Determinants of block matrices

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.

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If you write "x det y" in TeX, you see $x det y$, but if you write "x\det y", you see $x\det y$. Two differences, at least: "$\det$" is not italicized, and proper spacing before and after $\det$ appears. –  Michael Hardy Dec 13 '12 at 23:34
I appreciate all the help on part 2, but your solutions were too advanced for me. But I managed to solve it on my own. First I scaled the last $n$ rows by $i$ and the last $n$ columns by $i$. Then I used column/row-operations and the multilinearity of the derminant. I ended up with a matrix where I could use the above identity, which resulted in $$\det (-B+iA) \det (-B-iA) = \det (-B+iA) \overline{\det (-B+iA)} = \left| \det (-B+iA) \right|^2$$ And this is obviously positive. –  Elzee Dec 14 '12 at 11:52

Here's a hint for (1): think about scaling the first $n$ rows by $-i$.

Still thinking about (2). I haven't worked this out, but I think considering the signed permutations way of calculating the determinant, and pairing $\sigma$ with $\sigma^{-1}$ might be helpful.

edit: Yeah, okay, just got back, have something kinda similar to levap. A matrix of the form $D$ is exactly a complex linear one, under the identification $\mathbb{R}^{2n} \rightarrow \mathbb{C}^n$, where $\mathbb{R}^{2n}$ has ordered basis $x_1, \ldots, x_n, y_1, \ldots, y_n$ and $\mathbb{C}^n$ $z_i = x_i + i y_i$.

With this, one knows as an operator on $\mathbb{C}^n$ one can write it in Jordan canonical form. If one has a Jordan block: $$\begin{pmatrix} \lambda & - & \ldots \\ & \lambda & \ldots \end{pmatrix}$$with basis $w_i$, setting $w_i = x'_i + iy'_i$, in the ordered basis $x_1', y_1', x_2', y_2' \ldots$, it's easy to check that $D$ has determinant $|\lambda|^r$, where $r$ is the size of the Jordan block.

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I managed (1) now, thanks to your hint. I got no time left for (2) now, I'll have to try that later. Thank you. –  Elzee Dec 13 '12 at 23:10
no worries, glad to help. I'll post (a hint) if I figure it out –  uncookedfalcon Dec 13 '12 at 23:11
1. Consider $\begin{pmatrix}I\\&zI\end{pmatrix}C\begin{pmatrix}I\\&\bar{z}I\end{pmatrix}$ for some complex numbers $z$ of modulus 1.
2. 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.
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This might look highly unmotivated if you are unfamiliar with the notion of a complex linear structure on a vector space. Under the identification of $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ given by $$(z_1, ..., z_n) = (x_1 + iy_1, ..., x_n + iy_n) \to (x_1, ..., x_n, y_1, ..., y_n)$$ the complex matrices $A + iB \in \mathrm{GL}_n(\mathbb{C})$ are identified with the matrices of the form $D$ in $\mathrm{GL}_{2n}(\mathbb{R})$. This matrices are complex linear and so commute with "multiplication by $i$", which corresponds under the identification to multiplication by $J_0$.

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