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Ilan Ben-Bassat Omri Weinstein
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Outline
Why spectral gap? Undirected Regular Graphs. Directed Reversible Graphs. Example (Unit hypercube). Conductance. Reversible Chains.
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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.
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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||
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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
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So?...
vvPIvLv )(
vPvv vvvPv )1(
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IntuitionHow could eigenvalues and mixing time be connected?
111100111100 ...)...( Nt
Ntt
NNtt vPavPavPavavavaPvP
1111110111111000 ...... Nt
NNt
Nt
NNtt vavaavavava
t1
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Spectral Gap and Mixing Time
The spectral gap determines the mixing rate:
Larger Spectral Gap = Rapid Mixing
max1
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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)(
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Assume directed reversible graph (or general undirected graph).
We have no direct spectral analysis.
But P is similar to a symmetric matrix!
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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:
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vDvDS
vvDSDvSDD
vvP
T
TT
T
2/12/1
2/12/12/12/1 )(
S and P have the same eigenvalues.What about eigenvectors?
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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
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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
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Main Lemma:For every , we define:
So, for every we get:
U
})(
|)(),(|{max
, j
jjiP t
UjiU
0t
)(minmax
iIi
t
U
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So, we can get
1
0
)(N
i
iti
t ES
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min
maxmaxmax
|),(|
,
11max)(
tt
ji
t
ji
jjiP
UjiU j
jt
t
max
min
log
)log()(
Now, we can bound the mixing time:
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SummaryWe have bounded the mixing time forirreducible, a-periodic reversible
graphs.
Note:Reducible graphs have no unique
eigenvalue.Periodic graphs – the same (bipartite
graph).
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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
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..0.
....
....
1..0
1GA
Entry (i,j)
...),(
....
...),(
),(.),(
1
11
in
i
jj
vw
vw
vwvw
..0.
....
....
..
||
||2
2
2
V
VG IAGA
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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
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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