Let $A$ be the $N\times N$ adjacency matrix of a 1D grid graph (also called path graph?) of size $N\times 1$. this can be seen as a discrete line of $N$ connected vertices:
I know that the eigenvalues of $A$ can be represented as: \begin{equation} \lambda_k=2\cos \left(\frac{\pi k}{N+1}\right) \end{equation} for $k=1,\dots,N$.
What would be the eigenvectors of the adjacency matrix $A$?
$A$ is real and symmetric so I would expect to find a set of $N$ real orthonormal eigenvectors.
Edit By playing around, I found such a set: let $\vec{v}^{(k)}$ be the eigenvector associated to eigenvalue $\lambda_k$. Then we have: \begin{equation} v_j^{(k)}=\sqrt{\frac{2}{N+1}}\sin\left(\frac{\pi j k}{N+1}\right) \end{equation} satisfying the orthonormal relations: $\sum_j v_j^{(k)}v_j^{(l)}=\delta_{kl}$.
Is there any intuitive explanation behind this? If we had added weights to our graph, would the eigensystem be preserved up to some proportional term per node?