Linear Transformations between 2 non-standard basis of Polynomials 
If 
  $$ 
A = \begin{pmatrix}  1 & -1 & 2 \\
-2 & 1 &-1 \\ 1 & 2 & 3  \end{pmatrix} 
$$ 
  is the matrix representation of a
  linear transformation $T : P_3(x) \to P_3(x)$  with respect to
  bases $\{1-x,x(1-x),x(1+x)\}$ and $\{1,1+x,1+x^2\}$. Find T.

While i have worked with transforming non-standard to standard basis, this is the first one i am encountering with transformation between 2 non-standard polynomial basis. I am not sure if i am working out rightly.
$T[1-x]    = 1(1) -2(1+x) +1(1+x^2)$
$T[x(1-x)] = -1(1) +1(1+x) +2(1+x^2)$
$T[x(1+x)] =  2(1) -1(1+x) +3(1+x^2)$
Therefore, $T[a(1-x)+b(x(1-x))+c(x(1+x))] = (a-b+2c)(1) + (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2)$
Is this fine ?
 A: I also suspect the real question is to find the image of $a+bx+cx^2$ in the canonical basis.
I would do it in a formal way first: denote $X, Y$, &c. the column vectors of  polynomials in the canonical basis, $X_1, Y_1$, &c. their column vectors in the first basis and $X_2, Y_2$, &c. their column vectors in the second basis.
We're given the matrix $A$ of a linear transformation $T$ from $(P_2(x),\mathcal B_1)$ to $(P_2(x),\mathcal B_2)$, i.e. we have  a matrix relation
$$Y_2=AX_1$$
and asked for the matrix of this same linear transformation from $(P_2(x),\mathcal B_\text{canon})$ to itself, i.e. we're asked for the matrix $T$ such that
$$Y=TX.$$
Now that's easy, given the change of basis matrices:
$$P_1=\begin{bmatrix}\!\!\begin{array}{rrc}
1&0&0\\-1&1&1\\0&\:\llap-1&1
\end{array}\end{bmatrix},\qquad P_2=\begin{bmatrix}
1&1&1\\0&1&0\\0&0&1
\end{bmatrix}.$$
We have $Y=P_2Y_2$, $X=P_1X_1$, so
$$Y=P_2Y_2=P_2AX_1=(\underbrace{P_2AP_1^{-1}}_T)X.$$
There remains to find the inverse of $P_1$, which is standard by row reduction.
A: Everything you've written is correct, although I suspect the problem is asking you to find $T[a + bx + cx^2]$.  
A: What you did is fine, but now you have to compute $T[\alpha+\beta x+\gamma x^2]$ for arbitrary $\alpha,\beta,\gamma\in\mathbb R$. In order to do that, solve the equation$$\alpha+\beta x+\gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$\left\{\begin{array}{l}a-b+2c=\alpha\\-2a+b-c=\beta\\a+2b+3c=\gamma.\end{array}\right.$$
