How do I find $[T]_E$ when given this linear transformation $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$? In this question I have been asked to find $[T]_E$.
I was given that $T: R_3[x] \rightarrow R_3[x]$ is a linear transformation 
defined as: $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$,
$p(x),p(x)',p(x)'' \in R_3[x]$ and $E = \{ 1,x,x^2 \}$ is the basis of the transformation.
For solving this question I went to the definition of matrix that represents a transformation that says:
$[T]_E = [[Te_1],[Te_2],[Te_3]]$
Now, my problem is how do I map each vector in E with this transformation  (first time seeing transformation with derivatives)
Thank you in advance.  
 A: Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$  Compute
$$\begin{align}
Te_1 & = T(1)\\
 & = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\\
 & = 2\\
 & = \color{#08F}2 \cdot 1 + \color{#08F}0 \cdot x + \color{#08F}0 \cdot x^2\\
 & = \color{#08F}2 e_1 + \color{#08F}0e_2 + \color{#08F}0e_3,\\
[Te_1]_E & = {\begin{bmatrix}\color{#08F}{2\\ 0\\ 0}\end{bmatrix}};\\\\
Te_2 & = T(x)\\
 & = (1 + 2x^2)x'' + (1 - 2x)x' + x + 0\\
 & = 1 - x\\
 & = \color{#08F}1 \cdot 1 \color{#08F}{- 1} \cdot x + \color{#08F}0 \cdot x^2\\
 & = \color{#08F}1 e_1 \color{#08F}{- 1}e_2 + \color{#08F}0e_3,\\
[Te_2]_E & = {\begin{bmatrix}\color{#08F}{1\\ -1\\ 0}\end{bmatrix}};\\\\
Te_3 & = T(x^2)\\
 & = (1 + 2x^2)(x^2)'' + (1 - 2x)(x^2)' + x^2 + 0^2\\
 & = 2 - 2x + x^2\\
 & = \color{#08F}2 \cdot 1 \color{#08F}{- 2} \cdot x + \color{#08F}1 \cdot x^2\\
 & = \color{#08F}2 e_1 \color{#08F}{- 2}e_2 + \color{#08F}1e_3,\\
[Te_3]_E & = {\begin{bmatrix}\color{#08F}{2\\ -2\\ 1}\end{bmatrix}}.\end{align}$$
Thus,
$$[T]_E = {\begin{bmatrix}
2 &  1 &  2\\
0 & -1 & -2\\
0 &  0 &  1\end{bmatrix}}.$$
