Determinant of an $n\times n$ complex matrix as an $2n\times 2n$ real determinant If $A$ is an $n\times n$ complex matrix. Is it possible to write $\vert \det A\vert^2$ as a $2n\times 2n$ matrix with blocks containing the real and imaginary parts of $A$?
I remember seeing such a formula, but can not remember where. Any details, (and possibly references) for such a result would be greatly appreciated.
 A: Let $A=B+iC$, where $B$ and $C$ are $n\times n$ real matrices, and let
$$
\widetilde{A}=\left(
\begin{array}{rr}
B & -C \\
C &  B
\end{array}
\right).
$$
Then $\det\widetilde{A}=|\det A|^2$.
Proof.
$$
\begin{align*}
\det\widetilde{A}&=\det\left(
\begin{array}{cc}
B+iC & -C+iB \\
   C &     B 
\end{array}
\right)=
\det\left(
\begin{array}{cc}
B+iC & 0   \\
   C & B-iC 
\end{array}
\right)=
\det
\left(
\begin{array}{ll}
A &  0 \\
C & \overline{A}
\end{array}
\right)\\
&=
(\det A)(\det\overline{A})=(\det A)(\overline{\det A})=|\det A|^2.
\end{align*}
$$
A: Write $A=A_1+iA_2$ where $A_1$ and $A_2$ are real matrices. Let 
$$B:=\pmatrix{A_1&iA_2\\iA_2&A_1}.$$
We have 
$$\det B=\det\pmatrix{A_1+iA_2&iA_2\\A_1+iA_2&A_1}=\det\pmatrix{I&iA_2\\I&A_1}\cdot\det\pmatrix{A_1+iA_2&0\\0&I},$$
and 
$$\det\pmatrix{I&iA_2\\I&A_1}=\det\pmatrix{I&iA_2\\0&A_1-iA_2}$$
hence 
$$\det B=\det(A_1-iA_2)\det(A_1+iA_2)=\det A\cdot\det\bar A=|\det A|^2.$$
A: Davide's answer tells most of the story, in particular giving the proof for the determinant, but not quite all of it, so I want to supplement it with a couple of remarks.
I think that it is more common to replace Davide's matrix $B$ with a real matrix. This can be achieved by conjugating it with matrix of the block form 
$$D=\pmatrix{I&0\cr0&iI\cr},$$ when
$$D^{-1}BD=\pmatrix{A_1&-A_2\cr A_2 &A_1\cr}.$$
Because conjugation preserves the determinant, Davide's calculation tells that here we also have
$$\det(D^{-1}BD)=\det(B)=|\det A|^2.$$
Further conjugating (shuffling rows and columns) allows us to replace each and every complex entry $z=a+bi$ with a $2\times2$ block
$$
(z)=\pmatrix{a&-b\cr b&a\cr}.
$$
Doing it this way makes it clear that if $A$ represents a linear mapping $T$ from
$V=\mathbf{C}^n$ to itself with respect to basis $v_1,v_2,\ldots,v_n$, then
$D^{-1}BD$ represents the same mapping $T$, when we view $V$ as a real vector space of dimension $2n$ and use the basis $v_1,v_2,\ldots,v_n,iv_1,iv_2,\ldots,iv_n.$
The extra shuffling I talked about would reorder this latter basis to $v_1,iv_1,\ldots$.
A geometric interpretation of this is that $\det B$ gives the scaling of Lebesgue measure (or hypervolumes) of a box $K=\prod_{i=1}^{2n}[c_i,d_i]$ of real dimension $2n$ under $T$:
$$
\det (B)=\frac{m(T(K))}{m(K)}.$$
$\det A$ does the same, but because the coordinates are complex there, we need to use $|\det A|^2$ to get the scaling right. This is seen already in the complex plane, where multiplication by $a+bi$ multiplies the areas of rectangles by a factor of $a^2+b^2$.
