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Questions related to the spectrum of graphs, defined using one of the possible variants of the discrete Laplace operator or Laplacian matrix. See https://en.wikipedia.org/wiki/Discrete_Laplace_operator
5
votes
A graph spectra problem?
The adjacency matrix of the product is $A_1 \otimes J + I \otimes A_2$, where $J$ is the all ones matrix of size $n = |V(G_2)|$ and $I$ is the identity matrix of size $m = |V(G_1)|$. The two matrices …
4
votes
1
answer
89
views
Strongly/distance regular graphs over $\mathbb{Z}_2^n$ with the same parameters
I am wondering if there is a known example of a pair of non-isomorphic graphs $G$ and $H$ that are both Cayley graphs for $\mathbb{Z}_2^n$ (for some $n$) and are both distance regular and have the sam …
2
votes
Accepted
Co-spectral fractional isomorphic graphs with different Laplacian spectrum
EDIT: According to your link, if two graphs are cospectral with a common equitable partition, then they have cospectral complements. But this implies that they have cospectral Seidel matrices (see The …
9
votes
Spectrum of orthogonality graph (2)
If $G$ is a Cayley graph for $\mathbb{Z}_2^n$ with connection set $C \subseteq \mathbb{Z}_2^n \setminus \{0\}$, then for each element $a \in \mathbb{Z}_2^n$ there is an eigenvector $v$ given by
$$v_x …
6
votes
1
answer
558
views
Eigenvalue inequality for regular graphs
I recently proved an inequality relating some of the eigenvalues of a regular graph with each other, and I was wondering if it is already known. I was unable to find it online, and a quick skim throug …
10
votes
0
answers
225
views
Cospectral mate of rhombic dodecahedron
I am wondering if the following pair of cospectral graphs was previously known.
The rhombic dodecahedron graph looks like this (graph6 string: 'M?????rrAiTOd_YO?'):
As far as I know, it was previously …
4
votes
How do eigenvalues of combinatorial Laplacian relates to automorphisms in graphs?
I think you can more or less only go in one direction here: a large amount of symmetry can imply few eigenvalues. Intuitively, this makes sense because if $f: V(G) \to \mathbb{R}$ is an eigenvector fo …