# Homework Help: Linear Algebra - Linear Transformation of a polynomial

1. Dec 5, 2011

### Jonmundsson

1. The problem statement, all variables and given/known data
Let $h: \mathbb{P_2} \rightarrow \mathbb{P_2}$ represent the transformation $h(p(x)) = xp'(x) + p(1-x)$ for every polynomial $p(x) \in \mathbb{P_2}$. Find the matrix of h with respect to the standard basis $\{1, x, x^2\}$

2. Relevant equations

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

3. The attempt at a solution
Let $p(x) = ax^2 + bx + c$ then $p'(x) = 2ax + b$ and $p(1-x) = a(1-x)^2 + b(1 - x) + c = ax^2 - 2ax + a + b - bx + c = ax^2 - 2ax - bx + a + b + c$.

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

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

Am I on the right track?

2. Dec 5, 2011

### 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, $x^2$.

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 $p(x)= x^2$, then $p'(x)= 2x$ and $p(1- x)= 1- 2x+ x^2$. $T(p)= x(2x)+ 1- 2x+ x^2= 3x^2- 2x+ 1$.

3. Dec 5, 2011

### Jonmundsson

I see. Thank you for the help.