1-Jonathan Spectral Graph Theory

8/11/2019 1-Jonathan Spectral Graph Theory

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OutlineIntroduction

Matrix Tree TheoremPageRank and metrics of centrality

Spectral Graph Theory and You:Matrix Tree Theorem and Centrality Metrics

Jonathan Gootenberg

March 11, 2013

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OutlineIntroduction


Outline of Topics

1 IntroductionMotivationBasics of Spectral Graph TheoryUnderstanding the characteristic polynomial

2 Matrix Tree TheoremPreliminary concepts

Proof of Matrix Tree Theorem

3 PageRank and metrics of centrality

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OutlineIntroduction


MotivationBasics of Spectral Graph TheoryUnderstanding the characteristic polynomial

Why use spectral graph theory?

The eigenvalues of a graphs adjacency matrix (and others) canreveal important information about the community structure.The number of matrix treesThe most central nodes

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OutlineIntroduction



Easily calculate all spanning trees

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Find most central nodes in a network

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Some denitions: Babys rst matrix

Denition (Adjacency Matrix A)

Given an undirected graph G = ( V , E ), the adjacency matrix of G is the n n matrix A = A(G ) with entries aij such that

aij =1, {v i , v j } E 0, otherwise

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Some denitions: Babys rst matrix

G = A(G ) =0 1 0 11 0 1 10 1 0 1

1 1 1 0

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Putting the spec in Spectral

Denition (Spectrum of a graph)

The spectrum of graph G is the set of eigenvalues of A(G ), alongwith their multiplicities. If the eigenvalues of A(G ) are 0 > 0 > . . . > s 1 with multiplicites m( 0), . . . , m( s 1) then

Spec G = 0 1 s 1

m( 0) m( 1) m( s 1)

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A(G ) =

0 1 0 1

1 0 1 10 1 0 11 1 1 0

Spec G =12 1 + 17 12 1 17 1 0

1 1 1 1

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A(G ) =

0 1 0 1

1 0 1 10 1 0 11 1 1 0

Spec G =12 1 + 17 12 1 17 1 0

1 1 1 1

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What does the spectrum of a graph tell us?

Denition (Characteristic Polynomial)The characteristic polynomial of G , (G , ) is

(G , ) = det( I A)= n + c 1 n 1 + c 2 n 2 + . . . + c n

c 1 =i

i

= Tr( A)

= 0

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Outline M ti ti


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IntroductionMatrix Tree Theorem

PageRank and metrics of centrality




(G , ) = det( I A)= n + c 1 n 1 + c 2 n 2 + . . . + c n

c 1 =i

i

= Tr( A)

= 0

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Outline Motivation


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(G , ) = det( I A)= n + c 1 n 1 + c 2 n 2 + . . . + c n

c 1 =i

i

= Tr( A)

= 0

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Outline Motivation


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(G , ) = det( I A)= n + c 1 n 1 + c 2 n 2 + . . . + c n

c 1 =i

i

= Tr( A)

= 0

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Outline Motivation


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Characteristic Polynomial (G , ) = n + c 1 n 1 + c 2 n 2 + . . . + c n

You can express c i in terms of the principal minors of A

Principal MinorThe principal minor det(A)J , J , J

{1, . . . , n

} is the determiniant

of the subset of A the same subset J of columns and rows

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OutlineI d i Motivation


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Characteristic Polynomial

(G , ) = n

+ c 1n 1

+ c 2n 2

+ . . . + c n

You can express c i in terms of the principal minors of A

c i (1) i is the sum of the principal minors of size i i c i (1)

i

= |J |= i det(A)J , J

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OutlineI t d ti Motivation


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Basics of Spectral Graph TheoryUnderstanding the characteristic polynomial


Consider c 2

All non-zero principal minors are of the form0 11 0

Therefore,

c

2 =

|E

|

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OutlineIntroduction Motivation


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Consider c 3

The non-trivial principal matrices of size 3 are

0 1 01 0 00 0 0

,0 1 11 0 01 0 0

,0 1 11 0 11 1 0

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Consider c 3

0 1 01 0 00 0 0

= 0 ,0 1 11 0 01 0 0

= 0 ,0 1 11 0 11 1 0

= 2

Therefore, c 3 is twice the number of triangles in G

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OutlineIntroduction Preliminary concepts


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Preliminary conceptsProof of Matrix Tree Theorem

The incidence matrix

Denition (Incidence Matrix S )Given an undirected graph G = ( V , E ) with

V = {1, . . . , n}, E = {e 1, . . . , e m } the incidence matrix of G is then m matrix S = S (G ) with entries S ij such that

S ij =1, e j ends at i

1, e j starts at i

0, otherwise

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The incidence matrix

G = S (G ) =1 0 0 1 0

1 1 0 0 10 1 1 0 00 0 1

1

1

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y pProof of Matrix Tree Theorem

The incidence matrix implies connectivity of the graph

Theoremrank(S) = n - |number of connected components of G |

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y pProof of Matrix Tree Theorem


Theoremrank(S) = n - |number of connected components of G |Reorder the graph such that

S =

S 1 0S 2

...... . . .0 S c

where c is the number of connected components of G We wish toshow the rank of each component is ni 1

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Theoremrank(S) = n - |number of connected components of G |For a given connected component S i , suppose we have a

summation over the rows s j n i

j =1

j s j = 0

1 0 0 1 0

1 1 0 0 10 1 1 0 00 0 1 1 115/19


f f


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Theoremrank(S) = n - |number of connected components of G |Consider a row s k , k = 0

1 0 0 1 0

1 1 0 0 10 1

1 0 00 0 1 1 1

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OutlineIntroduction

M i T ThPreliminary conceptsP f f M i T Th


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Theoremrank(S) = n - |number of connected components of G |Each non-zero column in s k has a corresponding rows l = k = l

1 0 0 1 0

1

1 0 0 1

0 1 1 0 00 0 1 1 -1

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OutlineIntroduction

Matri Tree TheoremPreliminary conceptsProof of Matri Tree Theorem


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Theoremrank(S) = n - |number of connected components of G |Since the graph is connected, all j = 1, and

n i

j =1s j = 0

Therefore, the rank of S i is ni

1

Remark-S Since S i has rank ni 1, we can remove an arbitrary row to createS i without loss of information

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OutlineIntroduction

Matrix Tree TheoremPreliminary conceptsProof of Matrix Tree Theorem


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Theorem

rank(S) = n - |number of connected components of G |Remark-S Since S i has rank ni 1, we can remove an arbitrary row to createS i without loss of information

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OutlineIntroduction

Matrix Tree TheoremPreliminary conceptsProof of Matrix Tree Theorem


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Matrix Tree Theorem

In the following proof, we will try all selections of n 1 edges anduse the determinant to see if the resulting subgraph is connected.We use create the matrix that is the combination of the columnsof the incidence matrix S : Q = S S T

1 0 0 1 0

1

1 0 0 1

0 1 1 0 0

1 1 00 1 10 0

11 0 00 1 0

=2 1 0

1 3

1

0 1 2

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OutlineIntroductionMatrix Tree Theorem



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Matrix Tree Theorem


1 0 0 1 0

1

1 0 0 1

0 1 1 0 0

1 1 00 1 10 0

11 0 00 1 0

=2 1 0

1 3

1

0 1 2

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Matrix Tree Theorem


1 0 0 1 0

1

1 0 0 1

0 1 1 0 0

1 1 00 1 10 0

11 0 00 1 0

=2 1 0

1 3

1

0 1 2

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Matrix Tree Theorem

TheoremThe number of spanning trees of a graph G is equal to

det( Q ), where Q = S

S

T

Graph Laplacian Q Q is referred to as the graph laplacian, and can also be expressedas

Q = D Awhere D is the degree matrix

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Matrix Tree Theorem


det( Q ), where Q = S S T

LemmaIf T = ( V , E ) is an directed tree rooted at n, we can order E suchthat e i ends at i.

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Matrix Tree Theorem



LemmaIf T = ( V , E ) is an directed tree rooted at n, we can order E suchthat e i ends at i.

Proof.Label edges such that e i := ( p (i ), i ), where p (i ) is the parent of i

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Matrix Tree Theorem



LemmaIf G = ( V , E ),

|E

| = n

1 is a directed graph that is not a tree,

det( S ) = 0

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Matrix Tree Theorem



LemmaIf G = ( V , E ), |E | = n 1 is a directed graph that is not a tree,det( S ) = 0Proof.If |E | = n 1 and G is not a tree, then it is not connected, andrank(S ) = rank( S ) n 2

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Matrix Tree Theorem



LemmaIf T = ( V , E ),

|E

| = n

1 is a tree with e i

E ending at i

V ,

det( S ) = 1

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Matrix Tree Theorem

LemmaIf T = ( V , E ), |E | = n 1 is a tree with e i E ending at i V ,det( S ) = 1Proof.By the results of the previous lemmas, we can order the verticessuch that p (i ) > i . Then

S =

1 0 1

.. .

...... . . . . . . 0 0 1

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Matrix Tree Theorem

Proof.We will use the linearity of the determinant to break down det(Q )

into a sum of determinants of subgraphs:

det( Q ) = det( S j )

where all S j are subgraphs with n

1 edges (Tr(S j ) = n

1).

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Formulation of PageRank

PageRank (PR) is a centrality metric which approximates a websurfer.

Jumps with probability q : q n . Like typing a URL

Surfs with probability 1q : j : j i f ( j )k out ( j ) . Like clicking a linkPageRank reccurance

f (i ) = q n + (1 q ) j : j i

f ( j )k out ( j )

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f (i ) = q n + (1 q ) j : j i

f ( j )k out ( j )

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f (i ) = q n + (1 q ) j : j i

f ( j )k out ( j )

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f (i ) = q n + (1 q ) j : j i

f ( j )k out ( j )

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Solving the PageRank reccurance

PageRank reccurance

f (i ) = q n

+ (1 q ) j : j i

f ( j )k out ( j )

Denition (Transition Matrix P )Given an undirected graph G = ( V , E ), the transition matrix of G is the n n matrix P = P (G ) with entries p ij such that

p ij = 1

k a ik , {v i , v j } E

0, otherwise

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PageRank reccurance

f (i ) =

q

n + (1 q ) j : j i f ( j )

k out ( j )

j : j i

f ( j )k out ( j )

= fP

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PageRank reccurance

f (i ) = q

n + (1

q )

j : j i

f ( j )

k out ( j )

j : j i

f ( j )k out ( j )

= fP

We can use the property i f (i ) = 1 = f 1T

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PageRank reccurance

f (i ) = q n

+ (1 q ) j : j i

f ( j )k out ( j )

j : j i

f ( j )k out ( j )

= fP

We can use the propertyi f (i ) = 1 = f 1T

f = q n

f 1T 1 + (1 q )fP = f q n

J + (1 q )P

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PageRank reccurance

f (i ) = q n

+ (1 q ) j : j i

f ( j )k out ( j )

j : j i

f ( j )k out ( j )

= fP

We can use the property i f (i ) = 1 = f 1T

f = q n

f 1T 1 + (1 q )fP = f q n

J + (1 q )P We can compute PageRank with the eigenvector!

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1-Jonathan Spectral Graph Theory

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