Consider the linear transformation T R3 rightarrow P2 such t

Consider the linear transformation T: R^3 rightarrow P_2 such that T(e_1) = 1, T(e_2) = l + x + x^2 and T(e_3) = ? + x^2. a) Find the general formula for T(a, b, c). b) Find the ker(T) and the nullity(T) c) Find the rank(G) and the Range(T)

Solution

(a) Since (a, b, c) = ae₁ +be2 +ce_3,hence T(a, b, c) = T ( ae₁ +be2 +ce₃ ) = a T(e₁) + b T(e₂) +c T(e₃) = a*1+b(1 +x +x²) + c(x + x²) = (b +c) x² +(b +c)x +(a+ b)

(b) Ker(T) = {v: T(v)= 0}. Let v = (a,b,c) where a,b, c are arbitrary real numbers. Then T (a,b, c) = (b +c) x² + (b +c)x +(a+ b). Thus, if (a, b, c) Ker (T), then b +c = 0 and a + b = 0 so that b = -a and c = -b = a. Then v = ( a ,-a, a) = a(1,-1,1). Thus Ker (T) = span { (1,-1,1)}. The nullity of T being the dimension of Ker(T) is 1.

(c) As per the rank-nullity theorem, the rank of T + the nullity of T = the dimension of R³ = 3. Since the nullity of T is 1, the rank of T is 3 - 1= 2. The range of T, is the set of values of T i.e. {T(v) P₂ : v R³ } = { (b +c) x² + (b +c)x +(a+ b): a, b, c R}. Now, if b +c = p and a +b = r, then the range of T = { px² + px + q ; p, q R}. Thus the range of T is the set of all 2^nd degree polynomials where the coefficients of x² and x are equal.