Eigenvalue location for chain graphs

No threshold graphs are cospectral

“…Proof: Let D be the matrix of H given by (1). We will prove the case n even, the case n odd follows a similar way.…”

“…, t l ) ∈ I n,l then t i ≡ n + i − l (mod 2). In other words, elements in I n,l are increasing sequences alternating even and odds numbers such that the last term has the same than parity n. For instance while I 6,4 = {(1, 2, 3, 4), (1,2,3,6), (1,2,5,6), (1,4,5,6), (3,4,5,6)}.…”

Vertex types in threshold and chain graphs

“…Let W be the eigenspace corresponding to λ. If for each x ∈ W , we have x(v) = 0, then v cannot be a downer vertex as for any x ∈ W , the vector x ′ obtained by deleting the v-th component, is a λ-eigenvector of G − v, and therefore we have The rest of the paper is organized as follows: in Section 2 we give some particular results about vertex types in threshold graphs, while in Section 3 we put focus on chain graphs, and among others we disprove Conjecture 3.1 from [1], which states that in any chain graph, every vertex is a downer with respect to every non-zero eigenvalue. Besides we point out that some weak versions of the same conjecture are true.…”

“…A question raises whether they can have neutral vertices. In [1] it is conjectured that this cannot be the case.…”

“…of P 4 -free graphs. Recall that threshold graphs are {P 4 , 2K 2 , C 4 }-free graphs, while chain graphs are {2K 2 , C 3 , C 5 }-free graphs -see [1,3] for more details. Note, if these graphs are not connected then (since 2K 2 is forbidden) at most one of its components is non-trivial (others are trivial, i.e.…”