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601-hw-weeks1-2

# 601-hw-weeks1-2 - AN18 Answers to Selected Odd-Numbered...

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Unformatted text preview: AN18 Answers to Selected Odd-Numbered Exercises 2 L _2 7. Not a vector space 9. Not a vector space _ I: _ - = 11' xk _ (3/00) I: 1 ] + (5/4“ I) 1 11. Not a vector space 13. A vector space 1 6(3)" _ [0(_1)k 119 15. Not a vector space 25. A vector space 4—, [ 3(3)‘ + 5(_1)’< . x4 = 62 ; 27. Not a vector space 29. A vector space 88571 _ _ x10 = ; the sequence {xk} has no limit 44288 Exercises 5.3, p. 373 and IIinl _) 0°. 1. Not a subspace 3. A subspace —3 —1 5. A subspace 7. Not a subspace 13. x. = —2(1)* 1 — 2(2)" 1 — 9. A subspace 11. Not a subspace -7 ~2 13. A subspace 15. Not a subspace k k 1 6 + 2(2) — 5H) 17- pm = —p.(x) + 31220:) - 2mm 5(—l)* 2 = —2—2(2)* —10(-l)“ : 19. A =(-1 —2.r)3. +(2+3x)Bz+xB3 —33., 2 14 + 40)" — 10(—1)* x arbitrary 33 2049 21. cos 2x = — sin2 x + cos2 x X4 = _44 ; X10 = —2060 ; 23. W = Sp{l,x2} l l 0 68 4100 25. In Exercise 2, W = Sp { [ :l , the sequence {xk} has no limit and llxk 1| —> oo. 0 0 0 2 l —201 000 000 15- x(t)=3321 —2e-' 1 v a [1] [1] [000][100][010] 0 —2 l 0 0 O 17. x(:) = -2 .. e2! _3 + e3' 2 [ 0 0 l :l} ; in Exercise 3, W = 2 l 0 16 3 S“:—1—11:l|:000:| P ~ . 21. =—.18; =— k 0' xk‘ll8(l)[|0]+ 000 [00 0 0 0 L(_18)k 10 . the limit is 16 3 [ ]' ; in EXCI'CiSC 5, “I = 118 ' -6 ’ 118 10 ' Sp{—l + x. —2 + x2}; in Exercise 6, W = Sp{l, —4x + x2}; in Exercise 8, 0 l 0 CHAPTER 5 W = Sp{1 — x2,x} 27. W = SpiBl, 82, E12. E13. 521. 523. 531, 532}. Exercises 5.2, p. 366 l- 0 —7 5 12 —22' 38 —l 0 0 —11 —3 —12 ’ —50 —6 —15 ’ Where 3, = 0 1 0 and 0 0 0 7 —21 28 —42 —7 -—l4 -1 0 0 3. e‘—23inx, ex—2sinx+3\/P—+—1, -2e‘—sinx+ 32= 0 0 0 3Jx2+l 0 O 1 5. c. = —2 + C3, c; = 3 — c3, C3 arbitrary 29. A = B+thereB = (A+AT)/2 andC = (A—AT)/2 Answers to Selected Odd-Numbered Exercises AN19 31. a) W = SplEn. E .3, E23}; 38. The set (u. v} is linearly dependent if and only if one of _1 0 0 _1 0 0 the vectors is a scalar multiple of the other. b) W = Sp 0 1 0 ' 0 0 0 , a) Linearly independent; 0 0 0 0 0 l b) Linearly independent; c) Linearly dependent; 0 —l 0 0 0 l d)L1nearly dependent; e) Linearly dependent 0 O l , 0 0 0 ; Exercises 5.5, p. 390 0 0 0 O 0 0 1- b) [Em 52:. 522, E“, 532, E33} is a basis for V1. [El]. En. £13, 522. E23, E33} is a basis for V2. 1 ' 0 0 0 1 c) dim(V) = 9, dim(V.) = 6, dim(Vz) = 6 e) W = Sp 0 0 0 , 0 0 1 , a” 0 0 0 0 0 0 0 0 3. V, n V2 = 0 €122 0 I athazpass 0 0 0 0 0 033 0 l 0 , arbitrary real numbers}; dim(Vl n V2) = 3 0 0 l 5. dim(W) = 3 7. dim(W) = 3 l 0 0 —l l 9. iii) The set S is linearly dependent. d) W = Sp 0 l 0 , 0 0 I , 11. ii) The set S does not span V. 0 0 0 0 0 13. iii) The set S is linearly dependent. 0 0 l l —l l 0 0 0 21.a)A= 0 1—2 :b) [5 21]T 0 0 O 0 0 l 33. x. =—6a +5b+37c+15d; l 1 l x2=3a—2b—17c-7d; 23. A”: 0 1 2 ; a) [7(1): x3=—a+b+5c+2d; x4=2c+d; 0 0 l C = —1231+632 — 83 — B4; D=SB,—3Bz+B3+B4 6+llx+7x2; b) p(x) = 4 +21: —x2; c) (10:) = 5+x; Exercises 5.4, p. 386 d) p(x) = 8 — 2x — x2 1' {[ _1 I],[ —l 0 :ll: —1 0 ]} ExercisesS.6,p.401 0 0 1 0 0 1 9. (my) = —3. m = J5. uyu = A 3. {512.521.522} 5. {1+x2,x—2xzf “x—YII=V l3 7. (M2, 11. (m) = 52. up" = 3%. qu| = 3J6. llP - (III = 2 9. [—9x + 3x2 + x3, 8x — 6):2 +x‘} 13.3)[2—1 3 2]T; b) [1 0—1 117; c) [2 3 O 0]T 15. Linearl y independent 13 15 17. Linearly dependent 17 19. Linearly dependent 21. Linearly independent . For (x, y) = xTy, the graph of S is the circle with equa- tion x2 + y2 = 1. For (x, y) = 4x1y. + xzyz, the graph of S is the ellipse with equation 4x2 + y2 = 1. .a.=7,a2=4 - q = (—5/3)P0 ‘ 5P1 — 4P2 19. P0 = l. p. = 16,172 = x2 —2. p3 = x3 — (l7/5)x, p4 = x4 - (31/7)x2 + 72/35 23- (mm. mm} 25- (Al. A2. A3) 25. p'(x) = (3/2)x2 — (3/5)x +1/20 27' 1'4 11 - 3F 27. p‘(x) z 0.84'1471pooc) — 0.467544p|(x) — 31. [a+b—2c+7d.—b+2c—4d,c-2d.d]T 0.4309201); (x) + 0.07882 p3(x) ANZO Answers to Selected Odd-Numbered Exercises 29. d) 730:) = 2x2 — 1, T3(x) = 4x3 — 3x, T401) = 8x4 — 8x2 +1, T5(x) = 16x5 — 20x3 + 5x Exercises 5.7, p. 410 1. Not a linear transformation . A linear transformation A linear transformation Not a linear transformation .a) 11 +1:2 +6x3; b) T(ao + 0.x + azxz) = (no + 2m) + (an + amt2 + (—a1 + am3 11. a) 8 +142: — 9x2; a b b)T(|: :|)=(a+b+2d)+ c d (—a+b+2c+d)x+(b —c—2d)x2 13. a) {2, 6x, 12x2} is a basis for 720'). b) Nullity(T) = 2 e) T[(ao/2)x2 + (01/6)X3 + (dz/12)x“] = 00 + a1x + a2}:2 15. N(T) = {a0 + a,x + azxzz a0 + 2a. + 4a: = 0}; R(T) = R1 17. b) M1) = [9}; 73(1) = V 19 rank(T) 3 2 l 0 ' nullity(T) 2 3 4 5 T cannot be one—to-one. rank(T) 3 2 1 0 21. nullity(T) 0 1 2 3 RU”) = ’P; is not possible. 27. b) Nullity(T) = 0: rank(T) = 4 sorry-u Exercises 58, p. 418 1. (S + T)(p) = p’(0) + (x + 2)p(X); (S + T)(x)= 1+ 2x +x2; (S + T)(x2) = 2x2 + x3 3. (H 0 DUB) = 1906) + (x + 2)p’(X) + 2p(0): (H o T)(x) = 2x + 2 5. b) There is no polynomial p in P; such that T(p) = x, so T"(x) is not deﬁned. 7. T"(e") = e"; T"(e2") = (l/2)e2‘; T"(e3") = (1/3)e3‘; T“(cze“r + be” + as”) = ae" + (lb/2)?”r + (c/3)e3" 9. T"(A) = AT 0 b ll.c) TO: (1 ]) =a+bx+r:.7:2’-'i-d.\t3 c Exercises 5.9, . 1. OOOOO @0000 1 O 0 O 0 GOO-— GOO—- OONO OUJOO booo do a. 9. a) [p15 = . a2 as 2a0 ao + 20. [T(p)]c = a. + 2a2 01/2 0 0 01/3 13. a) Q = COO— 0—OO OOH-O —069 15. 3 6 3 3 —l -1 0 0 OWOD l7. Door—N COO—'N OOH 00—N— OOOHN u—mo Ov—NOO OONAO AONO #NOOO 0930100 #00000 ...
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