Linear transformation $\mathbb R^3$ to $M^{2\times 2}$ Can you give me an hint on how to solve this?
Consider the linear transformation $S: \mathbb R^3(x) \to M^{2 \times 2}(\mathbb R)$ defined by
$$S(a_3x^3+a_2x^2+a_1x+a_0)=\begin{bmatrix} a_0 & a_1\\a_2 & a_3\end{bmatrix}$$
Consider in $\mathbb R^3(x)$ the basis $P=((x^3, x^2, x, 1))$ and the basis in $M^{2x2}(\mathbb R)$
$$\mathcal{N}=\left( \begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right)$$
What is the matrix $A$ that represents $S$ in relation to the basis $P$ and $\mathcal{N}$?
 A: The rule reads: In order to obtain a matrix $[S]$ for a given linear transformation $S$ from an $n$-dimensional vector space $X$ to another $m$-dimensional vector space $Y$ ($m=n=4$ in your case),  do the following:
First choose (independently) a basis  both in $X$ and in $Y$, and set up an "empty" matrix $[\ \ ]$ with $m$ rows and $n$ columns – an $(m\times n)$-matrix in short. (In your case the two bases have already been chosen.)
Then write the images of the $n$ basis vectors of $X$, expressed in terms of the basis of $Y$, one after the other into the the $n$ columns of the prepared empty matrix $[\ \ ]$.
I'll do this for the first basis vector $x^3$ of $X:=P_3[x]$ (polynomials in $x$ of degree $\leq3$). The monomial $x^3$ is characterized by $a_3=1$ and all other $a_k=0$, whence its image $S(x^3)$ is by definition of $S$ the matrix $\left[\matrix{0&0\cr 0&1\cr}\right]$, which happens to be the fourth basis vector of $Y:=M^{2\times2}$. It follows that $S(x^3)$ when expressed in terms of the four basis matrices of $Y$ appears as $(0,0,0,1)$, so that the first column of $[S]$ will be
$$\matrix{0\cr0\cr0\cr1\cr}\ .$$
