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The complex general linear group is a subgroup of the group of real matrices of twice the dimension and with positive determinant.

Let us decompose complex matrices $M$ as $M=A+iB$, where $A,B$ are real matrices. Now consider the correspondence $$f(A+iB)=\begin{pmatrix} A & -B \\ B & A\end{pmatrix}.$$

If $\det f(M)=|\det M|^2$ for square matrices, then we would have $GL(n,\mathbb C)\subseteq GL_+(2n,\mathbb R)$ with the identification $M\to f(M)$, which is an injective homomorphism. In other words, the complex general linear group would be a subgroup of the group of real matrices of twice the dimension and with positive determinant.

How is $\det f(M)=|\det M|^2$?

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  • $\begingroup$ What is $\mathrm{det}f(M)$? Is it $A^2 + B^2$? $\endgroup$ – user89987 Mar 25 '14 at 10:42
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Knowing that the determinant of a real matrix treated as a complex one is the same and that fundamental operations do not change the determinant we get: $$det \left(\begin{array}{} A & -B\\B & A\end{array}\right)=det\left(\begin{array}{} A-iB & -B\\B+iA & A\end{array}\right)= det\left(\begin{array}{} \bar{M} & -B\\i \bar M & A\end{array}\right)= det\left(\begin{array}{} \bar M & -B\\i \bar M -i \bar M & A +i B\end{array}\right)= det \left(\begin{array}{} \bar M & -B\\0 & M\end{array}\right)=|detM|^2$$

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