Question

i want answers of all Questions

Example. As another special case of examples we may regard the set R of all of n umber vector 1.4.6. Example. Yet another al l the vector space M of mx matrices of members of where m - NI. We will use M. horthand for M F ) and M. for M.(R) 1.4.9. Exercise. Let be the total real numbers. Define an operation of addition by y the maximum of u and y for all , VEV. Detine an operation of "scalar multiplication by a Dar for all a ER and eV. Prove or disprove: under the operations and the set V is a vector space 1.4.10. Exercise. Let V be the set of all real numbers ar such that I > 0. Define an operation of "addition" by TEX for all 1, y EV. Define an operation of "scalar multiplication" by for all a € R and EV. Prove or disprove: under the operations and the set V is a vector space. 1.4.11. Exercise. Let V be R?. the set of all ordered pairs ( y) of real numbers. Define an operation of "addition" by (u, v) ( y) = (u+1+1, + y +1) for all (ti, c) and (,) in V. Define an operation of "scalar multiplication" by a (x,y) = (ar, ay) for all a E Rand (I, EV. Prove or disprove: under the operations and the set V is a vector space. 1.4.12. Exercise. Let V be the set of all n x n matrices of real numbers. Define an operation of "addition" by ABB = (AB+BA) for all A, BEV. Define an operation of "scalar multiplication" by abA=0 for all a ER and AEV. Prove or disprover under the operations and the set V is n vector space. (If you have forgotten how to multiply matrices, look in any beginning linear algebra text.) 1.4.13. Proposition. If is a rector and a is a scalar, then ar = 0 if and only if a = 0 or r=0. In example 1.1.9 we saw how to make the family of equivalence classes of directed segments in the plane into an Abelian group. We may also define scalar multiplication on these equivalence classes by declaring that (1) if a > 0, then a PO = PR where P, Q, and R are collinear, P does not lie between Q and R, and the length of the directed seginent (PR) is a times the length of (P.2);

Answer 1

Ex 1.4.9 V-R, Hly maximum of x andy we will show this is not a group under 1 to conclude is not a vector spare for v to be an abelian group, assume e to be the identity Hence e Hy = y for all y eu definition of identity but e le-1 = maximum of e ande-1 So e He- e fet, Hence by contradiction, there is no additive identity inv. So is not a vector space under I and I Ex 1.4.10 VEIRO & [ily = xy , & Ex= xa we will show firstly v is an abelian group under a) associativity x [y] Z) = (x479) #2 = xyz 6) Identity is 1 as a ] 1 = x= 1 [Ex c) inverse is at as an t-aital d) commutattity holds as Ely = xy = 4 x = y a Now we will show distribitutivity a scalar multipulation het x, y EV, KER & A (x Hy) = x (xy) = (x y) = xx ya = xEx. & Ely = (x Ex) ] & Jy) Now let XEV, X, BER so (+B) x = xa+ pXXX X (REX) Aloso 1 Ex=x'=x and (LB) [] X = XFR = (x3) K & [] (REX) Hence vis a vector space under and Ex 1.4.11 v= PR² (uu) [h (x, y) = (1+x+1, 0+ 4+1) (y) = (xx, y) Let & = 2, so 2 7 ((u, v) (ay)) = 20 (U+x+1, u xy + 1) = (24 +2x 42, 24424 But [2000] D [20 (x,y) = (20,2V) (2x, 2y) = (u+2x+1, 2v424 + ) so they are not equal [distributivity fails] Hence Vis not a vector Space

EX 1.4.12 het V = Mn (IR) A HB = (AB+BA) & D A =0 In denote the identity matrix of order with Then AE In 1 (A. In + In A AS A . In A=In A = f (A+A) = ₂ x 2 A A for all A] Hence In has to be the identity, so for it to be a group is every element must have an inverse Let x be the inverse of zero matrix = 0 = 10 - so * In = x [110 = √(x 0 +0.x) = 0 = 0 a contradiction . So o has no inverse - Vis not a group so v is not a vector space under add [I and I