Suppose we have n particles connected by a string tied to fixed ends.

In matrix notation the equation of motion can be written as
$\begin{pmatrix}2\omega_o^2&-\omega_o^2&0&0&.&.&.&0\\ -\omega_o^2&2\omega_o^2&-\omega_o^2&0&.&.&.&0\\ 0&-\omega_o^2&2\omega_o^2&-\omega_o^2&0&0&...&0\\.\\.\\.\\ 0&0&0&.&.&-\omega_o^2&2\omega_o^2&-\omega_o^2\\ 0&0&0&.&.&.&-\omega_o^2&2\omega_o^2&\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\.\\.\\.\\x_{N-1}\\x_N\end{pmatrix}=\begin{pmatrix}\ddot x_1\\\ddot x_2\\\ddot x_3\\.\\.\\.\\\ddot x_{N-1}\\\ddot x_N\end{pmatrix}\tag{1}$
So, to solve it we have to diagonalize the matrix.

If my matrix is this:
$A=\begin{pmatrix}2\omega_o^2&-\omega_o^2&0&0&.&.&.&-\omega_o^2\\ -\omega_o^2&2\omega_o^2&-\omega_o^2&0&.&.&.&0\\ 0&-\omega_o^2&2\omega_o^2&-\omega_o^2&0&0&...&0\\.\\.\\.\\ 0&0&0&.&.&-\omega_o^2&2\omega_o^2&-\omega_o^2\\ -\omega_o^2&0&0&.&.&.&-\omega_o^2&2\omega_o^2&\end{pmatrix}$
This I have studied how to diagonalize. We can see that the matrix $A$ shares the same eigenvectors with the upper shift matrix (as their commutator is 0)which is given as:
$U=\begin{pmatrix}0&1&0&0&.&.&.&0\\ 0&0&1&0&.&.&.&0\\ 0&0&0&1&0&0&...&0\\.\\.\\.\\ 0&0&0&.&.&0&0&1\\ 1&0&0&.&.&.&0&0&\end{pmatrix}$
The eigenvalues of $U$ is given as $\omega^k$ where $\omega=e^{i\frac{2\pi}{N}}$ and $0\leq k< N$
The eigen vectors are given as $\begin{pmatrix}1\\ \omega^k\\\omega^{2k}\\...\\\omega^{(N-1)k}\end{pmatrix}$

But the matrix given in $(1)$ is different from $A$, then is there any way that we can diagonalize the $A$?