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Consider numbers of the type $n=2^m+1$. Prove that such an n is prime only if $n=F_K$ for some $ k \in N$, where $F_k$ is a Fermat Prime.

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.

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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

$$\det B=\det(A_1-iA_2)\det(A_1+iA_2)=\det A\cdot\det\bar A=|\det A|^2.$$

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 $a^2+b^2$.

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*}

$$

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