An Introduction to Spectral Graph Theory

Transcription

1 An Introduction to Spectral Graph Theory Mackenzie Wheeler Supervisor: Dr. Gary MacGillivray University of Victoria

2 Outline

3 Outline 1. How many walks are there from vertices v i to v j of length d?

4 Outline 1. How many walks are there from vertices v i to v j of length d? 2. How many labelled spanning trees of G exist?

5 Outline 1. How many walks are there from vertices v i to v j of length d? 2. How many labelled spanning trees of G exist? Graph Theory Review Define the Adjacency matrix A(G) Answer Question 1 Linear Algebra Review Define the Laplacian matrix L(G) Answer Question 2

6 Graph Theory Review Definition Two vertices v i and v j V (G) are said to be adjacent if {v i, v j } E(G).

7 Graph Theory Review Definition Two vertices v i and v j V (G) are said to be adjacent if {v i, v j } E(G). v 1 v 5 v 2 v 4 v 3

8 Graph Theory Review Definition Two vertices v i and v j V (G) are said to be adjacent if {v i, v j } E(G). v 1 v 5 v 2 v 4 v 3 Definition A walk in a graph G is a sequence of vertices {v 1, v 2,..., v k } such that v i is adjacent to v i+1 for all 1 i k 1. The length of the walk is k 1.

9 The Adjacency Matrix Definition The adjacency matrix A(G) of a graph G is defined by (A(G)) ij = { 1 v i v j E(G) 0 v i v j / E(G)

10 The Adjacency Matrix Definition The adjacency matrix A(G) of a graph G is defined by (A(G)) ij = { 1 v i v j E(G) 0 v i v j / E(G) Example Let G = C 5, then we have that A(C 5 ) =

11 Counting walks Question Given a graph G, how many walks are there from v i to v j of length d?

12 Counting walks Question Given a graph G, how many walks are there from v i to v j of length d? Proposition Let G be a graph with n vertices and adjacency matrix A(G), then the number of walks from v i to v j of length d in G is given by (A(G)) d ij.

13 The Adjacency Matrix Proof.

14 The Adjacency Matrix Proof. For d = 1, A d is A.

15 The Adjacency Matrix Proof. For d = 1, A d is A. Consider A d+1 = A d A

16 The Adjacency Matrix Proof. For d = 1, A d is A. Consider A d+1 = A d A Then a d+1 ij = n k=1 ad ik a kj.

17 The Adjacency Matrix Proof. For d = 1, A d is A. Consider A d+1 = A d A Then a d+1 ij = n k=1 ad ik a kj. a d ik a kj is the number of walks from v i to v j which are walks from v i to v k of length d, followed by a walk of length 1 from from v k to v j.

18 The Adjacency Matrix Proof. For d = 1, A d is A. Consider A d+1 = A d A Then a d+1 ij = n k=1 ad ik a kj. a d ik a kj is the number of walks from v i to v j which are walks from v i to v k of length d, followed by a walk of length 1 from from v k to v j. Therefore a n+1 ij = n k=1 ad ik a kj is to total number of walks from v i to v j of length d + 1.

19 The Adjacency Matrix 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

20 The Laplacian Matrix Definition The Laplacian matrix L(G) of a graph G is defined by deg(v i ) i = j (L(G)) ij = 1 i j and v i v j E(G) 0 otherwise

21 The Laplacian Matrix Definition The Laplacian matrix L(G) of a graph G is defined by deg(v i ) i = j (L(G)) ij = 1 i j and v i v j E(G) 0 otherwise Example Let G = C 5, then we have that L(C 5 ) =

22 Linear Algebra Review Definition Let A M n n (R) and let v R n be a nonzero vector. Then v is an eigenvector of A if there exists a scalar λ R, such that Av = λv. We say that λ is an eigenvalue of A with corresponding eigenvector v.

23 Linear Algebra Review Definition Let A M n n (R) and let v R n be a nonzero vector. Then v is an eigenvector of A if there exists a scalar λ R, such that Av = λv. We say that λ is an eigenvalue of A with corresponding eigenvector v. Proposition Let A M n n (R) be a symmetric matrix, then the eigenvalues of A are all real numbers.

24 Linear Algebra Review Definition Let A M n n (R), and let a ij denote the entry in the i th row and j th column. A is diagonally dominant if a ii j i a ij for all 1 i n.

25 Linear Algebra Review Definition Let A M n n (R), and let a ij denote the entry in the i th row and j th column. A is diagonally dominant if a ii j i a ij for all 1 i n. Example A =

26 Linear Algebra Review Proposition Let A be a symmetric, diagonally dominant n n matrix such that a ii > 0 for all a i n. Then all the eigenvalues of A are non-negative.

27 Linear Algebra Review Proposition Let A be a symmetric, diagonally dominant n n matrix such that a ii > 0 for all a i n. Then all the eigenvalues of A are non-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 the eigenvalues of L(G) as 0 = λ 0 λ 1 λ 2 λ n 1.

28 The Laplacian Matrix

29 The Laplacian Matrix Proposition Let G be a graph with Laplacian matrix L(G). Then λ = 0 is an eigenvalue of L(G) with v = (1, 1,..., 1) as a corresponding eigenvector.

30 The Laplacian Matrix Proposition Let G be a graph with Laplacian matrix L(G). Then λ = 0 is an eigenvalue of L(G) with v = (1, 1,..., 1) as a corresponding eigenvector. Proposition Let G be a connected graph with Laplacian L(G). Then λ = 0 is an eigenvalue of L(G) with multiplicity one.

31 Counting Labelled Spanning Trees

32 Counting Labelled Spanning Trees Definition A graph G is a tree if G is connected and contains no cycles.

33 Counting Labelled Spanning Trees Definition A graph G is a tree if G is connected and contains no cycles. Definition Let G be a graph with a subgraph T. T is a spanning tree of G if V (T ) = V (G) and T is a tree.

34 Counting Labelled Spanning Trees Definition A graph G is a tree if G is connected and contains no cycles. Definition Let G be a graph with a subgraph T. T is a spanning tree of G if V (T ) = V (G) and T is a tree. Example The Petersen Graph

35 Counting Labelled Spanning Trees Definition A graph G is a tree if G is connected and contains no cycles. Definition Let G be a graph with a subgraph T. T is a spanning tree of G if V (T ) = V (G) and T is a tree. Example A spanning tree of the Petersen graph

36 Counting Labelled Spanning Trees Question Given a graph G with vertices labelled {v 1, v 2,..., v n } how many labelled spanning trees of G exist?

37 Counting Labelled Spanning Trees Question Given a graph G with vertices labelled {v 1, v 2,..., v n } how many labelled spanning trees of G exist? Theorem (Kirchoff s Theorem) Let G be a connected graph with n 2 labelled vertices, and let 0 = λ 0 < λ 1 λ 2 λ n 1 be the eigenvalues of L(G). Then the number of spanning trees on G, t(g), is given by n 1 t(g) = det(l(g)[i]) = 1 n k=1. Where L(G)[i] denotes the matrix obtained from L(G) by deleteing the i th row and i th column. λ k

38 Counting Labelled Spanning Trees Proof Outline:

39 Counting Labelled Spanning Trees Proof Outline: We preceed by induction on V (G) + E(G) = n + m

40 Counting Labelled Spanning Trees Proof Outline: We preceed by induction on V (G) + E(G) = n + m When n + m = 3, the only connected graph is P 2

41 Counting Labelled Spanning Trees Proof Outline: We preceed by induction on V (G) + E(G) = n + m When n + m = 3, the only connected graph is P 2 P 2 L(P 2 ) = [ 1 ] 1 1 1

42 Counting Labelled Spanning Trees Proof Outline:

43 Counting Labelled Spanning Trees Proof Outline: Consider a graph V (G) + E(G) = n + m + 1

44 Counting Labelled Spanning Trees Proof Outline: Consider a graph V (G) + E(G) = n + m + 1 Let e = v i v j be an edge incident with the vertex v i

45 Counting Labelled Spanning Trees Proof Outline: Consider a graph V (G) + E(G) = n + m + 1 Let e = v i v j be an edge incident with the vertex v i Notice that t(g) = t(g e) + t(g \ e)

46 Counting Labelled Spanning Trees Proof Outline: Consider a graph V (G) + E(G) = n + m + 1 Let e = v i v j be an edge incident with the vertex v i Notice that t(g) = t(g e) + t(g \ e) v 1 v 2 v 2 v 4 v 3 (v 1, v 4 ) v 3

47 Counting Labelled Spanning Trees Proof Outline: Consider a graph V (G) + E(G) = n + m + 1 Let e = v i v j be an edge incident with the vertex v i Notice that t(g) = t(g e) + t(g \ e) v 1 v 2 v 2 v 4 v 3 (v 1, v 4 ) L(G \ e) = v 3

48 Counting Labelled Spanning Trees 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).

49 Counting Labelled Spanning Trees Corollary (Cayley s Formula) The number of labelled spanning trees on the complete graph K n is n n 2.

50 Counting Labelled Spanning Trees Corollary (Cayley s Formula) The number of labelled spanning trees on the complete graph K n is n n 2. Proof. The eigenvalues of L(K n ) are 0 and n with multiplicity 1 and n 1, respectively. Therefore, by Kirchoff s Theorem the number of spanning trees on K n is nn 1 n = n n 2.

51 References C. Godsil and G. Royle, Algebraic Graph Theory, Graduate Texts in Mathematics, A. E. Brouwer and W..H Haemers, Spectra of Graphs, Springer, 2011.