7

A graph is called walk regular if the number of closed walks starting from vertex $u$ of length $k$ does not depend on $u$. If $A$ is the adjacency matrix of the graph, this means that $A^k$ has equal diagonal values for every $k>0$.

I'm interested in a weaker condition, that there exists two vertices $u,v$ such that the number of closed walks of length $k$ is the same for $u$ and $v$. Are there results or references about this?

In other words, when does $[A^k]_{uu} = [A^k]_{vv}$ hold?

For example, it would be nice to have necessary/sufficient conditions for this property to hold. If there is a graph automorphism that maps $u\to v$ then certainly $u$ and $v$ have the same number of closed walks, though the existence of such an automorphism is not necessary (an example is the Folkman graph which is walk-regular but not vertex-transitive).

I'm aware that if the graph has $n$ vertices and $u,v$ have the same number of closed walks for $k=1,2,\ldots n-1$, then by Hamilton-Cayley's theorem it is true also for all $k\geq n$.

I'm asking in the case of a simple undirected graph, but if there are generalizations for directed and/or weighted graphs they're highly welcome.

askarmboze
  • 171
  • Have you tried to make use of Ihara's Zeta function? – draks ... Jul 02 '20 at 08:37
  • @draks... I did not know about it, thanks for the reference! Sadly, I do not see how to use it, since 1) it uses all the paths in a graph, and not those starting from only one vertex 2) it uses paths that do not pass through the same edge twice consecutively (going forward and then immediately backwards), while I'm counting also the paths that do. – askarmboze Jul 06 '20 at 12:00
  • So you ask if $(A^k){u,u}=(A^k){v,v}$? – draks ... Jul 07 '20 at 06:24
  • Yes, I'm looking for necessary/sufficient conditions on A for it to happen. – askarmboze Jul 09 '20 at 20:12

1 Answers1

3

Continuing my search through various articles, I found the answer: two such vertices are called cospectral vertices.

There is a result, that for an undirected graph $G$ with adjacency matrix $A$, the following are equivalent to $u,v$ cospectral:

  1. For every integer $k>0$, $[A^k]_{u,u}=[A^k]_{v,v}$.
  2. The graphs obtained removing a vertex, $G\setminus \{u\}$ and $G\setminus \{v\}$ have the same spectrum
  3. For every $\lambda$ eigenvalue of $A$, let $P$ be the spectral projector on the $\lambda$-Eigenspace of $A$ (since $A$ is symmetric, $P$ is orthogonal). The condition is then $P_{u,u} = P_{v,v}$ for every spectral projector.

Equivalently, the third condition can be reformulated through graph angles.

References:

Chris Godsil, Jamie Smith, "Strongly Cospectral Vertices" , on arXiv https://arxiv.org/abs/1709.07975
In this articles more equivalent conditions are given, and is studied a generalization of the concept.

Chris Godsil, Brendan D. McKay, "Feasibility Conditions for the existence of Walk-Regular Graphs" , Linear Algebra Appl., Vol. 30 (1980), pp. 51-61
This paper proves properties for walk-regular graphs, but its results can be applied also to cospectral vertices.

Dragoš Cvetković, Peter Rowlinson, Slobodan Simić, "Eigenspaces of Graphs", Cambridge University Press (1997)
A very good book, where can be found results concerning graph angles.

Allen J. Schwenk,"Almost all trees are cospectral" , New Directions in the Theory of Graphs, Academic Press (1973), pp. 275–307
I think this was the first paper to use cospectral vertices (although as in the above condition 2, so not explicitly).

askarmboze
  • 171