Proof. Define
$\varphi \colon R_n \to M_{n \times n}(\mathbb{C})$ by
$\displaystyle \varphi \left( \left( \begin{array}{cc} A & B \\ -B & A \end{array} \right) \right) = A+iB$ for
$A,B \in M_{n \times n}(\mathbb{R})$
Let $A,B,C,D \in M_{n \times n}(\mathbb{R})$ such that $\displaystyle \varphi \left( \left( \begin{array}{cc} A & B \\ -B & A \end{array} \right) \right) =\varphi \left( \left( \begin{array}{cc} C & D \\ -D & C \end{array} \right) \right)$ Then $A+iB=C+iD$ Therefore, $A=C$ and $B=D$ Hence, $\displaystyle \left( \begin{array}{cc} A & B \\ -B & A \end{array} \right) = \left( \begin{array}{cc} C & D \\ -D & C \end{array} \right)$ It follows that $\varphi$ is injective.
Let $Z \in M_{n \times n}(\mathbb{C})$ Then there exist $X,Y \in M_{n \times n}(\mathbb{R})$ such that $X+iY=Z$ Since $\varphi \left( \left( \begin{array}{cc} X & Y \\ -Y & X \end{array} \right) \right)=X+iY=Z$ it follows that $\varphi$ is surjective.
Let $A,B,C,D \in M_{n \times n}(\mathbb{R})$ Then
$\begin{array}{rl} \displaystyle \varphi \left( \left( \begin{array}{cc} A & B \\ -B & A \end{array} \right) + \left( \begin{array}{cc} C & D \\ -D & C \end{array} \right) \right) & \displaystyle =\varphi \left( \left( \begin{array}{cc} A+C & B+D \\ -B-D & A+C \end{array} \right) \right) \\ \\ & =A+C+i(B+D) \\ \\ & =A+iB+C+iD \\ \\ & \displaystyle =\varphi \left( \left( \begin{array}{cc} A & B \\ -B & A \end{array} \right) \right) + \varphi \left( \left( \begin{array}{cc} C & D \\ -D & C \end{array} \right) \right) \end{array}$
and
$\begin{array}{rl} \displaystyle \varphi \left( \left( \begin{array}{cc} A & B \\ -B & A \end{array} \right) \left( \begin{array}{cc} C & D \\ -D & C \end{array} \right) \right) & \displaystyle =\varphi \left( \left( \begin{array}{cc} AC-BD & AD+BC \\ -AD-BC & AC-BD \end{array} \right) \right) \\ \\ & =AC-BD+i(AD+BC) \\ \\ & =(A+iB)(C+iD) \\ \\ & \displaystyle =\varphi \left( \left( \begin{array}{cc} A & B \\ -B & A \end{array} \right) \right) \varphi \left( \left( \begin{array}{cc} C & D \\ -D & C \end{array} \right) \right) . \end{array}$
It follows that $\varphi$ is an isomorphism. 
