eigenvectors of random graphs: nodal domains james r. lee university of washington yael dekel and...
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eigenvectors of random graphs: nodal domains
James R. LeeUniversity of Washington
G(n;p)
Yael Dekel and Nati LinialHebrew University
preliminaries
Random graphsedges present with probability prandom d-regular graph
Adjacency matrix
EigenvectorsA (non-zero) function f : V R is an eigenvector of G ifthere exists an (eigenvalue) for which
for every x 2 V, where (x) is the set of neighbors of x.
eigenvaules of random graphs/discrete matrices
- Let’s arrange the eigenvalues of matrices so that
Much is known about the large eigenvalues of random graphs, e.g. G(n,½)
Wigner semi-circle law Füredi-Komlos
more recently, the small (magnitude) eigenvalues of random non-symmetric discrete matrices
Rudelson 06, Tao-Vu 06, Rudelson-Vershynin 07 (Littlewood-Offord estimates)
TRACE
MET
HOD
Litvak-Pajor-Rudelson-Tomczak-Jaegermann 05 for singular values of rectangular matrices
… but significantly less is understood about the eigenvectors.
spectral analysis
In many areas such as machine learning and computer vision, eigenvectors ofgraphs are the primary tools for tasks like partitioning and clustering.[Shi and Malik (image segmentation); Coifman et. al (PDE, machine learning); Pothen, Simon and Lou (matrix sparsification)]
Heuristics for random instances of NP-hard problems, e.g. - Refuting random 3-SAT above the threshold - Planted cliques, bisections, assignments, colorings
(x1 _ ¹x7 _ ¹x9) ^(x3 _ x11 _ ¹x1) ^¢¢¢
are eigenvectors uniform on the sphere?
If we scale the (non-first) eigenvectors of G(n,½) so they lie on Sn-1, do they behave like random vectors on the sphere?
?For example, do we (almost surely) have…
or
X open problem:
Discrete version of “quantum chaos” (?)
are eigenvectors uniform on the sphere?
Nodal domains
If f : V R is an eigenvector of a graph G, then f partitions G into maximal connectedcomponents on which f has constant sign (say, positive vs. non-positive).
Graph with positive and non-positive nodes marked.
So this graph/eigenvector pair has 6 domains.
Our question:What is the nodal domain structure of the eigenvectors of G(n,p)?
If we scale the (non-first) eigenvectors of G(n,½) so they lie on Sn-1, do they behave like random vectors on the sphere?
Observation: If we choose a random vector on Sn-1 and a random graph, then almost surely the number of domains is precisely 2.
nodal domains
- If fk is the kth eigenvector of G, then a discrete version [Davies-Leydold-Stadler] of Courant’s nodal domain theorem (from Riemannian geometry) says that fk has at most k nodal domains.
observations:
- If G has 2N nodal domains, then it has an independent set of size N, hence N=O(log n)/p.
theorem:Almost surely, every eigenvector of G(n,p) has 2 primary nodal domains,along with (possibly) O(1/p) exceptional vertices.
Experiments suggest that there are at most 2 nodaldomains even for: p À 1p
n
nodal domains
theorem:Almost surely, every eigenvector of G(n,p) has 2 primary nodal domains,along with (possibly) O(1/p) exceptional vertices.
Experiments suggest that there are at most 2 nodaldomains even for: p À 1p
n
pro
bab
ilit
y o
f excep
tion
al vert
ex
number of nodes
can be a delicate issue:In the combinatorial Laplacian of G(n,½),
exceptional vertices can occur(it’s always the vertex of max degree
in the largest eigenvalue)
nodal domains
theorem:
main lemma (2-norm can’t vanish on large subsets):Almost surely, for every (non-first) eigenvector f of G(n,p) and every subset of vertices SWith |S|¸(1-)n, we have ||f|S||2 ¸ p() where p() ! 1 as ! 0 andp() > 0 for < 0.01. (The point is that p() is independent of n.)
follows from…
x
Almost surely, every eigenvector of G(n,p) has 2 primary nodal domains,along with (possibly) O(1/p) exceptional vertices.
the main lemma and LPRT
main lemma:Almost surely, for every (non-first) eigenvector f of G(n,p) and every subset of vertices SWith |S|¸(1-)n, we have ||f|S||2 ¸ p() where p() ! 1 as ! 0 andp() > 0 for < 0.01. (The point is that p() is independent of n.)
Consider p = ½ and |S|=0.99n.
S
z
VnS
¯¯¯¯¯¯
X
y2VAyzf (y)
¯¯¯¯¯¯· j¸ f (z)j
¯¯¯¯¯¯
X
y2V(1 ¡ 2Ayz)f (y) ¡
X
y2Vf (y)
¯¯¯¯¯¯· 2j¸ f (z)j
¯¯¯¯¯¯
X
y2V(1 ¡ 2Ayz)f (y)
¯¯¯¯¯¯. j¸ f (z)j
X
z2S
¯¯¯¯¯¯
X
y2V(1 ¡ 2Ayz)f (y)
¯¯¯¯¯¯
2
. j¸ j2 ¢kf jSk22
the main lemma and LPRT
main lemma:Almost surely, for every (non-first) eigenvector f of G(n,p) and every subset of vertices SWith |S|¸(1-)n, we have ||f|S||2 ¸ p() where p() ! 1 as ! 0 andp() > 0 for < 0.01. (The point is that p() is independent of n.)
Consider p = ½ and |S|=0.99n.
S
z
VnS
kB (f j ¹S)k2 .p
nkf jSk2
B is i.i.d. The above inequality yields
but this almost surely impossible (even taking a union bound over all S’s)
Let B be the 0:99n £ 0:01n bipartite§ 1 adjacency matrix between S and ¹S.
lower bounding singular values
Want to show that for a (1+) n £ n random sign matrix B,
Want to argue that is often large for i.i.d. signs {1, …, n}
The vectors and have very different behaviors.
As 0, need a very good understanding of “bad” vectors. For eps>1, this is easy (Payley-Zygmund, Chernoff, union bound over a net) For 0<eps<1, this requires also a quantitative CLT (for the “spread” vectors) [LPRT] For eps=0, requires a deeper understanding of the additive structure of the coordinates Tao-Vu 06 showed that this is related to the additive structure of the coordinates, e.g. whether (rescaled) coordinates lie in arithmetic progression. (See Rudelson-Vershynin for state of the art)
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