I have $n$ particles on a circle with the Hamiltonian \begin{equation} H = \sum_{n=1}^N \frac{p_n^2}{2m} + \frac{1}{2}m\omega^2 \sum_{n=1}^N (x_{n+1}-x_n)^2 \end{equation}
I need to find the energy eigenvalues for this system. The potential can be written in terms of a Circulant matrix \begin{equation} V = \frac{1}{2} m\omega^2 X^T\underbrace{\begin{pmatrix} 2 & -1 & 0 &\cdots &0 & -1\\ -1 & 2 & -1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots& \ddots &\vdots & \vdots \\ -1 & 0 & 0 &\cdots & -1 & 2 \end{pmatrix}}_{N\times N \text{ matrix } C}X \end{equation}
Where $X =\begin{pmatrix}x_1 & x_2 & \cdots & x_n \end{pmatrix}$. I now have to diagonalize the matrix $C$. I know that the eigenvectors of a circulant matrix are given as \begin{equation} X^{[q]}= \begin{pmatrix} 1 & \exp \frac{2\pi q}{N} & \exp \frac{4 \pi q}{N} &\cdots & \exp\frac{2(n-1)q\pi}{N} \end{pmatrix} \end{equation} And the corresponding eigenvalues of $m\omega^2 C$ are \begin{equation} \lambda_q = 4 m \omega^2 \sin \frac{\pi q}{N} \end{equation}
But these eigenvectors are not real and I am therefore having trouble to get the eigenenergies of the system. Can someone help me with this problem?