Define an $\mathbb R$-linear isomorphism $f_n : \mathbb C^n \to \mathbb R^{2n}$ by $(x_1 + iy_1, x_2 + iy_2, \dots) \mapsto (x_1, y_1, x_2, y_2, \dots )$. Let $A \in M^n (\mathbb C)$ and define $R_A(v) = v \cdot A$. Furthermore define an injective homomorphism $\rho_n : M^n(\mathbb C) \to M^{2n}(\mathbb R)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -b_{ij} & a_{ij} \end{array}$ if $A_{ij}=(a_{ij} + i b_{ij})$. It is an exercise in Tapp's book to show that the following diagram commutes: $\require{AMScd}$ \begin{CD} \mathbb C^n @>\displaystyle f_n>> \mathbb R^{2n}\\ @V \displaystyle R_A V V\# @VV \displaystyle{R_{\rho_n(A)}} V\\ \mathbb C^n @>>\displaystyle f_n> \mathbb R^{2n} \end{CD}
I proved it by calculating
$f_n(R_A(v))_i = (\sum_k u_k a_{ki} -w_kb_{ki}, \sum_k u_kb_{ki} + w_k a_{ki})$
and
$R_{\rho_n(A)}(f_n(v))_i = (\sum_k u_k a_{ki}-w_k b_{ki}, \sum_ku_k b_{ki}+w_ka_{ki})$
(by slight abuse of notation)
Is there a nicer way to show this? I don't quite get the point of the exercise if it boils down to a battle of computing indices.
HINT: \begin{eqnarray} (x+iy)\cdot (a+ib) &=& (x a - y b) + i (x b + y a) \\ \left( \begin{array}{cc} x,&\!\!\!\!y\\ \end{array} \right)\cdot \left( \begin{array}{cc} a&b \\ -b&a \\ \end{array} \right)&=& \left (\begin{array}{cc} x a - y b,&\!\!\!\!x b + y a \end{array} \right) \end{eqnarray}