Ch6 - Solutions to Homework 12 Math 110 Fall 2006 Prob 6.1.1(a True directly by definition(b True this is also required by the definition(c False

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Solutions to Homework 12. Math 110, Fall 2006. Prob 6.1.1. (a) True, directly by definition. (b) True, this is also required by the definition. (c) False, it is linear in the first component and conjugate linear in the second component. (d) False, there are in fact infinitely many inner products. (e) False, it also holds in infinite- dimensional spaces (see the proof, which does not use finite-dimensionality). (f) False, every rectangular matrix has a conjugate transpose (defined the usual way). (g) False when x is fixed. (h) True when x runs over the whole space. Prob 6.1.3. We have: h f, g i = Z 1 0 te t dt = ( t - 1) e t ± ± 1 0 = 1 , k f k 2 = Z 1 0 t 2 dt = t 3 3 ± ± ± ± 1 0 = 1 3 , so k f k = 1 3 , k g k 2 = Z 1 0 e 2 t dt = e 2 t 2 ± ± ± ± 1 0 = e 2 - 1 2 , so k g k = r e 2 - 1 2 , k f + g k 2 = Z 1 0 ( t + e t ) 2 dt = Z 1 0 ( t 2 + 2 e t + e 2 t ) dt = 3 e 2 + 11 6 , so k f + g k = r 3 e 2 + 11 6 . Since e 2 > 7, we can confirm that the Cauchy-Schwarz inequality holds: |h f, g i| = 1 q e 2 - 1 6 . Likewise, k f + g k = r 3 e 2 + 11 6 1 3 + r e 2 - 1 2 = k f k + k g k . The intermediate inequality can be checked numerically or by squaring both sides: 1 3 + e 2 - 1 2 + 2 r e 2 - 1 6 = 3 e 2 - 1 6 + 2 r e 2 - 1 6 3 e 2 - 1 6 + 2 = 3 e 2 + 11 6 .
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This note was uploaded on 03/06/2008 for the course MATH 110 taught by Professor Valdimarsson during the Spring '08 term at UCLA.

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Ch6 - Solutions to Homework 12 Math 110 Fall 2006 Prob 6.1.1(a True directly by definition(b True this is also required by the definition(c False

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