math171a_hw3_sol

# math171a_hw3_sol - Math 171A Homework 3 Solutions...

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Math 171A Homework 3 Solutions Instructor: Jiawang Nie February 1, 2012 1. (4 points) For a given nonzero matrix A and nonzero vector b , assume that b may be written as b = b R + b N , where b R Range ( A ) and b N Null ( A T ). (a) Show that b R and b N are unique. Solution: By contradiction, assume there are two decompositions, i.e. b = b R + b N = ˆ b R + ˆ b N for which b R , ˆ b R Range ( A ) and b N , ˆ b N Null ( A T ) and we want to show that b R = ˆ b R and b N = ˆ b N . First we note that by above we have b R ˆ b R = ˆ b N b N since Range ( A ) and Null ( A T ) are vector subspaces, we have that b R ˆ b R Range ( A ) and b N ˆ b N Null ( A T ) and by the above equality we have b R ˆ b R , b N ˆ b N Range ( A ) Null ( A T ) Now if we can just show that the only vector which lies in the intersection of Range ( A ) and Null ( A T ) is the zero vector, we will have shown that b R = ˆ b R and b N = ˆ b N . To do so, lets suppose we have some z Range ( A ) Null ( A T ). Since z Range ( A ) there exists some x such that Ax = z . Then we have, || z || 2 = || Ax || 2 = x T A T Ax = x T A T z = 0 since z Null ( A T ) implies A T z = 0. Since the norm of a vector is zero if and only if the vector itself is zero, we have that z = 0. This proves our claim. (b) Show that b T R b N = 0. Solution: To show that

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## This note was uploaded on 03/19/2012 for the course MATH 171a taught by Professor Staff during the Spring '08 term at UCSD.

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math171a_hw3_sol - Math 171A Homework 3 Solutions...

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