Find a transformation basis for the lin. transformation $\mathcal P_2\rightarrow \mathcal P_2; \space p\mapsto(q\cdot p)'$ 
Let $q=x+1$ and let $\mathcal P_2$ be the vector space of real
  polynomials of degree two or less.
Determine a transformation matrix for the linear transformation $\phi$ with respect to the basis $(1,x,x^2)$
$\phi: \mathcal P_2\rightarrow \mathcal P_2; \space  p\mapsto(q\cdot
p)'$

I wasn't sure how to approach this so I just let $p$ equal:
$$ax^2+bx+c$$
$$\implies (p \cdot q)'=[(ax^2+bx+c)(x+1)]'=(ax^3+ax^2+bx^2+bx+cx+c)'$$
$$=3ax^2+2ax+2bx+b+c=3ax^2+(2a+2b)x+(b+c)$$
Would my transformation matrix look like this:
$$\begin{pmatrix}b+c\\2(a+b)\\3a\end{pmatrix}\text{?}$$
 A: No. The matrix must be $3\times 3$. If the basis is $\{x^2,x,1\}$ it would be
$$\begin{pmatrix}3&0&0\\2&2&0\\0&1&1\end{pmatrix}$$
If it is not clear yet, just multiply this matrix by the vector
$$\begin{pmatrix}a\\b\\c\end{pmatrix}$$
and see what happens.
A: You should come back to the definition of the transformation matrix $M=(m_{ij})$ written in the basis $(e_1,\dots,e_n)$, where $$\displaystyle \phi(e_j)=\sum_{i=1}^n m_{ij} e_i$$
In your example, the basis $(e_1,e_2,e_3)$ is equal to the three polynomials $(1,x,x^2)$. So let's try to find $\phi(1)$. You have $$\phi(1)=(q.1)^\prime=((x+1).1)^\prime=1=1.1+0.x+0.x^2$$ Hence the first colunmn of your matrix is $$\begin{pmatrix}1\\0\\0\end{pmatrix}$$
You can follow from there and find the $3\times 3$ matrix requested which is $$\begin{pmatrix}1&1&0\\0&2&2\\0&0&3\end{pmatrix}$$
You will verify that for a polynomial of degree $2$, $p(x)=a.1+b.x+c.x^2$ you have $$\phi(p)=\begin{pmatrix}1&1&0\\0&2&2\\0&0&3\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=(a+b).1+(2b+2c).x+3c.x^2$$
Also take care that the matrix to be found depends on the order of $1,x,x^2$. The matrix of $\phi$ in the basis $(1,x,x^2)$ is not the same in the basis $(x^2,x,1)$!
