Eigenvalues of some block matrix Let $\mathbf{M}$ be $2n\times 2n$ square block matrix $$\mathbf{M}=\left[\begin{array}{c|c}
\mathbf{0}&\mathbf{A}\\
\hline
\mathbf{B}&\mathbf{0}
\end{array}\right].$$Here, $\mathbf{A}$ is $n\times n$ square matrix with $\{a,a,\dots,a\}$ as the main diagonal, $\{-a,-a,\dots,-a\}$ as the upper secondary diagonal, and all the other entries are zeros. Similarly, $\mathbf{B}$ is $n\times n$ square matrix with $\{-b,-b,\dots,-b\}$ as the main diagonal, $\{b,b,\dots,b\}$ as the lower secondary diagonal, and all the other entries are zeros. That is, for $n=4$ this matrix would be $$\mathbf{M}=\left[\begin{array}{c c c c|c c c c}
0&0&0&0&a&-a&0&0\\
0&0&0&0&0&a&-a&0\\
0&0&0&0&0&0&a&-a\\
0&0&0&0&0&0&0&a\\
\hline
-b&0&0&0&0&0&0&0\\
b&-b&0&0&0&0&0&0\\
0&b&-b&0&0&0&0&0\\
0&0&b&-b&0&0&0&0
\end{array}\right].$$How to prove that $\mathbf{M}$ has all distinct pure imaginary eigenvalues when both $a$ and $b$ are positive? I don't know which property/principle/theorem to be used here.
 A: The following is a proof that the eigenvalues of $M$ are purely imaginary. 
If
$$
\begin{bmatrix}
0 & A \\
B & 0 
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} 
=
\lambda 
\begin{bmatrix}
x \\
y
\end{bmatrix},
$$
with $x$ and $y$ partitioned comformably with the matrices $A$ and $B$, respectively, then it can be shown that $(AB)x = \lambda^2 x$ via a simple substitution. Thus, $\lambda^2$ is an eigenvalue of $AB$, which (seemingly) has the tridiagonal form
$$
\begin{bmatrix}
-2ab & ab     &        &        &    \\
ab   & -2ab   & ab     &        &    \\
     & \ddots & \ddots & \ddots &    \\
     &        & ab     & -2ab   & ab \\
     &        &        & ab     & -ab
\end{bmatrix}.
$$
Since $AB$ is real and symmetric, it follows that the eigenvalues of $AB$ are real. The Gershgorin theorem asserts that every eigenvalue of the matrix $AB$ lies in the union of the disks $D(-2ab,ab)$, $D(-2ab, 2ab)$, and $D(-ab,ab)$, where, $D(c,r)$ denotes the disk centered at $c \in \mathbb{C}$ and radius $r>0$. But the disk $D(-2ab, 2ab)$ contains the other two disks. Thus, every eigenvalue of $AB$ is non-positive. 
However, a straightforward proof-by-induction shows that $\det AB = (-ab)^n = (-1)^n a^n b^n$. Thus, $AB$ is nonsingular, i.e., zero can not be an eigenvalue of $AB$ and every eigenvalue of $AB$ must be negative. 
Since $\lambda^2$ is an eigenvalue of $AB$, it follows that $\lambda^2 < 0$, which can only occur if $\lambda$ is purely imaginary.       
A: for a simple proof, I'd suggest starting with the special case that $a=b$, then
$\mathbf{M}=\left[\begin{array}{c|c}
\mathbf{0}&\mathbf{A}\\
\hline
\mathbf{B}&\mathbf{0}
\end{array}\right] = \left[\begin{array}{c|c}
\mathbf{0}&\mathbf{A}\\
\hline
-\mathbf{A}^T&\mathbf{0}
\end{array}\right]$   so $\mathbf M$ is skew symmetric and it immediately follows that all eigenvalues of $\mathbf M$ are purely imaginary (i.e. have zero real component).  
A fairly standard result for bipartite graphs tells us that the eigenvalues of $\mathbf M$ are given by the (multi)set of positive square roots and negative square roots of the eigenvalues $\big(\mathbf{AB}\big)$. (If writing in polar form, select angles in $[0, 2\pi)$ for uniqueness.)   
One way of proving this is to observe that for odd powers of k, we have
$\text{trace}\big(\mathbf M^k\big) = \text{trace}\big(\mathbf M^{2r+1}\big)= 0$ 
and for even powers of $k$ we have
$\text{trace}\big(\mathbf M^k\big) = \text{trace}\big(\mathbf M^{2r}\big)= 2\cdot\text{trace}\Big(\big(\mathbf {AB}\big)^{r}\Big)$ 
if you place the positive and negative square roots of $\big(\mathbf {AB}\big)$ in a 2n by 2n diagonal matrix $\mathbf D$ then we have
$\text{trace}\big(\mathbf M^k\big) = \text{trace}\big(\mathbf D^k\big)$
for all natural numbers $k$, which proves the claim.  
Note: in our special case of a=b the eigenvalues of interest are positive and negative square roots of the spectrum of $-\mathbf A^T \mathbf A$ which is a negative definite matrix with determinant $= (-1)^n\det\big(\mathbf A\big)^2 = (-1)^n a^{2n} \neq 0$.     
This special case implies the general case because we simply re-scale all eigenvalues by $\sqrt{\frac{b}{a}} = \text{positive constant}$  which maps the purely imaginary eigenvalues of the special case to the purely imaginary eigenvalues of the general case.  
