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6.1 (5,16,12) 6.2(9,11,22)

6.1 (5,16,12) 6.2(9,11,22) - Math 431 Assignment 11...

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Math 431 - Assignment 11 Solutions Section 6.1 5 . Let A = 1 i - i 2 . Note that A = A t . We have (a) h x + z, y i = ( x + z ) Ay * = xAy * + zAy * = h x, y i + h z, y i . (b) h cx, y i = ( cx ) Ay * = c ( xAy * ) = c h x, y i . (c) h x, y i = xAy * = x A y * = xA t y t = ( yAx * ) t = yAx * = h y, x i (d) If x = ( a, b ) = ( α + βi, γ + ) C , then h x, x i = ( a - ib, ia + 2 b ) a b = | a | 2 - ib a + ia b + 2 | b | 2 = | a | 2 - i (2 iIm ( b a )) + 2 | b | 2 = | a | 2 + 2 Im ( b a ) + 2 | b | 2 = ( α 2 + β 2 ) + 2( αδ - βγ ) + 2( γ 2 + δ 2 ) = ( α + δ ) 2 + ( β - γ ) 2 + γ 2 + δ 2 0 Moreover, h x, x i = 0 if and only if α = - δ , β = γ and γ = δ = 0. Therefore, h x, x i > 0, if x 6 = 0. Also, (1 - i, 2 + 3 i ) 1 i - i 2 2 - i 3 + 2 i = (1 - i, 2 + 3 i ) 2 - i + 3 i - 2 - 2 i - 1 + 6 + 4 i = (1 - i, 2 + 3 i ) 2 i 5 + 2 i = 2 i + 2 + 10 + 4 i + 15 i - 6 = 6 + 21 i 16(b) . Answer : No. Recall that C [0 , 1] stands for the vector space of real - valued functions f defined on the closed interval [0 , 1]. Consider the following real valued continuous function f ( x ) = ( 0 , if 0 x 1 / 2 , 2 x - 1 , if 1 / 2 x 1 .
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Then f is in V , it is not the zero element of V , but h f, f i = Z 1 / 2 0 f ( t ) 2 dt = Z 1 / 2 0 0 dt = 0 . 22 . Let x = n i =1 a i v i , y = n i =1 b i v i and z = n i =1 c i v i . Then (a) h x + z, y i = n i =1 ( a i + b i ) c i = n i =1 a i c i + n i =1 b i c i = h x, y i + h z, y i . (b) h cx, y i = n i =1 ( ca i ) b i = c (∑ n i =1 a i b i ) = c h x, y i . (c) h x, y i = n i =1 a i b i = n i =1 a i b i = n i =1 b i a i = h y, x i (d) h x, x i = n i =1 a i a i = n i =1 | a i | 2 0 for all x
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