An Introduction to Spectral Graph Theory - Web hostingweb.uvic.ca/~wheelerm/FPslides.pdf · Graph Theory Review De nition Two vertices v i and v j 2V(G) are said to be adjacent if

An Introduction to Spectral Graph Theory

Mackenzie WheelerSupervisor: Dr. Gary MacGillivray

University of [email protected]

Outline

1. How many walks are there from vertices vi to vj oflength d?

2. How many labelled spanning trees of G exist?

I Graph Theory Review

I Define the Adjacency matrix A(G )

I Answer Question 1

I Linear Algebra Review

I Define the Laplacian matrix L(G )

I Answer Question 2

Outline





I Answer Question 1



I Answer Question 2

Outline





I Answer Question 1



I Answer Question 2

Outline





I Answer Question 1



I Answer Question 2

Graph Theory Review

DefinitionTwo vertices vi and vj ∈ V (G ) are said to be adjacent if{vi , vj} ∈ E (G ).

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DefinitionA walk in a graph G is a sequence of vertices {v1, v2, . . . , vk} suchthat vi is adjacent to vi+1 for all 1 ≤ i ≤ k − 1. The length of thewalk is k − 1.

Graph Theory Review


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Graph Theory Review


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The Adjacency Matrix

DefinitionThe adjacency matrix A(G ) of a graph G is defined by

(A(G ))ij =

{1 vivj ∈ E (G )

0 vivj /∈ E (G )

Example

Let G = C5, then we have that

A(C5) =

0 1 0 0 11 0 1 0 00 1 0 1 00 0 1 0 11 0 0 1 0


DefinitionThe adjacency matrix A(G ) of a graph G is defined by

(A(G ))ij =

{1 vivj ∈ E (G )

0 vivj /∈ E (G )

Example


A(C5) =

0 1 0 0 11 0 1 0 00 1 0 1 00 0 1 0 11 0 0 1 0

Counting walks

QuestionGiven a graph G , how many walks are there from vi to vj of lengthd?

Proposition

Let G be a graph with n vertices and adjacency matrix A(G ), thenthe number of walks from vi to vj of length d in G is given by(A(G ))dij .

Counting walks

QuestionGiven a graph G , how many walks are there from vi to vj of lengthd?

Proposition

Let G be a graph with n vertices and adjacency matrix A(G ), thenthe number of walks from vi to vj of length d in G is given by(A(G ))dij .


Proof.

I For d = 1, Ad is A.

I Consider Ad+1 = AdA

I Then ad+1ij =

∑nk=1 a

dikakj .

I adikakj is the number of walks from vi to vj which are walksfrom vi to vk of length d , followed by a walk of length 1 fromfrom vk to vj .

I Therefore an+1ij =

∑nk=1 a

dikakj is to total number of walks

from vi to vj of length d + 1.


Proof.



I Then ad+1ij =

∑nk=1 a

dikakj .



∑nk=1 a




Proof.



I Then ad+1ij =

∑nk=1 a

dikakj .



∑nk=1 a




Proof.



I Then ad+1ij =

∑nk=1 a

dikakj .



∑nk=1 a




Proof.



I Then ad+1ij =

∑nk=1 a

dikakj .



∑nk=1 a




Proof.



I Then ad+1ij =

∑nk=1 a

dikakj .



∑nk=1 a




Corollary

Let G be a graph with e edges and t triangles, then

1. tr(A(G )2) = 2e

2. tr(A(G )3) = 6t

The Laplacian Matrix

DefinitionThe Laplacian matrix L(G ) of a graph G is defined by

(L(G ))ij =

deg(vi ) i = j

−1 i 6= j and vivj ∈ E (G )

0 otherwise

Example


L(C5) =

2 −1 0 0 −1−1 2 −1 0 00 −1 2 −1 00 0 −1 2 −1−1 0 0 −1 2


DefinitionThe Laplacian matrix L(G ) of a graph G is defined by

(L(G ))ij =

deg(vi ) i = j

−1 i 6= j and vivj ∈ E (G )

0 otherwise

Example


L(C5) =

2 −1 0 0 −1−1 2 −1 0 00 −1 2 −1 00 0 −1 2 −1−1 0 0 −1 2

Linear Algebra Review

DefinitionLet A ∈ Mn×n(R) and let v ∈ Rn be a nonzero vector. Then v isan eigenvector of A if there exists a scalar λ ∈ R, such thatAv = λv . We say that λ is an eigenvalue of A with correspondingeigenvector v .

Proposition

Let A ∈ Mn×n(R) be a symmetric matrix, then the eigenvalues ofA are all real numbers.


DefinitionLet A ∈ Mn×n(R) and let v ∈ Rn be a nonzero vector. Then v isan eigenvector of A if there exists a scalar λ ∈ R, such thatAv = λv . We say that λ is an eigenvalue of A with correspondingeigenvector v .

Proposition

Let A ∈ Mn×n(R) be a symmetric matrix, then the eigenvalues ofA are all real numbers.


DefinitionLet A ∈ Mn×n(R), and let aij denote the entry in the i th row andj th column. A is diagonally dominant if

|aii | ≥∑j 6=i

|aij |

for all 1 ≤ i ≤ n.

Example

A =

7 1 0 2 −1−1 6 −1 0 00 −1 3 1 00 0 −1 2 −12 0 0 −1 4


DefinitionLet A ∈ Mn×n(R), and let aij denote the entry in the i th row andj th column. A is diagonally dominant if

|aii | ≥∑j 6=i

|aij |

for all 1 ≤ i ≤ n.

Example

A =

7 1 0 2 −1−1 6 −1 0 00 −1 3 1 00 0 −1 2 −12 0 0 −1 4


Proposition

Let A be a symmetric, diagonally dominant n × n matrix such thataii > 0 for all a ≤ i ≤ n. Then all the eigenvalues of A arenon-negative.

Corollary

Let G be a graph with Laplacian L(G ). The eigenvalues of L(G )are all nonnegative real numbers. Therefore, we may list theeigenvalues of L(G ) as 0 = λ0 ≤ λ1 ≤ λ2 ≤ · · · ≤ λn−1.


Proposition

Let A be a symmetric, diagonally dominant n × n matrix such thataii > 0 for all a ≤ i ≤ n. Then all the eigenvalues of A arenon-negative.

Corollary

Let G be a graph with Laplacian L(G ). The eigenvalues of L(G )are all nonnegative real numbers. Therefore, we may list theeigenvalues of L(G ) as 0 = λ0 ≤ λ1 ≤ λ2 ≤ · · · ≤ λn−1.


Proposition

Let G be a graph with Laplacian matrix L(G ). Then λ = 0 is aneigenvalue of L(G ) with v = (1, 1, . . . , 1) as a correspondingeigenvector.

Proposition

Let G be a connected graph with Laplacian L(G ). Then λ = 0 isan eigenvalue of L(G ) with multiplicity one.


Proposition


Proposition



Proposition


Proposition


Counting Labelled Spanning Trees

DefinitionA graph G is a tree if G is connected and contains no cycles.

DefinitionLet G be a graph with a subgraph T . T is a spanning tree of G ifV (T ) = V (G ) and T is a tree.

Example




Example




Example




Example

The Petersen Graph




Example

A spanning tree of the Petersen graph


QuestionGiven a graph G with vertices labelled {v1, v2, . . . , vn} how manylabelled spanning trees of G exist?

Theorem (Kirchoff’s Theorem)

Let G be a connected graph with n ≥ 2 labelled vertices, and let0 = λ0 < λ1 ≤ λ2 ≤ · · · ≤ λn−1 be the eigenvalues of L(G ). Thenthe number of spanning trees on G , t(G ), is given by

t(G ) = det(L(G )[i ]) =1

n

n−1∏k=1

λk

.Where L(G )[i ] denotes the matrix obtained from L(G ) by deleteingthe i th row and i th column.


QuestionGiven a graph G with vertices labelled {v1, v2, . . . , vn} how manylabelled spanning trees of G exist?

Theorem (Kirchoff’s Theorem)

Let G be a connected graph with n ≥ 2 labelled vertices, and let0 = λ0 < λ1 ≤ λ2 ≤ · · · ≤ λn−1 be the eigenvalues of L(G ). Thenthe number of spanning trees on G , t(G ), is given by

t(G ) = det(L(G )[i ]) =1

n

n−1∏k=1

λk

.Where L(G )[i ] denotes the matrix obtained from L(G ) by deleteingthe i th row and i th column.


Proof Outline:

I We preceed by induction on |V (G )|+ |E (G )| = n + m

I When n + m = 3, the only connected graph is P2

P2

L(P2) =

[1 −1−1 1

]


Proof Outline:



P2

L(P2) =

[1 −1−1 1

]


Proof Outline:



P2

L(P2) =

[1 −1−1 1

]


Proof Outline:



P2

L(P2) =

[1 −1−1 1

]


Proof Outline:

I Consider a graph |V (G )|+ |E (G )| = n + m + 1

I Let e = vivj be an edge incident with the vertex viI Notice that t(G ) = t(G − e) + t(G \ e)

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L(G \ e) =

2 −2 0−2 3 −10 −1 1


Proof Outline:



v4

v1 v2

v3 (v1, v4)

v2

v3

L(G \ e) =

2 −2 0−2 3 −10 −1 1


Proof Outline:


I Let e = vivj be an edge incident with the vertex vi

I Notice that t(G ) = t(G − e) + t(G \ e)

v4

v1 v2

v3 (v1, v4)

v2

v3

L(G \ e) =

2 −2 0−2 3 −10 −1 1


Proof Outline:



v4

v1 v2

v3 (v1, v4)

v2

v3

L(G \ e) =

2 −2 0−2 3 −10 −1 1


Proof Outline:



v4

v1 v2

v3 (v1, v4)

v2

v3

L(G \ e) =

2 −2 0−2 3 −10 −1 1


Proof Outline:



v4

v1 v2

v3 (v1, v4)

v2

v3

L(G \ e) =

2 −2 0−2 3 −10 −1 1


Proof Outline:

By standard manipulation of the determinant we get

det(L(G )[i ]) = det(L(G − e)[i ]) + det(L(G \ e)[j ])

= t(G − e) + t(G \ e), by induction hypothesis

= t(G ).


Corollary (Cayley’s Formula)

The number of labelled spanning trees on the complete graph Kn

is nn−2.

Proof.The eigenvalues of L(Kn) are 0 and n with multiplicity 1 and n− 1,respectively. Therefore, by Kirchoff’s Theorem the number ofspanning trees on Kn is nn−1

n = nn−2.


Corollary (Cayley’s Formula)

The number of labelled spanning trees on the complete graph Kn

is nn−2.

Proof.The eigenvalues of L(Kn) are 0 and n with multiplicity 1 and n− 1,respectively. Therefore, by Kirchoff’s Theorem the number ofspanning trees on Kn is nn−1

n = nn−2.

References

C. Godsil and G. Royle, Algebraic Graph Theory, GraduateTexts in Mathematics, 2001.

A. E. Brouwer and W..H Haemers, Spectra of Graphs,Springer, 2011.

Documents

An Introduction to Spectral Graph Theory - Web hostingweb.uvic.ca/~wheelerm/FPslides.pdf · Graph Theory Review De nition Two vertices v i and v j 2V(G) are said to be adjacent if