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# solhw1 - Solutions for Homework 1 Foundations of...

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Solutions for Homework 1 Foundations of Computational Math 1 Fall 2010 Problem 1.1 This problem considers three basic vector norms: . 1 , . 2 , . . 1.1.a Prove that . 1 is a vector norm. 1.1.b Prove that . is a vector norm. 1.1.c Consider . 2 . i Show that . 2 is definite. ii Show that . 2 is homogeneous. iii Show that for . 2 the triangle inequality follows from the Cauchy inequality | x H y | ≤ x 2 y 2 . iv Assume you have two vectors x and y such that x 2 = y 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 = αx and ˜ y = βy where α and β 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 = 0 then there exists an element ξ j = 0. Therefore, x 1 = Σ i | ξ i | | ξ j | 0 Therefore x 1 is definite. We have αx 1 = Σ i | αξ i | = Σ i | α || ξ i | = | α | Σ i | ξ i | = | α | x 1 1

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and therefore x 1 is homogeneous. We have x + y 1 = Σ i | ξ i + η i | Σ i | ξ i | + | η i | = Σ i | ξ i | + Σ i | η i | = x 1 + y 1 and therefore x 1 satisfies the triangle inequality. Max Norm: If x = 0 then there exists an element ξ j = 0. Therefore, x = max i | ξ i | | ξ j | 0 Therefore x is definite. We have αx = max i | αξ i | = max i ( | α || ξ i | ) = | α | max i | ξ i | = | α | x and therefore x is homogeneous. We have x + y = max i | ξ i + η i | max i ( | ξ i | + | η i | ) max i | ξ i | + max i | η i | = x + y and therefore x satisfies the triangle inequality. Two Norm: If x = 0 then there exists an element ξ j = 0. Therefore, x 2 2 = Σ i | ξ i | 2 | ξ j | 2 > 0 2
Therefore x 2 is definite. We have αx 2 2 = Σ i | αξ i | 2 = Σ i ( | α | 2 | ξ i | 2 ) = | α | 2 Σ i | ξ i | 2 = | α | 2 x 2 and therefore x 2 is homogeneous. The triangle inequality follows from the Cauchy inequality for the two norm: | x H y | ≤ x 2 y 2 as follows x + y 2 2 = x H x + y H y + 2 R e ( x H y ) = x 2 2 + y 2 2 + 2 R e ( x H y ) x 2 2 + y 2 2 + 2 | x H y | x 2 2 + y 2 2 + 2 x 2 2 y 2 2 = ( x 2 2 + y 2 2 ) 2 . So the true problem is to prove the Cauchy inequality. To so do assume that we have two vectors x and y such that x 2 = y 2 = 1 and x H y = | x H y | . For any two such vectors we have x - y 2 2 = ( x - y ) H ( x - y ) = 2 - 2 x H y 0 Therefore | x H y | ≤ 1 = x 2 y 2 . To generalize to any two nonzero vectors ˜ x and ˜ y note that there must exist complex scalars α and β such that ˜ x = αx and ˜ y = βy where x and y satisfy the conditions above (see Lemma below). We have | ˜ x H ˜ y | = | αx H | = | αβ || x H y | | αβ | x 2 y 2 = | α | x 2 | β | y 2 = αx 2 βy 2 = ˜ x 2 ˜ y 2 3

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Lemma. Let ˜ x C n and ˜ y C n be such that ˜ x = 0 and ˜ y = 0 . There exists α C , β C , x C n and y C n such that x
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solhw1 - Solutions for Homework 1 Foundations of...

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