What is the Matrix corresponding to a Linear Transformation Given $T: P_2 \rightarrow P_3$ defined by:
$T(at^2 + bt +c) = (a-b+c)t^3 + (-a + 3b - 2c)t^2 +(-a-b)t +(2b-c)$
What is the corresponding Matrix of $T$? 
This is what I have: First I rewrite the transformation as follows:
$T_0 = 2b - c$
$T_1 = -a - b$
$T_2 = -a + 3b -2c$
$T_3 = a-b+c$
And I know $T_i = \sum \limits_{j=1}^n \mu_{ji} b_i$ where $b_i$ is the basis vector.
So my matrix $\mu$ is $\pmatrix{0&-1&-1&1\\
                                2&-1&3&-1\\
                                -1&0&-2&1}$


*

*Is this correct?

*Is my rewriting of $T$ correct?

 A: Whenever you are working out a problem such as this, the key fact to remember is always that a linear mapping's action is completely determined by it's action on a basis.
I am going to assume that you are using the standard basis vectors for $P_2$ and $P_3$, if not then it is a good exercise to modify the procedure below for your chosen basis. In this case, let us look at how the mapping acts on the basis vectors
$$T(1) = t^3 - 2t^2 - 1$$
$$T(t) = -t^3 + 3t^2 - t + 2$$
$$T(t^2) = t^3 - t^2 - t$$
In terms of the standard coordinates, this is then
$$T:\ \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}\mapsto\begin{pmatrix}-1 \\ 0 \\-2 \\ 1\end{pmatrix}$$
$$T:\ \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}\mapsto\begin{pmatrix}2 \\ -1 \\ 3 \\ -1\end{pmatrix}$$
$$T:\ \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}\mapsto\begin{pmatrix}0 \\ -1 \\ -1\\ 1\end{pmatrix}$$
Putting this all together gives you the standard matrix
$$[T] = \begin{pmatrix} -1 & 2 & 0 \\ 0 & -1 & -1 \\ -2 & 3 & -1 \\ 1 & -1 & 1\end{pmatrix}$$
