LinAlg4

# LinAlg4 - 4.1 SOLUTIONS Notes: This section is designed to...

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185 4.1 SOLUTIONS Notes : This section is designed to avoid the standard exercises in which a student is asked to check ten axioms on an array of sets. Theorem 1 provides the main homework tool in this section for showing that a set is a subspace. Students should be taught how to check the closure axioms. The exercises in this section (and the next few sections) emphasize n , to give students time to absorb the abstract concepts. Other vectors do appear later in the chapter: the space of signals is used in Section 4.8, and the spaces n of polynomials are used in many sections of Chapters 4 and 6. 1 . a . If u and v are in V , then their entries are nonnegative. Since a sum of nonnegative numbers is nonnegative, the vector u + v has nonnegative entries. Thus u + v is in V . b . Example: If 2 2  =   u and c = –1, then u is in V but c u is not in V . 2 . a . If x y = u is in W , then the vector x cx cc yc y  ==   u is in W because 2 () ()0 cx cy c xy =≥ since xy 0. b . Example: If 1 7 = u and 2 3 = v , then u and v are in W but u + v is not in W . 3 . Example: If .5 .5 = u and c = 4, then u is in H but c u is not in H . Since H is not closed under scalar multiplication, H is not a subspace of 2 . 4 . Note that u and v are on the line L , but u + v is not. u v L u+v 5 . Yes. Since the set is 2 Span{ } t , the set is a subspace by Theorem 1.

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186 CHAPTER 4 • Vector Spaces 6 . No. The zero vector is not in the set. 7 . No. The set is not closed under multiplication by scalars which are not integers. 8 . Yes. The zero vector is in the set H . If p and q are in H , then ( p + q )(0) = p (0) + q (0) = 0 + 0 = 0, so p + q is in H . For any scalar c , ( c p )(0) = c p (0) = c 0 = 0, so c p is in H . Thus H is a subspace by Theorem 1. 9 . The set H = Span { v }, where 1 3 2   =  v . Thus H is a subspace of 3 by Theorem 1. 10 . The set H = Span { v }, where 2 0 1 = v . Thus H is a subspace of 3 by Theorem 1. 11 . The set W = Span { u , v }, where 5 1 0 = u and 2 0 1     =       v . Thus W is a subspace of 3 by Theorem 1. 12 . The set W = Span { u , v }, where 1 1 2 0 = u and 3 1 1 4 = v . Thus W is a subspace of 4 by Theorem 1. 13 . a . The vector w is not in the set 123 {, , } vvv . There are 3 vectors in the set . b . The set Span{ , , } contains infinitely many vectors. c . The vector w is in the subspace spanned by if and only if the equation 11 2 2 33 xx x ++= w has a solution. Row reducing the augmented matrix for this system of linear equations gives 1243 1001 0121 0121 , 1362 0000 so the equation has a solution and w is in the subspace spanned by . 14 . The augmented matrix is found as in Exercise 13c. Since 1 248 1000 01 2 4 0 1 2 0 , 1367 0001 the equation x w has no solution, and w is not in the subspace spanned by .
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## This note was uploaded on 04/14/2008 for the course MATH 2700 taught by Professor Atim during the Spring '08 term at North Texas.

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LinAlg4 - 4.1 SOLUTIONS Notes: This section is designed to...

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