enumeration of the number of spanning trees in some ... - Toubkal

64CHAPTER 3.HOW TO COUNT THE NUMBER OF SPANNING TREES INGRAPHScolumn. By the matrix-tree theorem, the product n−1 ∏of spanning trees of G.i=1λ i is exactly n times the number□Remark 3.4.7 [21] Let G be an undirected (multi)graph with at least one vertex, andLaplace matrix L with eigenvalues 0 = λ 1 ≥ λ 2 ≥ ... ≥ λ n . Let L ∗ vu be the (v, u)-cofactorof L. Then the number of spanning trees of G equalsτ(G) = L ∗ vu = det(L + 1 n 2 J) = 1 n λ 2...λ n for any v, u ∈ V (G).(The (i, j)-cofactor of a matrix M is by definition (−1) i+j det M(i, j), where M(i, j) isthe matrix obtained from M by deleting row i and column j. Note that L ∗ vu does notdepend on an ordering of the vertices of G)For example, the graph G of valency k on 2 vertices has Laplace matrix L as follows:( ) k −kL =,−k kso that λ 1 = 0, λ 2 = 2k, and τ(G) = 1 2 .2k = k. If we consider the complete graph K n,then λ 2 = ... = λ n = n, and therefore K n has τ(G) = n n−2 spanning trees. This formulawas proposed by Cayley [24]. Remark 4.3.14 is implicit in Kirchhoff [65] and it is knownas the Matrix-Tree Theorem [21].We consider two examples, shown in Figure 3.7 and Figure 3.8.Example 3.4.8 Consider the graph in Figure 3.7Figure 3.7: Example graph, with Laplacian matrix and eigenvalues. Numbers near eachvertex indicate the chosen ordering. The total number of spanning trees can be seen tobe 8 by inspection, which matches with Kirchhoff’s theorem.