Problem:
Consider the matrix $$A = \begin{pmatrix} 0 & 3\\ -4 & 1 \end{pmatrix}.$$ Show that $A$ lies in a subring of Mat$_{2\times 2}(\mathbb{R})$ that is isomorphic to $\mathbb{C}$.
My idea:
We show that $A = CDC^{-1}$ where D is in the form of
$$D = \begin{pmatrix} a & -b\\ b& a\end{pmatrix}.$$
It is easy to see that all elements as such form a subring of $Mat_{2\times 2}(\mathbb{R})$ isomorphic to $\mathbb{C}$. Using the fact that $\det(A) = \det(D)$ and $tr(A) =tr(D)$ we obtain that $a = 1/2$ and $b = \sqrt{47/4}$. We can then search for $C$. Set
$$C = \begin{pmatrix} u_1 & u_2\\ u_3 & u_4\end{pmatrix},$$ and solve for $AC = CD$ we arrive at
$$\begin{pmatrix}1/2 & 4 & -\sqrt{47/4} & 0\\ 3 & 1/2 & 0 & \sqrt{47/4}\\ \sqrt{47/4} & 0 & 1/2 & 4\\ 0 & \sqrt{47/4} & -3 & -1/2\end{pmatrix}\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix} = \begin{pmatrix}0\\0\\0\\0\end{pmatrix}.$$ Since the left matrix has rank 2, the solution space is 2 dimensional, and we can find one such that $ad-bc\ne 0$.
Question:
Is my solution correct? What is a simpler proof?
The characteristic polynomial of $A$, which is also the minimal polynomial, is $p(\lambda) = \lambda^2 - \lambda + 12 = (\lambda - 1/2)^2 + 47/4$. So in any correspondence between the subring and the complex numbers, $A$ should correspond to a complex number $\omega$ with $(\omega - 1/2)^2 + 47/4 = 0$. We can take $\omega = \dfrac{1}{2} + \dfrac{\sqrt{47} }{2} i$. Take the subring to be the linear span (over $\mathbb R$) of $I$ and $A$, and the isomorphism from this to $\mathbb C$ is $aI + bA \mapsto a + b \omega$.