Consider $g(p)=xf(p)$ and linearity and construct the matrix for the transformation. I have the following assigment:
a) Consider $P_3$ and let $f$ be the derivative transformation of $p$, $f(p)=p'$. Argue that $f$ is linear. From $f$ applied on each of the basis vectors construct a matrix representing $f$ in the standard basis for $P_3$
So if $P_3$ we have  $p(x)=a_0+a_1x+a_2x^2+a_3x^3$ and the standard basis $\{1,x,x^2,x^3 \}$
Now we transform by deriving: $p(x)=a_1+2a_2x+3a_3x^2$ and the standard basis $\{1,x,x^2 \}$
Given $\frac{d}{dx}=T$ we have $T:\mathbb{R}^4 \rightarrow \mathbb{R}^3$
So is this linear? Yes it is because the definition of a linear function says for $f:D \rightarrow C$ we have $f(x+y)=f(x)+f(y),\ \ \forall x,y \in D$
So we have $$f'(x+y)=f'(x)+f'(y)$$
Now I will find the matrix represinting $f$
$$A \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix}= \begin{bmatrix} a_1 \\ 2a_2 \\ 3a_3\end{bmatrix}$$
$$A= \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 &3 \end{bmatrix}$$
b) Consider $g(p)=xf(p)$ and consider he same questions.
Am I correct in a)? Need a little help in b) 
 A: You are almost 100% correct in a), except for a small technicality: 

So is this linear? Yes it is because the definition of a linear function says for $f:D→C$ we have $f(x+y)=f(x)+f(y)$,  $∀x,y∈D$

You also have to prove that $f(ax)=af(x)$ for $a\in \mathbb{R}$.
For b), if you have a third degree polynomial and you differentiate it, then multiply it by $x$, you get a third degree polynomial. So your transformation $g$ will map $\mathbb{R}^4$ into what? 
After you figure that out, figuring out the coefficients of the matrix representing $g$ is basically the same process that you did for a).

Specifically, the matrix representing $g$ is
\begin{pmatrix}
   0   & 0   & 0       & 0       \\
   0   & a_1 & 0       & 0       \\
   0   & 0   & 2 * a_2 & 0       \\
   0   & 0   & 0       & 3 * a_3 
\end{pmatrix}
and it will map $\left\{1, x, x^2, x^3\right\}$ into $\left\{1, x, x^2, x^3\right\}$
We can check by taking some polynomial and seeing that our transformation gives us the same answer as the computation.
Take $p=1+x+x^2+x^3$. Then $f(p) = 1+2x+3x^2$ and $xf(p) = x + 2x^2 + 3x^3$. From there, we can tell that our matrix is correct, since if we pass it $(1,1,1,1)$ we get $(0,1,2,3)$.
