Question

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Rather than use the standard definitions of addition and scalar multiplication in R3, suppose these two operations are defined as follows. With these new definitions, is R3 a vector space? Justify your answers. (a) (x1, Y1, 21) + (x2, Y2, 22) = (x1 + x2, Y1 + y2, 21 + 22) c(x, y, z) = (cx, 0, cz) O The set is a vector space. O The set is not a vector space because the associative property of addition is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the associative property of multiplication is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied. (b) (X1,Y1, 21) + (x2, Y2, 22) = (0, 0, 0) c(x, y, z) = (cx, cy, cz) O The set is a vector space. O The set is not a vector space because the commutative property of addition is not satisfied. • The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the multiplicative identity property is not satisfied.

Answer 1

Let us first give the definition of the Vector space: Real vector space. A non-empty set V is said to form a real vector space (or a vector space over the field R) if (i) there is a binary composition (+) on V, called addition, satisfying the conditions V1. a + Be V for all a, ße V V2. a + B = B + a for all a, BE V V3. a + (B +Y) = (a + b) + y for all a, b, ye V V4. there exists an element in V such that a+=a for all ae V . [Additive Identity Property] V5. For each a in V there exists an element -a in V such that a + (-a) = 0 and (ii) there is an external composition of R with V, called ‘multiplication by real numbers' satisfying the conditions V6. cae V for all ce R, alla e V V7. c(da) = (cd)a for all c, de R, all a e V V8. с(a + b) = ca + cp for all CÉ R, all a, BE V [ Distributive Property] V9. (c + d)a = ca + da for all c, d e R and all a e V V10. 1 a=Q, 1 being the identity element in R. [ Multiplicative Identity Property] The elements of V are called vectors and the elements of R are called .scalars. R is said to be the ground field (or the field of scalars) of the : vector space V. Now, Here V=R

(a) Here & and is defined as.. Car Win7i) + (x2, 423+2) = (a + 22, Yrt Y2, 4412) c. (M1, 4,71) = (cal, 9, C71) Now, we absence that en ona definition of vector space multiplicative identity property vio kill have to satisfy lie 1.1=2 - ie y con, 4,7) = (1,4,7) but by on definition (n, Y, 7) = (0, 0,7) & (m 4,7) Hence it is not a vector spare. :. Care, 41,71)-+(22, 4.72) =(0,0,0) (1,4,7) = (2x, (9,07) Now, Here are obseave that on our definition of vector space Additive Identity propeaty (V4) must be satisfied. de a some OER3 such that (240) = d VLERS But according to ona definition if L=(2, 4 z) and O=(a, b, c) then (2, 4, 7) + (a,h,() = (0,0,0) & (nyt) Hence it ŕ not a rectan space.

Caxy, 4, 71)+(23) Y3372)= (a +72+9, Yyt Yz49, 217849) (2347)= (ea, cy, (7) Here, we obscrue that an oni definition of eectan space Distributive propeaty (V8) mgst be satisfied 1. i c(2+B) = C&A (P. Now, if L= (cai, 4,7) 1 and ß= (22, 92,72), them e{c^4,4)+(29,10,12) se (rast + 9, yyt ** 9, 2148z+9) ( econemote), e curteta), e (7z+ 9)) and e(m1,4,7) = (em, cy,, (71) elmo, 45,72) = Camp, C42, (72) ce cas, 40170) + (,1,A2) = (24,, (4,021) +(690,(42) (9) (enje cmt 9, Cypt(449, - Cart (22+9) met e{{m, 4yan)f (10,495% ne (2AB) & CLA(B! Hence it is not a creatar space.. -

B : (d) (an, 4071) +(22, Yoytz)= (21+22+8, 4,4 42+8, 277278) - and emy 4,7)=(eat 86-8, C1+868, C7+86-8) Now, we obscrue that S2 =(Me 117) - v. for, &, BE R²7 2+B E R L *B=6m2, 72297 V2. &,BER? = 2#B=(21f3+8, 4,44248, 2148278) 8 tds (22+ 34,+8, 4+4+8, 724948) . Hence Laß= 8+2 xa,ßER - ns. If de Can 4,41), B2 (M2, 72, 72) and 82 (1334323) . then dot (B+8)= (1992)+(2+93 +8, Y2+ 73 +8, 72 473 +8) mtastast 16, Yyt Y2 + Yst 16, 2rt2zt3tle) Similarly 643748= Cape menztle, yyt 4+%416, 7+2zt36/6) 1. Hence 2+13+8)= (2+3718. vu. § 68,-8,-8) E R3 such that Ca, 4,7 +68,-8,-0) = (-8+8, 7-8783 078+8) =M, 4.72. So. O= (-8,-8,8) exist in R

vs. for each Cm, 47) ER3 .. (-x-16, -4-16,746) ER such the cart) & (64-16, -9.16, -7-16) = Cann-lont8, y-y-1678 Z-7-1678 : : 6.8,-8,-8)= 8 Hence incesse of Cm, 417) os fa-16, Y. 16, 7-16) VO ER3 . V6 N (M; 4,7) ER3 (1,4177= Con +86-8, Cy+868,C2+26-8) : ..CLAR3 NafR3 . 17. C{d (0,4,7)2 = c(dat8d-8, dy+8d-8, d2+8d-8) y = dyt8d-8 Lz; = dz78d-8] Eccm +86-8, 64, +86-8, 674+86-87 (MYANM= dat8d-8

Eredm+86d-86 +86-8, edy +862–8C+86-8, i e dz+862–80+86–8) -Ceant&cd – 8, cay +86d-89cd77867-8) zed (m1 417) Tec(dd) =((d)& XXAR3 vs. Let Le Can, 9171) B= (m2) 72,72) 276= (22,4 +8, 8,44248, 244?2+8) So, c[d+1) = (cat(9378578678, Cyt(Y2786 +86-8, 0744 (72+84 +86-8) -Cemet.cm2 +8, Cueron , Copertinente (2= (cm, +86-8, C4, +868, C7, 48208) CB = cena +86-8, cod 808,672+86-8) Chec BP (cmpe cine +166-8, even C2+672 +166-8) - = C(2+B) fra ③ and 4

vg. (Ctd) (M, 47) = ( Ctd)n +8Cced)-8, (c+d)9+ 8(6+)-8). (+d) z4+ 8 (h+d) ~8)-D eCm, 4A) = Cent86-8, C9+86-8, (7-18 (-8) d (m17) = (ant8d-8, dy f8d-8, 47-782-8) C (79;47) 40,2)=Ctdn +8 Cetd)-8, <td)+96e4d9–8, (Std)ą = (Ctd) Cm, 47). such ther VID 1ER 19,47) (1.778.648, 4y+8-1-8.17 78-8) = (1/417) Hence it satisties all the ten properties Hence gt ha ventan space.