TITLE: Determinants of block matrices
QUESTION [6 upvotes]: Let $A,B \in \mathbb{R}^{n,n}$.
Now $C = \begin{pmatrix}  A & iB \\ -iB & A \end{pmatrix}$ and $D = \begin{pmatrix} A & B \\ -B & A \end{pmatrix}$.
Show that $\det(C) \in \mathbb{R}$ and $\det(D) \ge 0$.
I tried to transform $C$ so I could use $\det \begin{pmatrix} E & F \\ 0 & G \end{pmatrix} = \det(E)\det(G)$, but I didn't manage to. I have no clue how to show $\det(D) \ge 0$.
I'd rather have hints than fully-fledged solutions. Thanks.

REPLY [3 votes]: Consider $\begin{pmatrix}I\\&zI\end{pmatrix}C\begin{pmatrix}I\\&\bar{z}I\end{pmatrix}$ for some complex numbers $z$ of modulus 1.
As invertible matrices are dense in the matrix space and determinant is a continuous function in matrix entries, we may assume that $A$ is invertible. Using the block determinant formula, we get $\det D = \det A \det(A+BA^{-1}B) = (\det A)^2\det(I+A^{-1}BA^{-1}B)$. Now for any real matrix $X$, argue why $\det(I+X^2)$ must be nonnegative.