Condition for block symmetric real matrix eigenvalues to be real 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!
 A: this is not a complete answer, but it's a bit too long for a comment.
first notice that ${\rm Det}\;(\lambda-M)={\rm Det}\;(-\lambda-M)$ and ${\rm Det}\;(\lambda-M)={\rm Det}\;(\bar{\lambda}-M)$; it follows that
the eigenvalues $\lambda$ come in pairs $+\lambda,-\lambda$ and $\lambda,\bar{\lambda}$; if you fix the $a_i$'s and the $c_i$'s, and then follow the evolution of the eigenvalues with increasing $C$, starting from $C=0$, you will find that the eigenvalues all start off on the real axis, arranged symmetrically around the origin; Then at some critical value $C_0$ of $C$ a pair of eigenvalues meet at the origin, and take off in opposite directions along the imaginary axis.
To calculate this critical value of $C$, we demand that the determinant of $M$ vanishes; for the simple case $n=2$ this happens at
$$C_0=\frac{a_1 a_2}{c_1 c_2},$$
in agreement with your finding that all eigenvalues are real if $c_i{<}a_i$ and $C<1$.
For larger $n$ it remains to prove that $C_0>1$ if $c_i{<}a_i$ for all $i$.
