Consider an undirected ring graph of $n$-vertices. To be more specific, each node $i$ has edges $(i,i+1)$ and $(i,i-1)$. When all edges have weight $w_{i,j}=1$ then the eigenvalues and eigenvectors of the corresponding Laplacian are given by sinusoidal functions, see e.g. here
However, this construction and nice description of eigenvalues and eigenvectors seems to break as soon as we consider slightly different weights. Is there an intuition of why this happens? Is there a generalization to describe the eigenvalues and eigenvectors of a weighted ring graph?
Best Answer
The key point is that, depending on the graph weights, the Laplacian $\Delta$ may or may not commute with certain operators. When it's the case, then you're able to use that extra knowledge to infer some properties of the eigenvectors.
As long as the weights are all the same (say, $1$), then the graph has plenty of symmetries. So certain operators commute with the Laplacian $\Delta$ on the set of functions defined on the graph vertices. For instance, the Laplacian commutes with the "shift" operator $\tau$ defined by $$\tau f (i)=f(i+1) \text{ with node indices modulo }n$$ Likewise, it also commutes with the reflection operator, defined by $$\rho f(i)=f(-i) \text{ with node indices modulo }n$$ The commutativity implies that $\Delta$'s eigenspaces must be left invariant by those operators, leading to those nice "Fourier" eigenvectors.
Now, if you break those symmetries by tweaking the weights, there's no reason to expect any particular behavior for the eigenvectors. Unless your tweaking still preserves some symmetries.