Question

1. Why the following sets are not vector space? with the regular vector addition and scalar multiplication. a) V = {E: * > 0, y 20 with the regula b) V = {l*: *y 2 o} with the regular vector addition and scalar multiplication. c) V = {]: x2+y's 1} with the regular vector addition and scalar multiplication. 2. The set B = {1,1+t, t + t2 is a basis for P, the set of all polynomials with degree less than or equal to 2. Find the coordinate vector of p(t) = 5 + 2t + 3t2. 3. Let H = Span{ s ) and B = {1, U2, U3}. Show that B is a basis for Hand X is in H, and find the B-coordinate vector for X, for L-1 vi = 2 = 1, 1 , X = 03 = 1 4 (1) 4. Find the change-of-coordinates matrix from B to C. 1-31 C= C1 = T11

Answer 1

1 @ Vs Sri 7:«>, o. y so? det e ER c Love V, v [] then. evare & v so, Vin at closed anden scator multiplication leny v ={[%]:2y30 } then Blev [4] EV beat []+[4] = ¢v W v=$1[*] : 314517 [6] &V (JEV but [6] +6] - [ ] does not belongs to v Hence they are not vector Spaces. (2) 5+27 +3+* = 6(1) – 1(1+t) + 3(++) Hence Co-ordinate of PCA) = (6,-1,3) - (3) net ei vitez V2 + C3 Uz zo 349+48]+4/8) [3] Ci 20 - ei te, +6=0 ci-ez tez =0 = C2 20 € 20

Henee şv., V2, Vzf is lineanly independent Hence B= { , 9,9; } forms abosis for H now, 7 +01 Lt Hence Bevordinate welon -form X O set [6] = ai [$]+61 [3] >> a,-261 37 -521 +261 =5... 2013 3) 2,-3 ,61=-5. 1 + ] So, [81= -3 [s] -5 [?] Similarly B*] = [s] +2 [3] so, Change of co-ondinate matrix 2 Similan pal 1 -5 21
if you have any doubt please comment