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

linear-algebra complex-numbers matrices determinant

Solution 1:

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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