Báo cáo toán học: "Low Rank Co-Diagonal Matrices and Ramsey Graphs" doc

... ,is has rank at most k s , s ≤ ck 1/ , and co-diagonal matrix A has rank at most k 2ck 1/ , and this proves the theorem. Acknowledgment. The author is grateful to Noga Alon for his comments and ... constructions of Ramsey graphs. Keywords: composite modulus, explicit Ramsey- graph constructions, matrices over rings, co-diagonal matrices 1 Introduction In this work we ex...

Ngày tải lên: 07/08/2014, 06:20

... diagonal entry of E is 0 and A is complex (respectively, real), then by Corollary 2.4, rank( A) ≥ rank( B) + rank( B T ) ≥ 6 + 6 = 12 (respectively, rank( A) ≥ rank( B) + rank( B T ) ≥ 7 + 7 = 14). If ... zero-nonzero patterns C S 1 and C S 2 such that the minimum rank of matrices described by C S 1 is less over the complex numbers than over the real numbers 1 , and the minimu...

Ngày tải lên: 07/08/2014, 15:23

Ngày tải lên: 05/08/2014, 09:46

... R(K 3 ,K 3 ) > 5andR(C 4 ,C 4 ) > 5havethe adjacency matrices X and Y .ObservethatX + Y + I is the all-ones 5 × 5 matrix. We now construct four 26 × 26 adjacency matrices M i ,sothatM 1 and M 2 contain ... adjacency matrices M 1 and M 2 . We denote the vertices from the top of the matrices as 1, 2, 26. The graphs are showninFigure1. Vertices1−25 are the triangle-free const...

Ngày tải lên: 07/08/2014, 08:22

... Definitions and Terminology Let P be a locally finite, ranked partially ordered set with least element ˆ 0 and finitely many elements of each rank. For x, y ∈ P, we say that y covers x if x < y and ... exist sequences of constants {q n } n≥0 and {r n } n≥0 such that for any poset element x of rank n, DU(x) = q n UD(x)+r n x. Here, we introduce natural raising and lowering operato...

Ngày tải lên: 07/08/2014, 13:21

... 0-simplex and edge instead of 1 -simplex. Fo r more details and terminology concerning Graph Theory, see book of Diestel [4]. Proposition 1.3. Let M be a graph, N ⊆ M be a subgraph of M and f be ... SPAIN, 2007 and P.A.I. project 2008/FQM-189, SPAIN, 2008. the electronic journal of combinatorics 16 (2009), #R38 1 In this example m 0 = ∞, m 1 = ∞, b 0 = 1 and b 1 = 0. So, (I1) with p...

Ngày tải lên: 07/08/2014, 21:21

... ver tex set V = M ∪ T 1 ∪ T 2 of size n and containing all edges in the two trees and the expander on Q. We claim: 1. G is ( d n , 27 √ d)-quasirandom and has vertex degree in the range d(1 ± ... Lemmas 10 and 11, respec- tively. For these lemmas, we assume v(1) = v min = b > 0 and define S t and n t analogously: • S 1 = {1} • S t+1 = {i : i ∈ N(S t ) and v(i)  b(1 + c(t + 1)...

Ngày tải lên: 08/08/2014, 01:20

... than ε. On the other hand, for a sufficiently large i  j, the restrictions of σ(P 1 ) and σ(P 2 ) to ˆ S i are also longer than ε. Thus, the distance along C i between the first and the last vertex ... the first vertices of P i and Q 1 lie in the component of G − S i that contains x (or one of its rays if x is an end). The same is true for y and the last vertices of P i and Q 1 . Thus,...

Ngày tải lên: 08/08/2014, 01:20

... and undirected graphs G = (V, E) with vertex set V , edge set E, o rder n = |V |, and size m = |E|. The neighb ourhood of a vertex u ∈ V in the graph G is the set N G (u) = {v ∈ V | uv ∈ E} and ... G. Dominating and independent sets a r e among the most well-studied graph theoretical objects. The literature on this subject has been surveyed and detailed in the two books by Haynes, H...

Ngày tải lên: 08/08/2014, 12:22

... )are distinct and non-zero, so by the inductive hypothesis X and Y have full rank, and hence do not have zero as an eigenvalue. The spectrum of X ∪ Y is the union of the spectra of X and Y ,andsoX ∪ ... distinct and non-zero. Then Z has full rank (that is, rank equal to |V (Z)|). Proof. We will prove this by induction on |V (Z)|. Clearly it is true if | V (Z)| =2 because the on...

Ngày tải lên: 07/08/2014, 07:21