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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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up vote 13 down vote accepted

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

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Thanks you very much! – dernam Jul 3 '12 at 18:59
You are welcome! – Davide Giraudo Jul 3 '12 at 19:01

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 $a^2+b^2$.

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If $A$ and $B$ define the "same" linear map shouldn't $\det A = \det B$? – learner Jan 24 '15 at 4:27
@learner: No. For example multiplication by $2$ on $\Bbb{C}$ has determinant two (it's a 1x1 matrix $2I_1$), but when we view $\Bbb{C}$ as $\Bbb{R}^2$ it is a 2x2 matrix $2I_2$ with determinant $=4$. – Jyrki Lahtonen Jan 25 '15 at 9:04

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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This question already has a well accepted answer. You have contributed nothing new – Shailesh Nov 14 '15 at 13:42

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