Question 4.1.4: Define a mapping T: P3 → P2 by T (p(x)) = p′(x)where p′(x) i...
Define a mapping T : P_{3} → P_{2} by
T (p(x)) = p^{\prime } (x)
where p^{\prime } (x) is the derivative of p(x).
a. Show that T is a linear transformation.
b. Find the image of the polynomial p(x) = 3x³ + 2x² − x + 2.
c. Describe the polynomials in P_{3} that are mapped to the zero vector of P_{2} .
First observe that if p(x) is in P_{3} , then it has the form
p(x) = ax³ + bx² + cx + d
so that
T (p(x)) = p^{\prime } (x) = 3ax²+ 2bx + c
Since p^{\prime } (x) is in p_{2}, then T is a map from p_{3} into p_{2}.
a. To show that T is linear, let p(x) and q(x) be polynomials of degree 3 or less, and let k be a scalar. Recall from calculus that the derivative of a sum is the sum of the derivatives, and that the derivative of a scalar times a function is the scalar times the derivative of the function. Consequently,
T (kp(x) + q(x)) = \frac{d}{dx} (kp(x) + q(x))
= \frac{d}{dx} (kp(x)) + \frac{d}{dx} (q(x))
= kp^{\prime } (x) + q^{\prime } (x)
= kT (p(x)) + T (q(x))
Therefore, the mapping T is a linear transformation.
b. The image of the polynomial p(x) = 3x³ + 2x² − x + 2 is
T (p(x)) = \frac{d}{dx} (3x³ + 2x² − x + 2) = 9x²+ 4x − 1
c. The only functions in P_{3} with derivative equal to zero are the constant polynomials p(x) = c, where c is a real number.