More elegant proof of that this diagram commutes 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 $A_{ij}\mapsto \begin{array}{cc} a_{ij} &amp; b_{ij} \\ -b_{ij} &amp; a_{ij} \end{array}$ 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 @&gt;\displaystyle f_n&gt;&gt; \mathbb R^{2n}\\
    @V \displaystyle R_A V V\# @VV \displaystyle{R_{\rho_n(A)}} V\\
    \mathbb C^n @&gt;&gt;\displaystyle f_n&gt; \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) &amp;=&amp; (x a - y b) + i (x b + y a) \\
 \left(
\begin{array}{cc}
x,&amp;\!\!\!\!y\\
\end{array} \right)\cdot \left(
\begin{array}{cc}
a&amp;b \\
-b&amp;a \\
\end{array} \right)&amp;=&amp; \left (\begin{array}{cc}
x a - y b,&amp;\!\!\!\!x b + y a 
\end{array} \right)
\end{eqnarray}
HINT:
\begin{eqnarray}
(x+iy)\cdot (a+ib) &amp;=&amp; (x a - y b) + i (x b + y a) \\
 \left(
\begin{array}{cc}
x,&amp;\!\!\!\!y\\
\end{array} \right)\cdot \left(
\begin{array}{cc}
a&amp;b \\
-b&amp;a \\
\end{array} \right)&amp;=&amp; \left (\begin{array}{cc}
x a - y b,&amp;\!\!\!\!x b + y a 
\end{array} \right)
\end{eqnarray}