# Condition for block symmetric real matrix eigenvalues to be real

I have a (2nx2n) 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!

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## 1 Answer

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 $i$.

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Hi Carlo! Thank you for your reply and thank you for explaining why the eigenvalues come in pairs, I observed it numerically but did not write it down formally. However, there's a further complication I did not explain well, the coefficient $C$ is not a constant, but actually a $C_{ij}$ depending on the $i,j$ pair, so I cannot follow what you say about a critical value of $C$ (sorry, my fault, I was not precise enough in the question.) It holds anyway that $|C_{ij}|<1$ for any $i,j$. –  Giulia Dec 4 '12 at 16:44