So I have to deal with a square matrix with complex elements $A$ given by -
$$A_{ij}(k) = \underbrace{\frac{1}{2}(c_{ij}+c_ji)}_{\text{symmetric under interchange of $i$ and $j$}} \exp(Ik(i-j))$$
Where $I=\sqrt{-1}$
Notice that the matrix is Hermitian iff $k$ is real.
The only symmetry i could meaningfully obtain constraints on the eigenvalue was -
$$A_{ji}(Re(k),-Im(k)) = \left(A_{ij}(Re(k),Im(k))\right)^* $$
$$\implies A_{ij}(Re(k),Im(k)) = \left(A_{ji}(Re(k),-Im(k))\right)^* $$
$$\implies \sum_{ij}U^\dagger_{ai} A_{ij}(Re(k),Im(k)) V_{jb} = \sum_{ij}U^\dagger_{ai} \left(A_{ji}(Re(k),-Im(k))\right)^* V_{jb}$$
$$RHS = \left(\sum_{ij} V^\dagger_{bj} A_{ji}(Re(k),-Im(k)) U_{ia}\right)^* $$ $$\therefore \text{complex eigenvalues } a_n(Re(k),-Im(k)) = \left(a_n(Re(k),Im(k))\right)^* \implies a_n(k^*) = \left(a_n(k)\right)^*$$
Thus - $$ Re(a_n(k^*)) = Re(a_n(k)) $$ $$ Im(a_n(k^*)) = -Im(a_n(k)) $$
This is all I have till now. It would be nice if someone could prove some of the following observations I have made by plotting the values over a grid of k, with the constraint $Im(a_n(k)) = 0$.
While here the eigenvalues appear to be a complex function $a_n$ of a complex variable, if we manage to invert them such that we could talk of a multivalued function $k(a_0)$ given that $a_0\in \mathbb{R}$ -
- $\forall a_0\in \mathbb{R}$ there are exactly N $k_0 = $, where N is the dimension of the square matrix $A$.
- there are are saddle points at $k=\pm 2\pi$ and at $k=0$.
Note that these observations might not be generally true for such matrices - only for special cases!