Báo cáo toán học: "Low Rank Co-Diagonal Matrices and Ramsey Graphs" doc
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...
Báo cáo toán học: "Minimum rank of matrices described by a graph or pattern over the rational, real and complex numbers" doc
... 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...
Báo cáo toán học: "A Note on Two Multicolor Ramsey Number" doc
... 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...
Báo cáo toán học: "The Induced Subgraph Order on Unlabelled Graphs" doc
... 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...
Báo cáo toán học: "Discrete Morse inequalities on infinite graphs" docx
... 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...
Báo cáo toán học: "Entrywise Bounds for Eigenvectors of Random Graphs" docx
... 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)...
Báo cáo toán học: "Geodetic topological cycles in locally finite graphs" docx
... 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,...
Báo cáo toán học: "Random Procedures for Dominating Sets in Graphs" doc
... 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...
Báo cáo toán học: "The Rank of a Cograph" pptx
... )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...