Let $X \in M_n(\mathbb H)$ (Hermetian field).
It is possible to make $\mathbb H^n$ into a $2n$-dimensional vector space over $\mathbb C$, for example, by embedding $\mathbb C$ into $\mathbb H$ as the following subfields:
$\mathbb R1+\mathbb Ri$
$\mathbb R1+\mathbb Rj$
$\mathbb R1+\mathbb Rk$
where $i^2 = j^2 = k^2 = -1$
Here is what I need to do: I need to write out the corresponding matrix of $X$ as an element of $M_{2n}(\mathbb C)$, and if possible show that the trace of $X$ is independent of the choice.
Here is what I got so far and got stuck with:
For example, using the first of these embedding, it is possible to write $\mathfrak z = x+jy \in \mathbb H ^n$ with $x,y \in \mathbb C^n$, and likewise $C=A+jB \in M_n(\mathbb H)$, with $A,B \in M_n(\mathbb C)$.
If so, the maps $\mathfrak z \rightarrow \left( \begin{smallmatrix} a\\ c \end{smallmatrix} \right)$ and $C \rightarrow \left( \begin{smallmatrix} A&-\bar B\\ B&\bar A \end{smallmatrix} \right)$ identify $\mathbb H$$^n$ with $\mathbb C^{2n}$ and $M_n(\mathbb H)$ with the real subalgebra of $M_{2n}(\mathbb C)$ consisting of matrices $T$ such that $JT=\bar TJ$, where $J = \left( \begin{smallmatrix} 0&I\\ -I&0 \end{smallmatrix} \right)$.
How can I continue this and find the corresponding matrix of $X$ as an element of $M_{2n}(\mathbb C)$, and show that the trace of $X$ is independent of the choice?