Determinant of the real matrix representation of a $\mathbb{C}$-linear map If $f$ is a complex linear map on $\mathbb{C}^n$ with matrix $A=(a_{i,j})$ then, viewing $f$ as a real linear map on $\mathbb{R}^{2n}$, we find that the real $2n\times 2n$ matrix of $f$ has the form of a block matrix $(\alpha_{i,j})$ where 
$$\alpha_{i,j}=\begin{pmatrix}
Re(a_{i,j}) & -Im(a_{i,j}) \\
Im(a_{i,j}) & Re(a_{i,j}) 
\end{pmatrix}$$
that is, the real matrix has the form 
$$\mathcal{A}=\begin{pmatrix}
a & -b & c &-d & e &-f & \dots & \\
b & a & d & c & f & e & \dots & \\
g & -h & i & -j & k & -\ell & \dots & \\
h & g & j & i & \ell & k & \dots & \\
\vdots & & & & & & & \\
\end{pmatrix}$$
I was wondering if there is any way to take advantage of the symmetries of this matrix in order to compute its determinant. I know that the answer should be that $\det \mathcal{A}=|\det A|^2$ but I dont see how to obtain this by a straightforward computation. I tried doing a cofactor expansion but quickly got lost in the messiness of it. Any helpj would be greatly appreciated. Thanks!
 A: You can use the determinant formula for block matrices. If $\mathcal{A}$ is a $np \times np$-matrix made out of commuting $p\times p$-blocks, then the determinant of $\mathcal{A}$ is the same as the determinant of the $p \times p$-matrix obtained by computing the determinant of the block matrix considering its entries as the entries of an $n\times n$-matrix.
In your case we have $p=2$ and the blocks commute because complex numbers commute. Since each $2\times 2$-block of $\mathcal{A}$ corresponds to the complex number at the corresponding position in $A$, we already know the $2\times 2$-matrix of which we have to take the determinant: It is simply the $2 \times 2$-matrix representing the complex number $\det(A)$.
So if $\det(A) = a + ib$, then
$$\det(\mathcal{A}) = \det\begin{pmatrix}a & -b\\ b & a\end{pmatrix} = a^2 + b^2 = |\det(A)|^2$$
Edit:
It might be useful to have an example. For $n = p = 2$ the formula boils down to:
$$\det\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \det(AD - BC).$$
Note that the term $AD - BC$ is formally the determinant of a $2 \times 2$-matrix.
Suppose that $A$ is given by
$$A = \begin{pmatrix}1 & i \\ i & i\end{pmatrix}.$$
Then
$$\mathcal{A} = \left(\begin{array}{cc|cc}
  1 & 0 & 0 & -1 \\
  0 & 1 & 1 &  0 \\ \hline
  0 & -1& 0 & -1\\
  1 & 0 & 1 & 0
\end{array}\right)$$
and by the formula above we have
$$\det(\mathcal{A}) = \det\left( \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} -
\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}
\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \right)
= \det \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$$
On a formal level this is similar to computing the determinant of $A$:
$$\det(A) = ( 1 \cdot i - i \cdot i) = 1+i.$$
Also note that in your matrix representation of complex numbers:
$$\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \hat{=}\; 1+i.$$
