Question

Math 407 Homework 4 Name: 1. Why the following sets are not vector space? with the regular vector addition and scalar multiplication. b) v = {(7: «y 20} with the regular vector addition and scalar multiplication. with the regular vector addition and scalar multiplication. 2. The set B = {1,1+t, t+t?} 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+21+342 3. Let H =Span{ői, üz.us) and B = {üs, üz, is). Show that B is a basis for H and X is in H, and find the B-coordinate vector for X, for 4. Find the change-of-coordinates matrix from B to C. 3= {6 - 1.6 - []} c={a-[-s]. -=[7]}

Answer 1

o ovo {[:] : 220,w20] het [!]«v, W 60[1-6] #v det U .: v scalar multiplication is not is not closed under a vector space But (: 343-420) 75 -26-21820) 0)+ [:) -[:] #v8:4412- 2 v is v is not not closed under regular addition a vector space but [:] cv and filer [!+[:] - [1] ¢ v Ep+7=2>] > V is not v is closed under addition, not a vector space.

- {1,176, they is a leais for P. PD) = 5+ 2t +3+ Consider 5+ 2+ + 3 = a.l + b (1+t) + c (t++~) = a.lt 6.1 tbit + cott c. ta = (a+b). 1 +(6+c) t + cota equating the coefficiente of powers of t, o a+b=5, b+c=2,c=3 b- 2-C = 2-3--| As a = 5-5 = 5-61) = 5+1=6 Co-ordinate vector of P(1) is =