# Non-negative determinant of a block matrix

Here's the problem I've been stuck on for some time now.

Let $A,B \in M_n(\mathbb{R})$. Let $C= \begin{bmatrix} A & B \\ -B & A \\ \end{bmatrix}$ be a real $2n \times 2n$ matrix. Prove $\det(C) \geq 0$.

What I've tried so far are:

1. First I tried to write determinant of $C$ as the sum of $2^{2n}$ matrix determinants by expanding all rows of $C$ such that for each binary sequence of length $2n$ like $a = (a_1, a_2, ..., a_{2n})$, if $a_i = 0$ then the first $n$ entries of $i$-th row are zero and if $a_i = 1$ then the second $n$ entries of $i$-th row are zero. But couldn't come close to any answer.
2. Second we know that determinant is the product of eigenvalues. The characteristic polynomial of $C$ has real coefficients hence its complex roots come in conjugate pairs and have positive product. What remains is to prove that each negative eigenvalue has even multiplicity which I couldn't prove.

Any sort of hints and/or ideas are appreciated.

Hint. Start with $$\det{\begin{bmatrix}A & B \\-B & A\end{bmatrix}}=\det{\begin{bmatrix}A-iB& B+iA \\-B & A\end{bmatrix}}.$$
• $B+iA=i(A-iB)$. – user26857 May 16 '16 at 9:35
Let $\left( \begin{matrix} x \\ y \end{matrix} \right) \in \mathbb{R}^{2n}$ be an eigenvector with eigenvalue $\lambda$.
Then $\left( \begin{matrix} -y \\x \end{matrix} \right)$ is also an eigenvector with eigenvalue $\lambda$, so all eigenvalues have an even number of linear independent eigenvectors.
• @FedericoPoloni A generalized eigenvector of rank $2$ has an associated eigenvector of rank $1$, and you can switch $x$ and $y$ in both eigenvectors, to obtain a generalized an associated eigenvector of the same eigenvalue, thus still keeping an even number of eigenvalues. And the same for eigenvectors of higher rank. – Hetebrij May 16 '16 at 18:57