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!

  • $\begingroup$ A possibly useful step is that after applying a perfect shuffle permutation that matrix takes the form $\Pi^T M_n \Pi = \begin{bmatrix}B+A & -B \\ B & -A-B\end{bmatrix}$, where $A=diag(a_i)=A^T$ and $B = B^T= (b_{ij})$ with zeros on its diagonal. In particular, this matrix is Hamiltonian, which explains the symmetry in the eigenvalues already noticed by @Carlo. $\endgroup$ Jul 22, 2020 at 20:20

1 Answer 1


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 $i$.

  • $\begingroup$ 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 $i,j$. $\endgroup$
    – Giulia
    Dec 4, 2012 at 16:44

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