Solution09_121 - Solutions to PS 9 (Math 121) J. Scholtz...

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Solutions to PS 9 (Math 121) J. Scholtz January 6, 2007 Question 1: 6.1/20 Prove the Polar Identities: 1. || x + y || 2 + || x - y || 2 = h x + y,x + y i-h x - y,x - y i = h x,x i + h y,y i +2 h x,y i +2 h y,x i-h x,x i-h y,y i = 4 h x,y i . 2. || x + y || 2 + i || x + iy || 2 -|| x - y || 2 - i || x - iy || 2 = ( || x + y || 2 -|| x - y || 2 )+ i ( || x + iy || 2 -|| x - iy || 2 ) = 4 h x,y i . Question 2: 6.1/24d Show that || ( a,b ) || = max {| a | , | b |} is a norm. 1. Positive-definitness: clearly by definition it is positive. Also the only way max {| a | , | b |} = 0 is when ( a,b ) = (0 , 0). 2. || c ( a,b ) || = || ( ca,cb ) || = max {| ca | , | cb |} = | c | max {| a | , | b |} = | c ||| ( a,b ) || . 3. Note that | a + c | < | a | + | c | and | b + d | < | b | + | d | and so the inequality holds. Question 3: 6.1/25 If the inner product was defined using the polar identity and the definition of the norm from 24(d) then it would not be linear. It would be sufficient to show this on the pairs x = (1 , 0) and y = (2 , 2): h (1 , 0) , (2 , 2) i = 1 4 (9 - 4) = 5 4 6 = 2 h (1 , 0) , (1 , 1) i = 1 2 (4 - 1) = 3 2 Question 4: 6.2/2(i) I am not including the solution since it is just computation. However, if you find this any problems with this question do not hesitate to contact me.
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Solution09_121 - Solutions to PS 9 (Math 121) J. Scholtz...

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