MATH 218 Problems Solution Manual Chapter 3.2, Fall 2011

64 Chapter 3 Vector Spaces [ited LabedeLtade[addta Jon 22. a. Since the additive identity is : . b. Since the other nine vector space properties also hold, V is a vector space. L+t 1+ 0¢ 1 1 l+a : 2 |. Letu= | 2—a |. Then the additive inverse ig —u = 3 3+ 2a l-a I+¢ 1+ 0t 2+a |. b. Each of the ten vector space axioms is satisfied. ce. 0@| 2—t 2—0t 8-24 3+ 2t 3+ 2(O)t 24, Since S'is a subset of R? with the same stan- 25. Each of the ten Yector space axioms is satis- dard operations only vector space axioms (1) and fied. (6) need to be verified since the others are inher- ited from the vector space R°. If wy and we are in S, then let wi = au-+bv and wg = cu+dv, Then Wi + Wa = (a+c)u+(b+d)v and k(au + by) = (ka)u + (kb)v are also in S. 26. The set S is a plane through the origin in 27, Each of the ten vector space axioms is satis- R°, so the sum of vectors in S remains in S and a fied. scalar times a vector in S remains in S, Since the other vector space properties are inherited form the vector space R°, then $ is a vector space, c. Since 0© é = Ot = 8 , and : is the additive identity, then 0@v = 0. 23. a. The additive identity is 0 i i il wor 28, a. Since cos(0) = 1 and sin(0) = 0, the additive identity is A . The additive inverse of costa sinty . | cos(~ti) | _ [ cost; ; . conantic a Lon eran . is sin(~t1) = ~sinty | b. The ten required properties hold making V a vector space. c, The additive identity in this case is 0 not in V, so V is not a vector space. 29. Since (f +9)(0) = f(0) +9(0) = 1+1 = 2, then V is not closed under addition and hence is not a vector space, 30. Since ¢@ (dO f)(w) = c@ (f(a +d)) = f(a -+e+d) and (ed) © f(a”) = f(e + ed), do not agree for all scalars, V is not a vector space. 31, a, The zero vector is given by f(x +0) = a? and —f(@+t) = f(«@- 71). b. Each of the ten vector space axioms is satisfied. . Since cost and sint are not both 0 for any value of t, then , is Exercise Set 3,2 A subset W of a vector space V is a subspace of the vector space if vectors in W, using the same addition and scalar multiplication of V, satisfy the ten vector space properties. That is, W is a vector space, Many of the vector space properties are inherited by W from V. For example, if u and v are vectors in W, then they are also vectors in V, so that u@®v = v @ u. On the other hand, the additive identity may not be a vector in W, which is a requirement for being a vector space. To show that a subset is a subspace it is sufficient to verify that if u and v are in W and cis a scalar, then u + cv is another vector in W. 3.2 Subspaces . 65 For example, let g— at Ws t s,tER>, stt . 0 which is a subset of R*, Notice that if s = ¢ = 0; then the additive identity | 0 | for the vector space R® is 0 8.— 2t a— 2b also in W. Let u = t ,and v = b denote two arbitrary vectors in W. Notice that we have s+t a+b to use different parameters for the two vectors since the vectors may be different. Next we form the linear combination 8 Qt a~2b u-+cv = t +¢ b s+t a+b and simplify the sum to one vector. So 8 — 2t a— 2b (s — 2t) ++ (a — u+cvs t +c b = t+b s+t a+b (s +t) + (a+b) but this is not sufficient to show the vector is in W since the vector must be written in terms of just two parameters in the locations described in the definition of W. Continuing the simplification, we have that (s +a) ~ 2(t +b) u-+ev = . t+b (s +a) + (t +b) and now the vector u+ ev is in the required form with two parameters s--a and t+). Hence, W is a subspace of IR°. An arbitrary vector in W can also be written as s— Qt | 1 "= Q t =e] 0] +t] 1 s+t | 1 1 1 [ —2 and in this case | 0 | and 1 are linearly independent, so W is a plane in R°, The set W consists of 1 1 : 1] -2 all linear combinations of the two vectors | 0 | and t , called the span of the vectors and written 1 1 1 ~2 W = span oO], 1 1 1 The span of a set of vectors is always a subspace. Important facts to keep in mind are: ¢ There are linearly independent vectors that span a vector space. The coordinate vectors of R3 are a simple example. , ¢ Two linearly independent vectors can not span-R°, since they describe a plane and one vector can not span IR? since all linear combinations describe a line, 66 Chapter 3 Vector Spaces © Two linearly dependent vectors can not span R?. Let S$ = span { : , 3 } _ If v in in S, then there are scalars c; and cg such that veo[?]+a[ 3 ]=e[?]+0(-2[7]) -to-20[?] and hence, every vector in the span of $ is a linear combination of only one vector. 2 e A linearly dependent set of vectors can span a vector space. For example, let $ = { 1 ; : ; |}: 0 3 Since the coordinate vectors are in S, then span(S) = R?, but the vectors are linearly dependent since isJeelo]+9[t} In general, to determine whether or not a vectorv = | , | is inspan{uy,..., ux}, start with the vector , Un equation CyUy + CgUg + +++ + ChUK = V, and then solve the resulting linear system. These ideas apply to all vector spaces not just the Euclidean spaces. For example, if S = {A € Moxa| A is invertible}, then S is not a subspace of the vector space of 1 0] 0-1 | [3 ’ + ; : |= 4 a ; which is not invertible, To determine whether of not ; 7 | is in -1 0 12 0 1 | end | 11 fi 27, f -1 Of _ 7 2 -1 C1 — eg 2c, | | 2 ~1 alg 1 | +e 1 t}=[3 ‘| eg ate l=[i 1 |: The resulting linear system is c) ~ cz = 2, 2c, = —1,c9 = 1,¢c1 + cg = 1, is inconsistent and hence, the matrix is not in the span of the other two matrices, all 2 x 2 matrices. For example, the matrices and ‘ : | are both invertible, so are in S$, but the span of the two matrices , start with the equation MF Solutions to Exercises 1, tet | 9 | ana | ? Y Y2 | be two vectors in S and ca scalar. Then : +e 0 = 0 is in Y2 Yt S, so S is a subspace of R? ye Yi + cya (3/2)? +02 =9/4>1, 0 5, The set. is not a subspace of R?. If u= 1 2. The set $ is not a subspace of IR?, If u = many -( ; |. then u-+v = a ES. 3. The set $ is not a subspace of R*. If u = 2 v= ls y, thena+v = | , ES. 4, The set S is not ‘a subspace of R?. If u -|5 ] v= : , then u+v = 82 ¢ S since and ¢ = 0, then ev = , ¢ 8. set Sisas “i z y |_f tty Jog 6. The set S is a subspace since 30 + | ¢ 3y |- E 3(a + cy) és, 3.2 Subspaces ; 67 7. Since Cea Yt v1 + Cyr % | +e] yo | = | wo+ cya x3 Ys 3 + CYy3 and (a1 + cy1) + (a3 + ey3) = (#1 +23) + c(y1 + y3) = -2(c+1) = 2 if and only if = —2, S is not a subspace of R°, 8. Suppose 212223 = 0 and yiyays = 0, where 21 = 0, y3 = 0, and all other components are nonzero. Then (x1 +41)(@2 + y2)(%3 + ys) 4 0, so S is not a subspace. y 9. Since for all real numbers s,t,c, we have that 3 — Qt a — Qy (8 + cx) ~ 2(¢ + cy) 8 +e wv = s+ cx ; t+s ytau (t + ey) + (s+ ex) is in S, then S' is a subspace. x1 YL e+ yt 10, For any two vectors in S, we have | 2 | +] 2 |= 4 , which is not in S and hence, S is 3 ys 2+ Yy2 not a subspace. 11, If A and B are symmetric matrices and ¢ is a scalar, then (A+cB)* = At+cBt = A+cB, so S is a subspace. 13. Since the sum of invertible matrices may not be invertible, $ is not a subspace. 15. If A and B are upper triangular matrices, then A+ B and cB are also upper triangular, so S is a subspace, : 17. The set S is a subspace, 19. The set S is not a subspace since 2° — x = 0, which is not a polynomial of degree 3, . 21. If p(w) and q(x) are polynomials with p(0) = 0 and q(0) = 0, then (p + q)(0) = p(0) + 9(0) = 0, and (cq)(0) = eq(0) = 0, so S' is a subspace. 23. The set S is nota subspace, since for example (20 + 1) — (w? + 1) = 2”, which is not in S. 12. If A and B are idempotent matrices, then (A+B)? = A? + AB + BA-+ B? will equal A? + B? = A+B if and only if AB = —BA, so S is not a subspace. 14. If A and B are skew symmetric matrices and cis a scalar, then (A -+cB)' = A'+cBb=—A+ c(-B) = ~(A+ cB), so S$ is a subspace. 16. If A and B aré diagonal matrices and_c is a scalar, then A + eB is a diagonal matrix and hence, S' is a subspace. 18, If A = [: i |B = [¢ {| suet cod gi ih that a+c= 0 ande+g = 0, then A+ B= ate b+f ih c+g dt+h “ (a-+e)+(d+h) = (at+d)+(e+h) =0 and hence, S' is a subspace. 20, The set S' is not a subspace’ since (a? + 2%) — 2? = a, which is not a polynomial of even degree. 22, Yes, since ax? +/2(bx?) = (a + cb)a? € S. 24, The set S is a subspace (assuming the zero polynomial is in the set), 68 Chapter 3 Vector Spaces 25. The vector v is in the span of $ = {v1,v2,v3} provided there are scalars ¢1,¢2, and cy such that V = civi + C2V2 + c3vg. Row reduce the augmented matrix [v1 va vg | v]. We have that 1 -1 -1) 1 1 -1 -1) 1 1 -1 2 |—1 | reducestg| 0 1 0 1 ; 0 1 0 ] 1 0 0 3 ]-2 and since the linear system has a (unique) solution, the vector v is in the span. 26, Since —2 -~3 |, 9 the linear system has a (unique) solution and hence, the vector v is in the span, 27. Since 1-1 -1|-2 1-1-1) 1-1 2) 7 | reducestg} 0 1 0) 0 it 0 | -38 0 0 3 1 0 1 |-2 10 1 |-2 1 1-1/1 01 -2) 3 “arcs ; 0 2 —4! 6 reduces tg 00 0fo |? the matrix M is in the span, -l 1 -3] 5 00 0 0 28, Since 1 ou]. fio 4.44 1 1-1} 1 Jo sneos to | 2 2 72/0 0 2 -4] 2 | MEME To o oO fa]? -l 1 ~3} -3 00 070 the linear system is inconsistent and hence, the matrix M is not in the span. i 29. Since , c1(1 +2) +.¢2(n? — 2) + 03(8x) = 2a? — 6x — 11 if and only if (1 — 2¢2) + (ce, + 8¢3)a + con” = 2a? — 62-11, we have that c) = —7, cy = 2,¢3 = 3 and hence, the polynomial is in the span. 30. Since ea(1 + @) + c2(2? — 2) + cg(3a) = 8x? — 2 ~ 4 if and only if (cr — 2e2) + (er + 3e9)@ + cpa? = 3a® ~ 2 —4, we have that ce, = 2,c2 = 8,c3 = —1 and hence, the polynomial is in the span. 31. Since 2 1a -1 3 b + a —1.°3-1 0 | reduces to] 0 7)a+2b |, then span(S) = b a+c=0>, -2 —-lle 0 O}.ate c 32, Since . 12. 1a 12 1 a , a 1 3 2 | 6 | reducestg}] 0 1 1 —a+b , then span($) = b —5at3b+c=0>. 2 1 -1le 0 0 0] -Sa+3b+¢ ; e | | : 1 2 1-1 b | 33. The equation c; 1 0 + CQ 0 “y = ‘ 2 , leads to the linear system ¢ +c, = a, 2c, — cy = | b,c: = ¢,¢g = d, which gives span(S) = { ab oa? abe Rh. +b 2o=b 3 3.2 Subspaces 69 34, span(S) = { as ac ws > + ia fl fo) hanes 36. Since -4 2 2i/a 0 ~-1 1/6 1 0 lie -4 2 2 a reduces to} O -1 1 b : 0 0 2le+ja+t 4d the span is all polynomials of degree two or less. 38. a. abe x} a span(S) = b | 3a—b b, Since 2v1 — v2 = vg, the set S is linearly dependent. 1 -1 0 2 0 1 1 3-1 40. a. Since 2 L 35. span(S) = {ae* +be+¢e| a—c=0}. 37, a. T a span(S) = , b | a,be e} ; b, The set 9 is linearly independent. 39. a, span($) = R? b. The setS' is linearly independent. a 1-1 0 2 | a b | reducestoe| 0 2 1-3) ~-2a+b ; ¢c 0 0 -1 5 [8a~2b+e | every vector in R® is a linear combination of the vectors in S and hence. span(S) = R°. b, The set $ is linearly dependent since there are four vectors in R°, 41, a, span(S) = R° linearly dependent and span(T’) = R°, R’, 42, a, span(S) = { 3 b. The set S is linearly dependent, d, The set H is linearly independent and we still have span(H) = c, The set Tis also a,d,c€ R} b. The set S' is lmearly independent, c. The set T is also linearly independent and span(T’) = Moy. 43, a, span(S) = Pz b, The set S is linearly dependent. c, 222 +3245 = 2(1) — (w — 8) + 2(a? + 22) da The set T is linearly independent and span(T’) = P3. 44, a, Since 2814 — ty 28 — te $1 , 82 ti te te Sy 82 28 —t then S isa subspace. b. Since = 8 —s 2 —1 c. Since the vectors ; d. S¢ Rt 2(81 + 82) — (t1 + ete) _ 81 + C8 = ti + cla €§, —(81 + 82) 2 ~1 2 -1 1 0 : 1 0 0 +t l , then S$ = span 0 ; 1 = 0 : ~1 0 70 Chapter 3 Vector Spaces 45. a., b, Since 8 -1 0 "1 0 s~5é¢ | =s} 1 | +¢] -5 |, then S=span 1 },}] -5 : 28 + 3t 2 3 2 3 Therefore, S is a subspace. c, The vectors found in part (b) are linearly independent. d. Since the span of two linearly independent vectors in R® is a plane, then $ 4 R® 46. a. The subspace S$ consists of all matrices ‘ such that ~a—2b+¢+d=0. b. From part (a) not all matrices can be written as a linear combination of the matrices in S and hence, the span of S' is not equal to Mzx2. _c. The matrices that generate the set S are linearly independent. 47. Since A(x + cy) = 1 +e 1 =|! if and only if c= 0, then S' is not a subspace, 2 2 2 48, If u and v are in S, then A(u+ cv) = Au+cAv =0+0=0 and hence $ is a subspace, 49. Let B,, Bg € S. Since A commutes with By and Bo, we have that A(By + cB) = AB, +cABg = BjA+ ¢(B2A) = (By, + cBg)A and hence, By + cB, € S and S is a subspace. 50. Let wi = ur + v1 and we = uz + v2 be two elements of S, Let c be a scalar. Then Wi + cW2 = U1 + Vi + C(Ug + V2) = (uy + cug) + (vi + eva). Since S'and T are subspaces, then uy +cug € S and v1 + eve ET. Therefore, wi + cwe € 9 +T and hence, S+T is a subspace. 51. Let w ¢ $+ 7, so that w = u+v, where u € S, and v € T. Then there are scalars C1y+++yCm and di,...,dy such that w = an Guy + YL, divi. Therefore, w € span{uy,...,Um,V1,...,Vn} and we have shown that $+ 7 C span{uy,...,Um,Vi,...;Vn}- Now let w € span{ui,...,Um,Vi,..., Vn}, 80 there are scalars c1,... +Cm4n such that w = cpuy +e + CmUm + Cm41V1 + +++ + Cm4nVn, which is in $+. Therefore, span{ui,:..,Um,Vi,..- Vn CS+T. 52. a. Since a aly d -d}_[a+kd -a~kd])_ [| a+kd -(a+kd) b oe e ff bt+e ct+f | [i b+e c+f is in S, then S' is a subspace, Similarly, T is a subspace. b. The sets $ and T are given by _ 1-1 0 0 0 0 _ 1 0 0 1 0 0 sommn({ SELL a) [8 $]}e-omm(l s ShL8 8hL ED) 5+1=span{| fi -1]_f 1 0],fo 0] _fo1]. But | j 5 |=[-1 otis o|-[o 5 | 0 1 so on | or ws —— Bo o° uo — oo Ko os — i en oo os —s oo QOr ws a,