Search Results

I'll add some thoughts partially in response to Igor's answer, in that while I agree that cospectral graphs are intrinsically interesting, I think there is a bit more to it than that. Many authors (i …

I think this is false. Here is a graph and some code in Sage to compute the sum of elements of $AADAAD$ which appears to be odd. I tried to prove it for a while, failed, so then decided to try so …

So here is a family of graphs that satisfies your requirements.... is this the only family? Let $X_1$ be the graph consisting of $n-1$ disjoint copies of $K_2$. Then the spectrum of $X_1$ is $$\under …

This is a limited partial answer to the question ruling out the case where the graph is regular and has four distinct eigenvalues, so the spectrum is $\{k, (\sqrt{2})^a, (-\sqrt{2})^a, -k\}$. van Dam' …

The simplest source of cospectral graphs is lists of strongly regular graphs, lots of which are easily available from Ted Spence's web page at http://www.maths.gla.ac.uk/~es/srgraphs.php. Otherwise y …

Maybe I'll summarise everything from the comments as an answer. Firstly, a perfect matching $M$ of $K_{n,n}$ can be identified with a permutation of the set $[n] = \{1,\ldots,n\}$ simply by numbering …

Yes, I believe that it will have simple spectrum for d >= 3 and it feels like something that should have been proved, though I can't actually find it. There is a loose association between automorphis …

If you are interested in graphs (not digraphs), then the elements of the connection set must come in pairs, so you are only looking at subsets $$ C \subseteq \{\pm1, \pm2, \pm3, \pm4, \pm5\}. $$ Mor …

Here is the graph as requested from the comments. Take the cuboctahedron and then assign colours to the edges as follows: There are 8 triangles and 6 quadrilaterals in this graph, and it takes a mome …

This is a question that seems like it should have been studied before, but for some reason I cannot find much at all about it, and so I am asking for any pointers / references etc. The determinant of …

Yes, this is true, but I don't know a reference, so here's a proof (I think). Let $$ R(A, x) = \frac{x^T A x}{x^T x} $$ be the Rayleigh quotient. We know that for a symmetric (in fact Hermitian) matr …

I was looking for some work on half-transitive graphs when I stumbled across this question, to which the answer is "No" (at least in the original form where you want to bound the multiplicity by 2). H …

This is not an answer to the original question, just an answer to the question in the comments about "how did you calculate the characteristic polynomial?", which I couldn't fit into a comment. Actual …

Here is an alternative (more combinatorial) proof to the one linked to in my comment. Suppose that the digraph $D$ has a vertex of in-degree zero, which we may assume is vertex $1$. Then letting $\var …

This is false. Run this code in SageMath; you can do this at sagecell.sagemath.org if you do not have SageMath already installed on your own computer. h = DiGraph('DKCYW?') print(h.laplacian_matrix(). …