Define T P2 rightarrow R3 by Tp plp2p3f evaluated T on the

Define T: P_2 rightarrow R^3 by T(p) = (p(l),p(2),p(3)f. & evaluated T on the polynomial x^2 - 1 Show that T is a linear transformation. Find the matrix of T with respect to the basis 1.x, x^x3 for P_3 and the standard basis C_1,e_2,e_3 for R^3. Let v = (a, b, c) in R^3 and define a polynomial q(x) = a(x - 2)(x - 3)/2 - b(x - 1)(x - 3) + c(x - 1)(x - 3)/2. Find T(q). What can you conclude about im(T)? Use a theorem from the course to calculate dim(ker(T)) Find a basis for ker(T)

Solution

Solution: Given T : P₃ \\to R³ defined by T(p) = ( p(1), p(2), p(3) )^T.

(a) Here p(x) = x² -1, so p(1) = 1² -1= 1 - 1 =0, p(2) =2² - 1 =3, p(3) = 3² - 1= 8

Therefore T(p) = T(x² - 1) = ( p(1), p(2), p(3) )^T = ( 0, 3, 8 )^T.

(c) Matrix of T: Let basis of P₃ be {1, x, x², x³ } and standard basis for R³ be{e₁, e₂, e₃} .

Then T(1) = (1, 1, 1)^T = 1.e₁ +1.e₂ + 1.e₃  

T(x) = (1, 2, 3)^T = 1.e₁ + 2.e₂ + 3.e₃

T(x²) = (1, 4, 9)^T = 1.e₁ + 4.e₂ + 9.e₃

T(x³) = (1, 8, 27)^T = 1.e₁ + 8.e₂ + 27.e₃

Therefore matrix of T with respect to the given bases is

(d) Here v = (a, b, c)^T in R³ and q(x) = a(x-2)(x-3)/2 - b(x-1)(x-3) + c(x-1)(x-3)/2.

So q(1) = a(1-2)(1-3)/2 - b(1-1)(1-3) + c(1-1)(1-3)/2 = a.

q(2) = a(2-2)(2-3)/2 - b(2-1)(2-3) + c(2-1)(2-3)/2 = b - c/2

q(3) = a(3-2)(3-3)/2 - b(3-1)(3-3) + c(3-1)(3-3)/2 = 0

Therefore T(q) = ( q(1), q(2), q(3) )^T = (a, b-c/2, 0)^T .

1	1	1	1
1	2	4	8
1	3	9	27