Question: 1,2,3,8,10,14 Please SECTION 4.3 Exercises In Exercises 7-18, W is not
ID: 3115468 • Letter: Q
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Question: 1,2,3,8,10,14 Please
SECTION 4.3 Exercises In Exercises 7-18, W is not a subspace of the vector space. Verify this by giving a specific example that violates the test for a vector subspace (Theorem 4.5) 7. W is the set of all vectors in R3 whose third component is -1 B W is the set of all vectors in R2 whose second component is 1 9. W is the set of all vectors in R2 whose components are In Exercises 1-6, verify that W is a subspace of V. In each case, assume that V has the standard operations. 1,'w = {(x1, x2, x3, 0): x1, x2, and x3 are real numbers } 2) W = {(x, y, 2x-3y): x and y are real numbers} V=R3 is the set of all 2 x 2 matrices of the form rational numbers. 0) W is the set of all vectors in R whose components are integers. 11. W is the set of all nonnegative functions in C-oo, oo) 12, W is the set of all linear functions ax + b, a0, in 0 Cl V = M2.2 4, W is the set of all 3 × 2 matrices of the form al 13. W is the set of all vectors in R3 whose components are atb 0 nonnegative. 14. W is the set of all vectors in R3 whose components ane 0 V = M3.2 5. Calculus W is the set of all functions that are continuous on [O, 1]. V is the set of all functions that are integrable on [O, 1]. 6. Calculus Wis the set of all functions that are differentiable on [O, 1]. Vis the set of all functions that are continuous on [O, 1] Pythagorean triples. 15. W is the set of all matrices in M with zero determinants. 16. W is the set of all matrices in Mn such that A2-A. 17. W is the set of all vectors in R2 whose second component is the cube of the first.Explanation / Answer
0
a1
b1
0
and Y =
0
a2
b2
0
be 2 arbitrary elements of W and let k be an arbitrary real scalar. Then X+Y =
0
a1+a2
b1+b2
0
Thus, X+Y
0
ka1
kb1
0
Thus, kX W, so that W is closed under scalar multiplication. Also, 0 apparently belongs to W (when a,b= 0, X = 0). Hence W is a vector space, and therefore a subspace of V = M2,2.
8. Let X = (a,1) and y = (b,1) be 2 arbitrary elements of W. Then X+Y =(a+b,2). Thus, X+Y W so that W is not closed under vector addition. Hence W is a not vector space, and therefore not a subspace of V =R2.
10. Let X = (a,b), where a,b are integers, be an arbitrary element of W and let k be an arbitrary real scalar. Then kX =(ka,kb) which may not W when k is not an integer. Thus, W is not closed under scalar multiplication. Hence W is a not vector space, and therefore not a subspace of V =R2.
14. Let X = (a1,b1,c1), where c12 = a12+b12 and Y = (a2,b2,c2), where c22 = a22+b22 be 2 arbitrary elements of W. Then X+Y = (a1,b1,c1)+ (a2,b2,c2) = (a1+a2,b1+b2,c1+c2). However, (c1+c2)2 = c12+ c22+2c1 c2 = a12+b12+ a22+b22 +2(a12+b12)1/2( a22+b22)1/2 and (a1+a2)2 +( b1+b2)2 = a12+a22 +2a1a2 + b12+b22 +2b1b2= a12+b12+ a22+b22+2a1a2 +2b1b2 = c12+ c22+2a1a2 +2b1b2(c1+c2)2. Hence X+Y W so that W is not closed under vector addition. Hence W is a not vector space, and therefore not a subspace of V =R3.
0
a1
b1
0
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