Matrix of Linear Transformation Involving Derivatives? Here's a problem I'm working on: 
Find the matrix of T with respect to the standard bases $P_{3}$ and $\mathbb{R}^{3}$: 
$T(p(x)) = \left( \begin{array}{cc} p'(0) \\ p(0) \\ p(0) - p'(0)\end{array} \right)$
So I'll list the steps that I've been taking and hopefully someone will be able to tell me what I'm doing wrong. So the first thing I did was prove that the transformation was linear, which wasn't too bad. Now since I know that the transformation is linear I can make use of the theorem that says that every linear transformation can be written in the form $T(x) = Ax$ where $A$ is the coefficient matrix and $x$ is a vector. 
Now I believe the standard bases for the polynomial in my example is $1, x, x^{2}, x^{3}$, so I assumed I could do the following $A = ( T(1) \space \space T(x) \space \space T(x^{2}) \space \space T(x^{3}) )$ but here's when I start to get confused. The problem definition says that each function is being forced to evaluate zero, so knowing what the basis's are doesn't matter, you only need to know that there are four (correct me if I'm wrong about there being 4 standard basis). 
Next, so if I take the derivative of $p(x)$ I'd get some polynomial, but if I input zero I'd just get some constant. Does this mean that each of the four columns of my coefficient matrix are just going to be unique constants? Thanks in advance for any help offered. 
 A: To compute the matrix of a linear Transformation you have to take the base of the domain (here: $1, x, x^2, x^3$) , compute the images of these basis vectors and write the images as a linear combination of the given basis in the codomain (here: $e_1$, $e_2$, $e_3$). Then the coefficients of these vectors form a column of the matrix. That is your matrix is a $3 \times 4$-matrix whose entries are real numbers.
A: Let $p(x)=\sum_{k=0}^3 p_k x^k$. Then $T(p) = \begin{bmatrix} p_1 \\ p_0 \\ p_1-p_0 \end{bmatrix}$.
Let $A$ be the matrix of $T$ with the standard basis $1,x,x^2,x^3$ of $P_3$ and $e_1,e_2,e_3$ of $\mathbb{R}^3$ ($e_k$ is the vector with 1 in the $k$th place and 0 everywhere else. Then the above shows that, if $p =(p_0,p_1,p_2,p_3)^T \in \mathbb{R}^4$, then
$$Ap = T(\sum_{k=0}^3 p_k x^k) = 
\begin{bmatrix} p_1 \\ p_0 \\ p_1-p_0 \end{bmatrix}$$
Since $A = \begin{bmatrix} Ae_1 & Ae_2 & A e_3 & A e_4 \end{bmatrix}$, we obtain $A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\-1 & 1 & 0 & 0 \end{bmatrix}$.
