Question

5. Give an example of a basis for R3 which does not contain any of the vectors 00-0 and Justify your answer. 6. Show that {1, x - 1, (x - 1)2} spans P2 by showing that its span contains (1, x,x2}.

Answer 1

1 6 Let u,= (1,2,0) V₂ = (0,1,1) & V = (1,0,3) be vectors in TR² We show {v, U₂, Vaz is a basis of R² first we will show these are linearly independent. for some scalars L, BYEIR we have Lu, + B + ruz=0 (1,2,0) + B(0,1,1) + 8 (1,0,3) = (0,0,0) > (+r, 2x +B , B + 3%) = 10,0,0) Xtr=0_1 r = -2 2x+8 = o-li ß=-22 so putting these inli) B + 38 = otill - 24-32=0 - 5x=0 B = -24 = 2x0=0; (B=0 r = -2 = -0 ; [8=0 lie. all of x=8=8=0 ; 80 V, V2, V3 is linearly independent, Next, we have to show {u,, U2, Uz } generates! TR lie. for (X,Y,Z) ER & Scalars 2,8,8 EIR (X,Y,Z) = 2 (1,2,0) + B 10,1,1) + 8 (1,0,3) - 0 (x,y,z) = (x+r, 2x+B , B +30) Ltr=x r= x-2 22+8=y » ß=4- 24 B+38=2 with above two (4-22) + 3 (2-x)=2

y-22 +38- 32 =2 -52 = Z -Y-3 a = + (3x+4 75 ) 3x+4-2 B=9-2x = y 2 (3x+7-2). 54-62-24 +22 -6x+3y+22 & r = x-x= x -3x+Y-Z - 5x-(3x+7-2) 5x-3x-y+Z 5 re- 2xy +Z Hence egh (1) can be written as (28,2)= 32+9-2) (12,0) + (-6%+34 +22) (0,,1) + (2x-9 + 2) (1,0,3) Thus {u, U2, U3} generates IR AS {1, ₂, V3} are linearly independent and generates R² so it is basis of R², Also it does not contains e,= (1,0,0) =(0,1,0) & ez = (0,0.1).