Q3. Consider the vector space P, consisting of all polynomials of degree at most two together...

23.

Where Griven S = {Pich), B(A) 13,6}. Pi (e) = -t 45 = ~34² - 30 +5. P₂ (t) a Let us Considue the equation GP, (d) + C2 P ₂ (2) 20, a, (2 tlR. (-4'45) + CR(-34-34 +5) - (-4-3 (2) t² +(-3 (2) t + equating to efficient (5 4 + 5C, of both sides we have, - G-362 - 3 C2 = -> (220 59 th (2=0. from 6 & 6, pulling (2=0 we get, G=0 ire G=0 56=0 te G-0.

& G b, (A) + (2 P₂ (t) = 0 7 G = (2=0 = {2, (2) Bl} is linevely independent.

over The dimension of the rectorspace P2 IR is 3 as it's basis is { 1, t, th} also since s is lineavily independent, of s spans P2 dim of P₂ over IR will be a which is not possible, then s cannot span P2 yes , Let W = span{ b, (4), P, (t)} Then dim W = 2. By the extense on theorem we can extend the linearly independent set 26, (t), B (t)} to a basis of P2 there must enest a polynomial P₂ (t) in P2 sit ?P, (t) , B (t), P₃ basis for the vector space P2. ie is a B (A) is L:1R² IR? defined by - 342 un-242 U₂ - 4 EIR² & atir pet, 2 " ) 12 Uz utare, w L Then, Uqtare, L ta U2 = ~3 (Uq+drez) (utory) - 2 (u24xV2) (Uztaro) - (uitare),

MODUL -34₂ + 2 (-302) (u, -21.) + a(, -213) (uz -U1) + 6 (-21 -BU2 -3202 u, 2u₂ 42 201-2202 2 Uz-u nel 2 L Uz + x L ( 2 ) 2 :. L is a linear transformation.

6 (3) 0) 00:04 4()6) (0-0) - .. Standered matrix A of. L 3 A - -2 -1

we know L(x) = AX . "L -4 0 -3 4 - - 2 - 1 3x2 2x1 9 -9 - 4-6 +4+3 - 10 7 No Let be defined T : IR→IR? by T(*, ") = (x1, xy), which is not linear. as I (0,0) = (1,0) if I was linear, then T(0,0) = (0,0).