Question

Let T: P2 → P3 be the transformation that maps a polynomial p(t) into the polynomial (t-2)p(t).

Answer 1

(a) Applying T to p(t) = 2 - t + t ² gives

T ( p(t) ) = (t - 2) (2 - t + t ²) = -4 + 4t - 3t ² + t ³

(b) T is a linear transformation if for any p(t) and q(t) in P₂ and complex scalars a and b, the image of any linear combination of p and q is equal to the linear combination of the images of p and q. In other words,

T ( a p(t) + b q(t) ) = a T ( p(t) ) + b T ( q(t) )

Let

p(t) = α₀ + α₁ t + α₂ t ²

q(t) = β₀ + β₁ t + β₂ t ²

Compute the images of p and q :

T ( p(t) ) = (t - 2) (α₀ + α₁ t + α₂ t ²)

… = -2α₀ + (α₀ - 2α₁) t + (α₁ - 2α₂) t ² + α₂ t ³

Similarly,

T ( q(t) ) = -2β₀ + (β₀ - 2β₁) t + (β₁ - 2β₂) t ² + β₂ t ³

Then

a T ( p(t) ) + b T ( q(t) ) = a (-2α₀ + (α₀ - 2α₁) t + (α₁ - 2α₂) t ² + α₂ t ³) + b (-2β₀ + (β₀ - 2β₁) t + (β₁ - 2β₂) t ² + β₂ t ³)

… = c₀ + c₁ t + c₂ t ² + c₃ t ³

where

c₀ = -2 (a α₀ + b β₀)

c₁ = a (α₀ - 2α₁) + b (β₀ - 2β₁)

c₂ = a (α₁ - 2α₂) + b (β₁ - 2β₂)

c₃ = a α₂ + b β₂

Computing the image of a p(t) + b q(t) would give the same result; just multiply it by t - 2 and expand. This establishes that T is indeed linear.

(c) Find the image of each vector in the basis for P₂ :

T (1) = (t - 2) × 1 = t - 2

T (t ) = (t - 2) t = t ² - 2t

T (t ²) = (t - 2) t ² = t ³ - 2t ²

Then

$T=\begin{bmatrix}-2&0&0\\1&-2&0\\0&1&-2\\0&0&1\end{bmatrix}$