the matrix representing T with respect to the basis: B = { ± 1 2 , ± 2 1 } . Solution: (1) Let A be a matrix s.t. T ( ~x ) = A~x . Then A = ( T ( ~ e 1 ) T ( ~ e 2 ) ) On the other hand, A ± 3 1 1 1 = ± 2 0 - 4 2 by the given two conditions. It follows that A = ± 2 0 - 4 2 ¶± 3 1 1 1 - 1 = ± 1 - 1 - 3 5 Therefore T ( ~ e 1 ) = ± 1 - 3 ,T ( ~ e 2 ) = ± - 1 5 . (2) T ( ~x ) = T ± x 1 x 2 = A ± x 1 x 2 = ± x 1 - x 2 - 3 x 1 + 5 x 2 (3) T ± 1 2 = ± - 1 7 = 5 ± 1 2 - 3 ± 2 1 , and T ± 2 1 = ± 1 - 1 = - 1 ± 1 2 + 1 ± 2 1 . Therefore, [ T ] B = ± 5 - 1 - 3 1 . 1
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2. A = - 1 0 1 - 3 0 1 - 4 0 3 (1) Find the characteristic polynomial of A . (2) Find the eigenvalues of A . (3) Find a basis for each eigenspace of A . Solution: (1) f A ( λ ) = det ( λI - A ) = det λ + 1 0 - 1 3 λ - 1 4 0 λ - 3 = λ ( λ - 1) 2 . (2) f A ( λ ) = λ ( λ - 1) 2 = 0 . Therefore eigenvalues of A are 0 , 1. (3) E 0 = ker (0 I - A ) = ker 1 0 - 1 3 0 - 1 4 0 - 3 = ker 1 0 0 0 0 1 0 0 0 = span 0 1 0 , E 1 = ker (1 I - A ) = ker 2 0 - 1 3 1 - 1 4 0 - 2
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This note was uploaded on 04/01/2010 for the course MATH 110.201 taught by Professor Ha during the Spring '08 term at Johns Hopkins.

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Summer 2002 done - (15 pts)1. Let T : R2 R2 be a linear...

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