Computing Matrix Representation of a Linear Transformation I'm having some trouble figuring out this question:
Let $V=P_3(\mathbb{R})$ and $W=P_4(\mathbb{R})$. 
Let $D:W\to V$ be the derivative mapping $D(p)=p'$.
Let $Int:V\to W$ be the integration mapping $Int(p)=\int_0^x p(t)dt$.
Let $\alpha=\{1,x,x^2,x^3\}$ and $\beta=\{1,x,x^2,x^3,x^4\}$ be the standard bases in $V$ and $W$.
Compute : $[D]_{\alpha\beta}$ ,  $[Int]_{\beta\alpha}$ , $[DInt]_{\alpha\alpha}$ , $[IntD]_{\beta\beta}$.
I'm not sure how to go about finding the matrices that represent these transformations. Any help at all would be appreciated.
My reasoning is that $[D]_{\alpha\beta}$ would equal:
\begin{Bmatrix} 
      1 & 0 & 0 & 0 & 0\\ 
      0 & 1 & 0 & 0 & 0\\ 
      0 & 0 & 1 & 0 & 0\\
      0 & 0 & 0 & 1 & 0
   \end{Bmatrix}
Since if you multiply that matrix by 
$\beta=\{1,x,x^2,x^3,x^4\}$
You would end up with 
$\alpha=\{1,x,x^2,x^3\}$
 A: Let's apply the transformation $D$ to each element of the given basis $\beta=\{1,x,x^2,x^3,x^4\}$ of $W=P_4(\mathbb{R})$ in the order they are listed, and write each result with respect to the given basis $\alpha=\{1,x,x^2,x^3\}$ of $V=P_3(\mathbb{R})$.
$D(1)=(1)'=0=0\cdot1+0\cdot x+0\cdot x^2+0\cdot x^3$, therefore the first column is $\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$.
$D(x)=(x)'=1=1\cdot1+0\cdot x+0\cdot x^2+0\cdot x^3$, therefore the second column is $\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$.
$D(x^2)=(x^2)'=2x=0\cdot1+2\cdot x+0\cdot x^2+0\cdot x^3$, therefore the third column is $\begin{bmatrix}0\\2\\0\\0\end{bmatrix}$.
$D(x^3)=(x^3)'=3x^2=0\cdot1+0\cdot x+3\cdot x^2+0\cdot x^3$, therefore the fourth column is $\begin{bmatrix}0\\0\\3\\0\end{bmatrix}$.
$D(x^4)=(x^4)'=4x^3=0\cdot1+0\cdot x+0\cdot x^2+4\cdot x^3$, therefore the fifth column is $\begin{bmatrix}0\\0\\0\\4\end{bmatrix}$.
Thus the desired matrix is $[D]_{\alpha\beta}=\begin{bmatrix}0&1&0&0&0\\0&0&2&0&0\\0&0&0&3&0\\0&0&0&0&4\end{bmatrix}$.
