Linear Algebra - Linear Transformation of a polynomial

[Jonmundsson]

Homework Statement

Let [itex]h: \mathbb{P_2} \rightarrow \mathbb{P_2}[/itex] represent the transformation [itex]h(p(x)) = xp'(x) + p(1-x)[/itex] for every polynomial [itex]p(x) \in \mathbb{P_2}[/itex]. Find the matrix of h with respect to the standard basis [itex]\{1, x, x^2\}[/itex]

Homework Equations

Matrix A of transformation: [itex]{\bf A} = [T(e_1) \hspace{0.5em} T(e_2) \hspace{0.5em} \ldots \hspace{0.5em} T(e_n)][/itex]

The Attempt at a Solution

Let [itex]p(x) = ax^2 + bx + c[/itex] then [itex]p'(x) = 2ax + b[/itex] and [itex]p(1-x) = a(1-x)^2 + b(1 - x) + c = ax^2 - 2ax + a + b - bx + c = ax^2 - 2ax - bx + a + b + c[/itex].

Now we can rewrite [itex]h(p(x)) = xp'(x) + p(1-x)[/itex] as
[itex]h(p(x)) = x(2ax + b) + ax^2 - 2ax - bx + a + b + c[/itex]

This is as far as I have gotten. I'm guessing that I am supposed to put [itex]h(p(1)) = x(2a + b) + a - 2a - b + a + b + c = 2ax + bx + c[/itex] making the first column of h's matrix [itex]\left[ \begin{array}{c} c \\ 2a + b \\ 0 \end{array} \right] [/itex]. Process is repeated for [itex]h(p(x))[/itex] and [itex]h(p(x^2))[/itex]

Am I on the right track?

 

[HallsofIvy]

I think you are making a mistake in doing the "abstract" calculation of p(1-x). You are specifically asked to use the "standard" matrix, 1, x, [itex]x^2[/itex].

If p(x)= 1, then p'(x)= 0 and p(1- x)= 1. T(p)= xp'(x)+ p(1- x)= x(0)+ 1= 1.

If p(x)= x, then p'(x)= 1 and p(1- x)= 1- x. T(p)= x(1)+ 1- x= 2- x.

If [itex]p(x)= x^2[/itex], then [itex]p'(x)= 2x[/itex] and [itex]p(1- x)= 1- 2x+ x^2[/itex]. [itex]T(p)= x(2x)+ 1- 2x+ x^2= 3x^2- 2x+ 1[/itex].

 

[Jonmundsson]

I see. Thank you for the help.