HW9solutions_2

# HW9solutions_2 - MATH 115 SOLUTION SET 8-10 ANSWERS TO...

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MATH 115 SOLUTION SET 8-10 ANSWERS TO SELECTED PROBLEMS 1. Set 8 § 5.2 8. We always have dim E λ 2 1. On the other hand, since dim E λ 1 = n - 1, the algebraic multiplicity of λ 1 is at least n - 1. So the multiplicity m λ 2 of λ 2 is at most 1 (the two multiplicities must sum to n ). We have 1 m λ 2 dim E λ 2 1, from which we conclude dim E λ 2 = m and A is diagonalizable. 2. Set 9 § 6.1 16. Linearity is easy. For the property h f,g i = h g,f i , we check: h g,f i = 1 2 π Z 2 π 0 g ( t ) f ( t ) dt = 1 2 π Z 2 π 0 f ( t ) g ( t ) dt = h f,g i For the last property, suppose f H is nonzero. We claim h f,f i > 0. For suppose x 0 [0 , 2 π ] is a point with f ( x 0 ) 6 = 0; say | f ( x 0 ) | = δ . Then since f is continuous, there must be some subinterval I [0 , 2 π ] for which | f ( t ) | ≥ δ/ 2 whenever t I . We have 2 π h f,f i = Z 2 π 0 | f ( t ) | 2 dt Z I | f ( t ) | 2 dt Z I ± δ 2 ² 2 dt > 0 . 21. a) We have A * 1 = 1 2 ( A * + A ** ) = 1 2 ( A * + A ) = A 1 . The case of A 2 is similar.

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## This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.

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HW9solutions_2 - MATH 115 SOLUTION SET 8-10 ANSWERS TO...

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