If a graph $G$ has an [equitable partition](https://mathoverflow.net/questions/96858/complexity-of-equitable-partitions), then its charachteristic polynomial (for the adjacency matrix) has a divisor that can be seen as the characteristic polynomial (for the adjacency matrix) of a oriented and weighted, but smaller, graph $G'$. In other words, the spectrum of $G$ contains the spectrum of $G'$. Equality (without counting multiplicity) holds if the graph has further properties, in particular if it is walk regular (an [old result by Godsil and McKay](http://cs.anu.edu.au/~bdm/papers/WalkRegular.pdf)).


If $G$ only has an [almost equitable partition](https://mathoverflow.net/questions/122340/algorithmic-almost-equitable-partitioning), then the above assertion [still holds](http://www.sciencedirect.com/science/article/pii/S0195669806000072) if we replace "adjacency matrix" by "discrete laplacian". Is it known whether equality of spectra holds in special cases?



almost equitable partitions and spectra
I think the real problem is that the obvious notion of ``Laplacian walk-regular'' implies that the graph must be regular.


The proof of the ``old result'' can be modified to yield the following.
Let $L$ be the Laplacian of the graph $X$, let $v$ be a vertex in $X$ and
let $L\_v$ denote the matrix we get by deleting the $v$-row and $v$-column
from $L$. (So $L\_v$ is not a Laplacian.) Assume that if $\lambda$ is an eigenvalue
of $L$ with multiplicity $m\ge1$, then the multiplicity of $\lambda$ as an eigenvalue of $L\_v$ is $m-1$ (the least it can be). If $\pi$ is an almost equitable partition of $X$ in which $\{v\}$ is a cell, then each eigenvalue of $L$ occurs as an eigenvalue of $L\_v$.


This condition implies the old result, but I think it would be difficult to apply in practice.


