bounding the mixing time via spectral gap graph random walk seminar fall 2009

Ilan Ben-Bassat Omri Weinstein

Outline

Why spectral gap? Undirected Regular Graphs. Directed Reversible Graphs. Example (Unit hypercube). Conductance. Reversible Chains.

P- Transition matrix (ergodic) of anundirected regular graph.

P is real, stochastic and symmetric, thus: All eigenvalues are real. P has n real (orthogonal) eigenvectors.

P’s eigenvalues satisfy:

Why do we have an eigenvalue 1? Why do all eigenvalues satisfy ? Why are there no more 1’s? Why are there no (-1)’s?

1...1 1210 N

1||

Laplacian MatrixL = I – P

So, L is symmetric and positive semi definite.

L has eigenvalue 0 with eigenvector 1v.

n

ji

n

jijiij

n

jj

n

jiijji

n

iiijji

n

ii

ttt yyPyPyyyPyyyPyyIyyLyy1, 1,

2

1

2

1,1

2

1

2 0)(2

1)2(

2

1

Claim: The multiplicity of 0 is 1.2

1,

)(2

10 ji

n

jiij

t vvPLvv

So?...

vvPIvLv )(

vPvv vvvPv )1(

IntuitionHow could eigenvalues and mixing time be connected?

111100111100 ...)...( Nt

Ntt

NNtt vPavPavPavavavaPvP

1111110111111000 ...... Nt

NNt

Nt

NNtt vavaavavava

t1

Spectral Gap and Mixing Time

The spectral gap determines the mixing rate:

Larger Spectral Gap = Rapid Mixing

max1

As for P’s spectral decomposition, P has anorthonormal basis of eigenvectors.

We can bound by .

So, the mixing time is bounded by:

|| , jtjiP t

max

maxlog

log)(

Assume directed reversible graph (or general undirected graph).

We have no direct spectral analysis.

But P is similar to a symmetric matrix!

ProofLet be a matrix with diagonal entries .

Claim: is a symmetric matrix.

2/1D )(w

2/12/1 PDDS

j

jiijjiijjijcolumnilineji PDPDDPDDPDs

1)()()( ,

2/1,

2/12/1,

2/1)(

2/1)(

2/1,

i

ijjij Ps

1)( ,,

ijjjii PP ,, From reversibility:

vDvDS

vvDSDvSDD

vvP

T

TT

T

2/12/1

2/12/12/12/1 )(

S and P have the same eigenvalues.What about eigenvectors?

Still:

Why do all eigenvalues satisfy ? (same)

Why do we have an eigenvalue 1? (same)

Why is it unique?same for (-1).

As for 1: Omri will prove:

1210 ...1 N

1||

iii

jijiiji

y y

yyP

T 2

2

01

)(

min1

According to spectral decomposition ofsymmetric matrices:

Note:

1

0

)(1

0

)()(N

i

ii

N

i

iii EeeS

T

2/1)0()()()()( ;;02 DeEEEE Tiiji

Main Lemma:For every , we define:

So, for every we get:

U

})(

|)(),(|{max

, j

jjiP t

UjiU

0t

)(minmax

iIi

t

U

So, we can get

1

0

)(N

i

iti

t ES

min

maxmaxmax

|),(|

,

11max)(

tt

ji

t

ji

jjiP

UjiU j

jt

t

max

min

log

)log()(

Now, we can bound the mixing time:

SummaryWe have bounded the mixing time forirreducible, a-periodic reversible

graphs.

Note:Reducible graphs have no unique

eigenvalue.Periodic graphs – the same (bipartite

graph).

Graph ProductLet . The product

Is defined by and

2,1),,( iEVG iii

),(21 EVGGG

21 VVV ]}),([]),([|)),(),,{(( 12121221212121 EvvandwworEwwandvvwwvvE

0

1

2K(0,0)

(0,1) (1,1)

(0,1)

22 KK

222 ... KKKQn

..0.

....

....

1..0

1GA

Entry (i,j)

...),(

....

...),(

),(.),(

1

11

in

i

jj

vw

vw

vwvw

..0.

....

....

..

||

||2

2

2

V

VG IAGA

So, only the permutations that were counted for the determinant of AG1, will be counted here. We instead of we getSo,

)(k )(22 G

AI

The eigenvectors of Qn are We now re-compute every eigenvalue by: Adding n (self loops) Dividing by 2n (to get a transition matrix).

Now we get

And the mixing time satisfies:

},...2,1,0{ n

n2

11max

)()log()( 21 nOn