Matrix subring isomorphic to $\mathbb{C}$ 
I'm trying to prove that $S$, the subring of $M_2(\mathbb{R})$ generated by $\left(\begin{array}{cc}
r & 0 \\
0 & r \end{array} \right), r \in \mathbb{R} $ and $\left(\begin{array}{cc}
0 & 1 \\
-1 & 0 \end{array} \right)$ is isomorphic to $\mathbb{C}$. 

I thought of creating the map $f:\mathbb{C} \to S$ by $f(a+bi) = \left(\begin{array}{cc}
a & 0 \\
0 & a \end{array} \right) \left(\begin{array}{cc}
0 & 1 \\
-1 & 0 \end{array} \right)^b.$
But this doesn't seem like a very natural map, does anyone have any suggestions?
 A: Why not use $f(a + bi) = \begin{pmatrix}a&0\\0&a\end{pmatrix} + \begin{pmatrix}b&0\\0&b\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix} = \begin{pmatrix}a&b\\-b&a\end{pmatrix}$?
A: The first matrix you list "behaves" in $M_2(\mathbb{R})$ as $r$ does in $\mathbb{R}$ (if you add or multiply two such for different $r$'s). The second matrix "behaves" as $i$ does in $\mathbb{C}$ (calculate its square). So the counterpart of $a$ alone should be $\left(\begin{array}{cc}
a & 0 \\
0 & a \end{array} \right)$ and the counterpart of $bi$ alone should be $\left(\begin{array}{cc}
0 & b \\
-b & 0 \end{array} \right)$. This should be enough to finish on your own. 
P.S. Don't forget that you're treating $M_2(\mathbb{R})$ as a subring - so it has both addition and product of matrices. You were trying to only use products.
A: One of the possible constructions of $\mathbf{C}$ is to define its as the set of matrices of the form:
$$\begin{bmatrix}a&-b\\b&a\end{bmatrix}$$
This set of matrices is a subring of the ring of $2\times2$ matrices. The matrix $J=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ satisfies the relation:
$$J^2=-I\quad\text{(the unit matrix of dimension 2)}$$
and it turns out we obtain a commutative field. Further more, any such matrix can be written as 
$$aI+bJ$$
Thus if you start form another definition of $\mathbf C$, you can define an $\mathbf{R}$-linear mar from  $\mathbf C$ to $S$, sending $1$ to $I$ and $\mathrm i$ to $J$.
