# Proving $\det \left( \begin{smallmatrix} A & -B \\ B & A \end{smallmatrix} \right) =|\det(A+iB)|^2$

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

• What is $\mathrm{det}f(M)$? Is it $A^2 + B^2$? – user89987 Mar 25 '14 at 10:42

## 1 Answer

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: $$det \left(\begin{array}{} A & -B\\B & A\end{array}\right)=det\left(\begin{array}{} A-iB & -B\\B+iA & A\end{array}\right)= det\left(\begin{array}{} \bar{M} & -B\\i \bar M & A\end{array}\right)= det\left(\begin{array}{} \bar M & -B\\i \bar M -i \bar M & A +i B\end{array}\right)= det \left(\begin{array}{} \bar M & -B\\0 & M\end{array}\right)=|detM|^2$$