Determinant inequality for skew-symmetric 2×2 block matrices

Question

I've come across the following inequality:

Show that for every pair of square matrices with the same dimensions A and B, the following inequality holds:

$$\det\begin{bmatrix} A & B \\ -B & A \end{bmatrix}\ge0$$ Applying elementary operations would result in the following equalities: $$\det\begin{bmatrix} A & B \\ -B & A \end{bmatrix}=\det(A-iB)\det(A+iB)=\det(A^2+i(AB-BA)+B^2)$$ But I can't go any further.

Answer 1

The inequality holds only for real matrices (let $A=iI_3$ and $B=0$).
With the matrices being real, we have $A-iB=\overline{A+iB}$ (complex conjugate). As a result, we have: $$\det(A-iB)=\overline{\det(A+iB)}\implies\det(A-iB)\det(A+iB)=|\det(A+iB)|^2=|\det(A-iB)|^2$$ Finally: $$\det\begin{bmatrix} A & B \\ -B & A \end{bmatrix}=|\det(A+iB)|^2\ge0$$