MATH 218 Problems Solution Manual Chapter 3.1, Fall 2011

3.1 Definition of a Vector Space : 61 and since (21 +a) — (y1 + y2) + (41 + 22) = (wi — ya + 21) + (2 — yo + 2g) = O+0 = 0, the sum is also in V. Similarly, cu is in V for all scalars c. These are the only properties that. need to be verified to show that V is a vector space, @ Solutions to Exercises 1. In order to show that a set V with an addition and scalar multiplication defined is a vector space all ten properties in Definition 1 must be satisfied. To show that V is not a vector space it is sufficient to show any one of the properties does not hold. Since vy 22 Ly — to yo} @}] yo |=} ww ye zy 22 21 2Q and . Bip) By Bq —- 2 yo} OB) Mw | =] rw 22 BY 24 BY do not agree for all pairs of vectors, the operation ® is not. commutative, so V is not a vector space. 2. Notice that since scalar multiplication is defined in the standard way the necessary properties hold for this operation. Since the addition of two vectors is another vector, V is closed under ‘addition. Consider vy Y wm +y1— 1 : cx, + cy1 ~e co 22 1 @0] ye =c® wg+yo2-1 F{ c@g + cy2~-¢ &3 ys w3 + y3 ~ 1 ; cuz + cys —¢ but ay YI | ct Cy cx, + ey ~ 1 CO] ®2 |} OcO] yo | = | ex. | © | cy | = | cagt+eyo—-1 |, &3 Ys erg Cys x3 + Cyg — 1 so the expressions c@ (u @ v) and c®u@c@v do not agree for all pairs of vectors and scalars c. Therefore, V is not a vector space. , 3. The operation ® is not associative so V is not a vector space. That is, wy ‘| ae 23 421 + dae + 2a yi | ©} yy ®] ys | =|] 441 + 4ye + 2y3 2 2g |. £3 | 421 + 422 + 228 and - vy XQ x3 221 + 4aq + 4arg yw | ® y2 | @] ys =| 2y1+4y2+4y3 |, can 22, 23 221 +429 + 42g which do not agree for all vectors. 4, Since : vy ZQ T+ yt try te cO yi | @} yw =cO B+ Yo = ; zy 42 t3 + Y3 . but By Yi wy + yi + 2c CO] 2 | ©cO] yw | = wyry2e |, x3 ya ®+Yg 2 soc@(u+v) and ecO@u@cOv do not always agree. Chapter 3 Vector Spaces Yi + 2 (h]+E D8] @1 + LQ + 2&3 = and y+ ye + Ys ey x £3 le Le) 5. Let u = ° and —u = “* . Then u + y —Y (-u) = -u+u=0. rele R])- Yi Ye 21+ & o(a1 + te) yi + Y2 e(y1 + y2) = cay + cha =O "1 +e v2 . cy + cya VA Ya a e(a[2])=e[ &]=[ 2] ~eo I t+ 22 + 2g y+ Yo + y3 Ml 6. a b e f|_[ate b+f | [é alts i ]= [2 ah] em Moxa. 3, Since the associative property of addition holds for real numbers for any three 2 x 2 matrices My, + (Mo + M3) = (My + Mo) + M3. 5. Let u = [: i | and —u = |= - |: "Then u + (=u) = =0, . ab e f male ale[s af) =R| ore b+f |] _| kat+ke kb+kf ~ e+g d+h]|” | ke+kg kd+kh _ a b e f =e oo] eel e i]. 9. Since the real numbers have the associative property, LS al) ols a]. [2](2]- YL Yo 1+ BQ _ | % + ey _| % 4 Ly Yt Yo Yar V1 Yo yi 4, tet o=[ 9 | ond u= [9]. Then 0+u = 0 y u+0O=u, 6B.c ° = on | is a vector in R?, «ool 5] [65a] [54 ]- . a b ef | [ate b+f 22 ils Noles avn _|eta f+ “| gte h+d 4 c a ¢ 4, tet 0 =| } ene | i ‘|: Then O+u=u+0=u, 6.6/8 i f= [i kb cd ke kd is a matrix in Moxg. 8, Since real numbers have the distributive prop- . abl a b a b cen thed] eo gfe {eg ]ae[ sa. 3.1 Definition of a Vector Space 63 7. Since (c+ d)@ na Y y x z}| | @+e a+d |_| 2e+e+d eo[ 5 J+eo[ 5] =| y J+| y | 2y i |- [eter = does not equal Yy. for all vectors ) , then V is not a vector space, . . a c ate 8. a. Since | b | +] d | = | b+d |, the sum of two vectors in V is not another vector in V, then 1 1 2 V is not a vector space. b. If the third component always remains 1, then to show V is a vector space is equivalent to showing R? is a vector space with the standard componentwise operations. . 10. The set V with the standard operations is a vector space. 9, Since the operation @ is not commutative, then V is not a vector space. 12, The set V with the standard operations is a vector space, 11. The zero vector is given by 0 = ; . Since this vector is not in V, then V is not a vector space, 13. a. Since V is not closed under vector addition, then V is not a vector space. That is, if two matrices from V are added, then the row two, column two entry of the sum has the value 2 and hence, the sum is not in V, b. Each of the ten vector space axioms are satisfied with vector addition and scalar multiplication defined in this way. 14, Suppose A and B are skew symmetric, Since (A+B)! = Ab + Bl =—-A~B=-(A+B) and (cA)! = cA’ = —(cA), so the set: of skew symmet- ric matrices is closed under addition and scalar multiplication, The other vector space properties also hold and V is a vector space. 16. Suppose A and B are symmetric, ‘Since (A+ B)' = Ab+ Bt = A+B and (cA)! = cA! = cA, so the set of symmetric matrices is closed un- der addition and scalar multiplication. The other vector space properties also hold and V is a vector space. 15. The set of upper triangular matrices with the standard componentwise operations is a vec- tor space. 17. The set of invertible matrices is not a vector space. Let A= J and B = —I. Then A+ B is not invertible, and hence not in V. 18. Suppose A and B are idempotent matrices, Since (AB)® = ABAB = A?B? = AB if and only if A and B commute, the set of idempotent matrices is not a vector space. 19. If A and C are in V and k is a scalar, then (A+ C)B = AB + BC’= 0, and (kA)B = k(AB) = 0, so V is closed under addition and scalar multiplication. All the other required properties also hold since V is a subset of the vector space of all matrices with the same operations. Hence, V is a vector space. 20. The set V is closed under addition and scalar multiplication. Since V is a subset of the vector space of all 2 x 2 matrices with the standard operations, the other vector space properties also hold for V. So V is a vector space, : 1.0 0 1 A7!. b, If c= 0, then cA is not in V. Notice also that addition is not commutative, since AB is not always equal to BA, : 21. a, The additive identity is 0 = . Since A@ Aw! = AA7! = J, then the additive inverse of A is 64 : Chapter 3 Vector Spaces [otel=[ate)+ faze J= [setts Joero | the additive identity is 22. a. Since Oo 1 . b. Since the other nine vector space properties also hold, V is a vector space. c. Since 0© f = o | ms , , and , is the additive identity, then 0@ v = 0. 1+¢ 1+ 0¢ 1 l+a : 23. a, The additive identity isO = | 2]. Letu= 2—a |. Then the additive inverse is —u = 38 3+ 2a l-a L+¢t 1+0t 1 2+a |. b. Each of the ten vector space axioms is satisfied. c.0©| 2-t | = 2—0t = | 2 3 - 2a 3+ 2¢ 3+2(0)t 3 24, Since S'is a subset of R? with the same stan- 25. Each of the ten vector space axioms is satis- dard operations only vector space axioms (1) and fled. (6) need to be verified since the others are inher- ited from the vector space R°. If w1 and wg are in S, then let wy = aut+bv and we = cu+dv. Then Wi + We = (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 anda fied. scalar times a vector in S remains in 9. Since the | other vector space properties are inherited form the vector space R°, then S$ is a vector space. 28. a. Since cos(0) = 1 and sin(0) = 0, the additive identity is 1 0 is cos(—ti) | _ | cost, sin(—ti) | | sing, . The additive inverse of cost sin ty . b, The ten required properties hold making V a vector space. c. The additive identity in this case is ; . Since cost and sint are not both 0 for any value of ¢, then ; is not in V, so V is not a vector space. 29. Since (f +g)(0) = f(0) +9(0) = 1+1 = 2, then V is not closed under-addition and hence is not a vector space, 30. Since c@ (d@ f)(z) = ¢@ (f(a+d)) = f(w+e4+d) and (cd) @ f(z) = f(x + ed), do not agree for all scalars, V is not a vector space, t 31. a, The zero vector is given by f(z -+0) = #3 and —f(z +t) = f(a —t). b, Each of the ten vector space | axioms is satisfied, i 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 i verify that if u and v are in W and ¢ is a scalar, then u-+ ev is another vector in W,