Ma1bAnHW4Sol

Ma1bAnHW4Sol - CALIFORNIA INSTITUTE OF TECHNOLOGY...

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Unformatted text preview: CALIFORNIA INSTITUTE OF TECHNOLOGY Department of Mathematics Math 1b; Solutions to Homework Set 4 Due: February 4, 2008 1. For T L ( V ) let m ( T ) = m X ( T ) be the matrix of T with respect to X . We first calculate m ( f ) and m ( f 2 ). Now f ( e 1 ) = f (1 , 0) = (3 , 5) = 3 e 1 + 5 e 2 and f ( e 2 ) = f (0 , 1) = (2 , 1) = 2 e 1 + e 2 , so m ( f ) = 3 2 5 1 , and hence m ( f ) 2 = 19 8 20 11 . But by Theorem 16.16, m ( f 2 ) = m ( f ) 2 , so we have the answer to (2). Next f ( X ) = { (3 , 5) , (2 , 1) } is independent, so by Problem 3 on HW2, f ( V ) = V . Thus by Theorem 2C, N ( f ) = 0. Let g L ( V ) be the unique linear map with g ( e i ) = y i the i th member of Y , and B = m ( g ) the matrix for g . By Theorem 4.6: m Y ( f ) = B- 1 m ( f ) B Now g ( e 1 ) = e 1 + 2 e 2 and g ( e 2 ) = 2 e 1 + 3 e 2 , so B = m ( g ) = 1 2 2 3 and from Problem 4 on HW1: If B = a b c d then B- 1 = 1 ad- bc d- b- c a so B- 1 =- 3 2 2- 1 so m Y ( f ) = B- 1 m ( f ) B...
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This note was uploaded on 04/28/2010 for the course MATH 1B taught by Professor Aschbacher during the Winter '07 term at Caltech.

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Ma1bAnHW4Sol - CALIFORNIA INSTITUTE OF TECHNOLOGY...

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