Finding a transformation matrix with respect to a coordinate mapping 
Determine the transformation matrix of $T$ with respect to the
  coordinate mapping: 
$\xi(p) = (p(-1),p(0),p(1))$, and we define $T:\mathcal{P}_2 \to
> \mathcal{P}_2$, with $T(p(t))=p(t+1)$ where $\mathcal{P}_2$ is the vectorspace of polynomials of maximum degree 2 over a field $F$.

I'm not sure how to do this. The answer should be: 
$\begin{bmatrix}
   0 & 1 & 0 \\
   0 & 0 & 1 \\
   1 & -3 & 3
  \end{bmatrix}$
I know how to find the "standard" transformation matrix for $T$, by letting $T$ act on the standard basis $(1,t,t^2)$ of $\mathcal{P}_2$, then I can obtain a matrix which describes the linear transformation.
I know that a coordinate mapping/evaluation is isomorphism a from a vectorspace $V$, with dimension $n$ over a field $F$ to $F^n$, like this $ \xi: V\to F^n$, then $\xi_j =\pi_j\xi$, where $\pi_j:F^n \to F, (x_1,x_2,\dots,x_n) \mapsto x_j$. Since this is a linear transformation I can also find a matrix for this as above with $\xi(p) = (p(-1),p(0),p(1))$. 
I'm thinking that I should somehow should multiply the matrix for the transformation of $T$ and the transformation of coordinate mapping, but I can't get it to work so that I end up with the desired answer. It feels like I'm missing something (or rather a lot), I'm not sure how these should interact properly. Can someone help me out?
 A: Fix a basis, in this case you have $b_1(t)=1,b_2(t)=t,b_3(t)=t^2$.
Then you have $T(b_1) = t \mapsto 1 = b_1$, $T(b_2) = t \mapsto t+1 = b_1+b_2$ and
$T(b_3) = t \mapsto t^2+2t+1 = b_3+2b_2+b_1$.
Then you have $\xi (b_1) = (1,1,1)^T$, $\xi (b_2) = (-1,0,1)^T$ and $\xi (b_3) = (1,0,1)^T$.
You have $\xi (T(b_1)) = \xi (b_1)$, $\xi (T(b_2)) = \xi(b_1)+\xi(b_2)$ and $\xi (T(b_3)) = \xi(b_3)+2\xi(b_2)+\xi(b_1)$.
Now compute the matrix that maps the vectors $\xi(b_k) $ to the vectors $\xi(T(b_k))$. In fact, you can read off the matrix from the above (using the basis $\xi(b_k)$) as:
$T_\xi=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1
\end{bmatrix}$
Addendum: The matrix in the question is the same operator in a different basis, but it is not clear to me why this basis would be 'natural' in some sense (it is the controller canonical form, but deriving this is a minor pain).
A basis that 'works' is $\beta_1(t) = 1 -{3 \over 2} t + {1 \over 2} t^2$, 
$\beta_2(t) = 2t -t^2$, $\beta_3(t) = -{1 \over 2}t + {1 \over 2} t^2$. If $p \in {\cal P}_2$ and $x,y$ represent the coordinates in the bases $b_k, \beta_k$ respectively, then we have $x=S y$, where
$S = \begin{bmatrix} 1 & 0 & 0 \\
-{3 \over 2} & 2 & -{1 \over 2} \\
{1 \over 2} & -1 & {1 \over 2}
\end{bmatrix}$. The bases $\xi(b_k)$ and $\xi(\beta_k)$ are related in the same matter, and we have the matrix that maps the vectors $\xi(\beta_k) $ to the vectors $\xi(T(\beta_k))$ given by
$S^{-1} T_\xi S = \begin{bmatrix} 
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & -3  & 3
\end{bmatrix}$.
