Matrix subring isomorphic to $\mathbb{C}$

Question

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?

Answer 1

Why not use $f(a + bi) = \begin{pmatrix}a&0\\0&a\end{pmatrix} + \begin{pmatrix}b&0\\0&b\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix} = \begin{pmatrix}a&b\\-b&a\end{pmatrix}$?

Answer 2

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.

Answer 3

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 $J$.