I have a $2n \times 2n$ block symmetric matrix that in the simplest case ($n=2$) looks like:

$$ M_2 = \begin{bmatrix} a_1 & 0 & b_{1,2} & -b_{1,2}\\\ 0 & -a_1 & b_{1,2} & -b_{1,2}\\\ b_{1,2} & -b_{1,2} & a_2 & 0 \\\ b_{1,2} & -b_{1,2} & 0 & -a_2 \\\ \end{bmatrix} $$

All the elements are real. The general matrix has then this form:

$$ M_n = \begin{bmatrix} a_1 & 0 & b_{1,2} & -b_{1,2} & & b_{1,n-1} & -b_{1,n-1} & b_{1,n} & -b_{1,n}\\\ 0 & -a_1 & b_{1,2} & -b_{1,2}& \ldots & b_{1,n-1} & -b_{1,n-1} & b_{1,n} & -b_{1,n}\\\ b_{1,2} & -b_{1,2} & a_{2} & 0 & & b_{2,n-1} & -b_{2,n-1}& b_{2,n} & -b_{2,n} \\\ b_{1,2} & -b_{1,2} & 0 & -a_{2} & & b_{2,n-1} & -b_{2,n-1}& b_{2,n} & -b_{2,n}\\\ & \vdots & & & \ddots & &\vdots & \\\ b_{1,n-1} & -b_{1,n-1} & b_{2,n-1} & -b_{2,n-1} & \ldots & a_{n-1} & 0 & b_{n,n-1} & -b_{n,n-1}\\\ b_{1,n-1} & -b_{1,n-1} & b_{2,n-1} & -b_{2,n-1} & \ldots & 0 & -a_{n-1} & b_{n,n-1} & -b_{n,n-1}\\\ b_{1,n} & -b_{1,n} & b_{2,n} & -b_{2,n} & & b_{n,n-1} & -b_{n,n-1} &a_{n} & 0 \\\ b_{1,n} & -b_{1,n} & b_{2,n} & -b_{2,n} & \ldots & b_{n,n-1} & -b_{n,n-1} & 0 & -a_{n} \end{bmatrix} $$

Now, I am solving the eigenproblem numerically for various dimensions of $M$, and I always find the eigenvalues to be real for my values of $\{a_i\}$ and $\{b_{i,j}\}$.

I have the feeling that this is because in general the values $a_i$ on the diagonal are bigger than the off-diagonal elements $b_{i,j}$, but I would like to state a rule for this, because I want to be sure that in no case I will find complex eigenvalues.

Can anyone help me find out what is the condition for the eigenvalues of $M$ to be all real?

Thank you!

Note: To be a little more precise, the relation between the matrix elements is

$$b_{ij} = C_{ij}\frac{c_ic_j}{2\sqrt{a_i a_j}}$$

with $|C_{ij}|<1$ and $ c_i < a_i$. In the case $M_2$, where I can easily calculate the characteristic polynomial, I can show using this relation that eigenvalues are real. Maybe the higher dimension cases can be proved by induction? I tried but failed!