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Unformatted text preview: Solutions for Homework 1 Foundations of Computational Math 1 Fall 2010 Problem 1.1 This problem considers three basic vector norms: k . k 1 , k . k 2 , k . k . 1.1.a Prove that k . k 1 is a vector norm. 1.1.b Prove that k . k is a vector norm. 1.1.c Consider k . k 2 . i Show that k . k 2 is definite. ii Show that k . k 2 is homogeneous. iii Show that for k . k 2 the triangle inequality follows from the Cauchy inequality  x H y  k x k 2 k y k 2 . iv Assume you have two vectors x and y such that k x k 2 = k y k 2 = 1 and x H y =  x H y  , prove the Cauchy inequality holds for x and y . v Assume you have two arbitrary vectors x and y . Show that there exists x and y that satisfy the conditions of part (iv ) and x = x and y = y where and are scalars. vi Show the Cauchy inequality holds for two arbitrary vectors x and y . Solution: We consider each norm in turn. One Norm: If x 6 = 0 then there exists an element j 6 = 0. Therefore, k x k 1 = i  i   j  Therefore k x k 1 is definite. We have k x k 1 = i  i  = i   i  =   i  i  =  k x k 1 1 and therefore k x k 1 is homogeneous. We have k x + y k 1 = i  i + i  i  i  +  i  = i  i  + i  i  = k x k 1 + k y k 1 and therefore k x k 1 satisfies the triangle inequality. Max Norm: If x 6 = 0 then there exists an element j 6 = 0. Therefore, k x k = max i  i   j  Therefore k x k is definite. We have k x k = max i  i  = max i (   i  ) =   max i  i  =  k x k and therefore k x k is homogeneous. We have k x + y k = max i  i + i  max i (  i  +  i  ) max i  i  + max i  i  = k x k + k y k and therefore k x k satisfies the triangle inequality. Two Norm: If x 6 = 0 then there exists an element j 6 = 0. Therefore, k x k 2 2 = i  i  2  j  2 > 2 Therefore k x k 2 is definite. We have k x k 2 2 = i  i  2 = i (   2  i  2 ) =   2 i  i  2 =   2 k x k 2 and therefore k x k 2 is homogeneous. The triangle inequality follows from the Cauchy inequality for the two norm:  x H y  k x k 2 k y k 2 as follows k x + y k 2 2 = x H x + y H y + 2 R e ( x H y ) = k x k 2 2 + k y k 2 2 + 2 R e ( x H y ) k x k 2 2 + k y k 2 2 + 2  x H y  k x k 2 2 + k y k 2 2 + 2 k x k 2 2 k y k 2 2 = ( k x k 2 2 + k y k 2 2 ) 2 ....
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This note was uploaded on 07/25/2011 for the course MAD 5403 taught by Professor Gallivan during the Spring '11 term at University of Florida.
 Spring '11
 Gallivan

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