Let P be the set of polynomials in x of degree at most 3 P

Let P be the set of polynomials in x of degree at most 3: P = {ax^3 + bx^2 + cx + d | a, b, c, d elementof R]. We can add such polynomials together and multiply them by scalars to get more of the same kind, so we can think of P in the same way as we do a subspace of R^n. Seen this way, P has basis {1, x, x^2, x^3}. (a) Let D: P rightarrow P be the function defined by differentiation: D(p(x)) = p\'(x). Show that D is a linear transformation. (b) Find the kernel D. (c) Find the range of D.

Solution

(a). let p(x) and q(x) be 2 arbitrary elements of P and let k be an arbitrary scalar. Then D(p(x)+q(x)) = (p(x)+q(x))’ = p’(x)+q’(x) = D(p(x))+D(q(x)). Thus D preserves vector addition.Also, D(kp(x)) = (k(p(x))’ = kp’(x). Hence D preserves scalar multiplication also. Therefore, D is a linear transformation.

(b) Ker(D) = {p(x): D(p(x)) =0}. Further, D(p(x)) = p’(x). Hence, if D(p(x)) =0, then p’(x)= 0 and then p(x) = k, a constant. Thus, Ker(D) is span{1}.

Range(D)={q(x):q(x)=D(p(x) for some p(x) in P}. Further, if p(x) = ax³+bx²+cx+d, thenD(p(x)= p’(x)=3ax²+2bx+c. This means that every element of polynomials in x, of degree at most 2 will have a pre-image in P under D. Hence Range (D) = P₂, the set of all polynomials in x, of degree at most 2.