18.312: Algebraic Combinatorics Lionel Levine. Lecture 19
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1 832: Algebraic Combinatorics Lionel Levine Lecture date: April 2, 20 Lecture 9 Notes by: David Witmer Matrix-Tree Theorem Undirected Graphs Let G = (V, E) be a connected, undirected graph with n vertices, and let κ(g) be the number of spanning trees of G Definition (Laplacian matrix of undirected graph) The Laplacian matrix L of G is equal to D A, where d 0 D = 0 d n such that d i is the degree of vertex i, ie the number of edges incident to vertex i, and A is the adjacency matrix of G such that A = (a ij ), { if (i, j) E a ij = 0 else Theorem 2 (Matrix-Tree Theorem, Version ) κ(g) = n λ λ 2 λ n, where λ, λ 2,, λ n are non-zero eigenvalues of the Laplacian matrix L of G 2 Directed Graphs We can give another version of the Matrix-Tree Theorem for directed graphs First, we need to define spanning trees and Laplacian matrices for directed graphs Let Γ = (V, E) be a directed graph 9-
2 Definition 3 (Oriented spanning tree) An oriented spanning tree of Γ rooted at r V is a spanning subgraph T = (V, A) such that Every vertex v r has out degree 2 r has out degree 0 3 T has no oriented cycles Example 4 Consider the following directed graph: It has three oriented spanning trees: Definition 5 (Laplacian matrix of directed graph) The Laplacian matrix L of Γ is equal to D A, where d 0 D = 0 d n such that d i is the out degree of vertex i, ie #{j V (i, j) E}, and A is the adjacency matrix of Γ Theorem 6 (Matrix-Tree Theorem, Version 2) Let κ(γ, r) = #{oriented spanning trees of Γ rooted at r} 9-2
3 and L r be the Laplacian matrix of Γ with the row and column corresponding to vertex r crossed out Then κ(γ, r) = det L r where L r is the Laplacian matrix L with row and column r removed Example 7 Consider the directed graph from the previous example: Then we see that and so and Then D = A = L = L r = det L r = = 3 which matches what we found in the previous example We will prove this version of the Matrix-Tree Theorem and then show that it implies the version for undirected graphs Proof: Reorder the vertices of Γ so that r is the nth vertex Then det L r = d d 2 d n (other terms), since L r has the d i s on the diagonal and either or 0 for the off-diagonal 9-3
4 entries d d 2 d n counts the number of subgraphs H of Γ such that each vertex v r has out-degree So we have that H = T C C k, where T is an oriented tree rooted at r and each C i is an oriented cycle Then det L r = σ S n sgn(σ)l,σ() L n,σ(n ) Let fix(σ) = {i σ(i) = i} Then we have det L r = sgn(σ) d i L i,σ(i) σ S n i fix(σ) i/ fix(σ) i/ fix(σ) L i,σ(i) is only non-zero when (i, σ(i)) E for all i / fix(σ) In this case, L i,σ(i) = ( ) n fix(σ) We wish to write i/ fix(σ) det L r = subgraphs H Γ where C H is if H is an oriented spanning tree and 0 otherwise Any permutation σ consists of fixed points and cycles A subgraph H = T C C k arises from σ if and only if the union of all cycles C i of H contains all vertices not fixed by H, which, in turn, is true if and only if T fix(σ) We can then conclude that C H = {σ S n T fix(σ)} C H, sgn(σ)( ) n fix(σ) Our goal is then to show that C H is when H is a tree and 0 otherwise When H is a tree, H = T and there are no cycles Then all vertices are in fix(σ) and σ is the identity permutation The sign of the identity permutation is and n points are fixed, so C H = Lastly, we need to show that C H = 0 if k, ie if H has a cycle For each C i, we can either choose C i fix(σ) or C i to be a cycle of σ Let i,, i l be the indices of the C i s that are formed from vertices in cycles of σ All other points must be fixed by σ, so This means that C H = {i,,i l } [k] sgn(σ) = ( ) ( C i )++( C il ) ( ) ( C i )++( C il ) ( ) C i ++ C il 9-4
5 So, C H = ( ) S = S [k] k l=0 ( ) k ( ) k l = 0 if k 3 Proof of the Matrix Tree Theorem, Version Now we will show that Version 2 of the Matrix Tree Theorem implies the version for undirected graphs Proof: Given undirected graph G, let Γ be the directed graph with edges (i, j) and (j, i) for every edge of G We first observe that there is a bijection between the set of oriented spanning trees of Γ rooted at r and the set of spanning trees of G We can take any oriented spanning tree of Γ rooted at r and get a spanning tree of G by disregarding the root and the orientation of the edges For any spanning tree T of G, we can get an oriented spanning tree of Γ by orienting edges along the unique path from each vertex to r Such a path exists because T is connected and is unique because T has no cycles Then n nκ(g) = κ(γ, r) Let L be the Laplacian matrix of Γ Then the characteristic polynomial of L is χ(t) = det (ti L) It is true that n det L r = ( ) n [t]χ(t), r= where [t]χ(t) is the coefficient of t in χ(t) r= 9-5
6 So, we have that n nκ(g) = det L r = ( ) n [t]χ(t) r n = ( ) n [t] (t λ i ), where the λ i s are eigenvalues of L and λ n = 0 i= = ( ) n ( ) n λ λ n = λ λ n Therefore, κ(g) = n λ λ n 2 Cayley s Theorem Theorem 8 (Cayley s Theorem) The number of trees on n labeled vertices is n n 2 Example 9 Consider trees containing 4 vertices There are 6 = total, 4 of the form and 2 of the form 4 Proof: Any tree on n vertices is a spanning tree of the complete graph K n, so we can apply Version 2 of the Matrix-Tree Theorem So, κ(k n ) = n λ λ n, where λ,, λ n 9-6
7 are the non-zero eigenvalues of the Laplacian matrix n n L = = ni J, n where J is the n n matrix of ones J has the ones vector as one of its eigenvectors The remaining n eigenvectors are of the form, so J has eigenvalues n, 0,, 0, with 0 having multiplicity n This implies that L has eigenvalues 0, n,, n, with n having multiplicity n So, κ(k n ) = nn n = nn 2 3 Eigenvalues of the Adjacency Matrix Let G be an undirected, connected graph with n vertices Let P l be the number of closed paths in G of length l: P l = #{(v 0, v,, v l, v l = v 0 ) (v i, v i+ ) E for i = 0,,, l } Theorem 0 P l = φ l + + φ l n, where φ,, φ n are the eigenvalues of the adjacency matrix A of G Proof: We observe that (A l ) ij = #{paths of length l from i to j} So, P l = (A l ) + (A l ) (A l ) nn = Tr(A l ) 9-7
8 Note that this holds for both directed and undirected graphs Since G is undirected, A is symmetric, which means that A is diagonalizable so there exists some S such that φ 0 0 SAS 0 φ 2 0 = 0 0 φ n So, P l = Tr(A l ) = Tr(SA l S ) = Tr((SAS ) l ) φ l φ l 2 0 = Tr 0 0 φ l n = φ l + + φ l n 9-8
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