A question about the determinant of a block matrix.

Question

Suppose that $A，B\in M_n(\mathbb{C})$. Is it right that $$\det\begin{pmatrix}A&B\\-\bar{B}&\bar{A}\end{pmatrix}\geq 0$$ If this is right, how can I prove it?

Answer 1

By a continuity argument, you may assume that $A$ is nonsingular. Let $C=A^{-1}B$. Then $$ \det\pmatrix{A&B\\ -\overline{B}&\overline{A}} =\det\left[\pmatrix{A&0\\ 0&\overline{A}}\pmatrix{I&C\\ -\overline{C}&I}\right] =|\det(A)|^2\det(I+C\overline{C}). $$ For any complex square matrix $C$, all eigenvalues of $C\overline{C}$ that are not real nonnegative must occur in conjugate pairs (see D.C. Youla, A normal form for a matrix under the unitary congruence group, Canad. J. Math. 13, 1961, pp.694-704). Hence $\det(I+C\overline{C})\ge0$ (see also Complex square matrices. Difficult proof.) and the conclusion follows.