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

3.4. Counting the number of spanning trees in graphs algebraically 633.4.2 Other algebraic methods for counting τ(G)Some other methods for counting τ(G) are as follows. Let A be an n × n matrix. Theeigenvalues of A are defined as the n roots of the characteristic polynomial det(λI − A).The spectrum of A consists of all eigenvalues of A. All eigenvalues of a real symmetricmatrix are real. For further detail and more explanation on properties and significance ofthe theory of spectra, we refer the reader to [5, 11, 21, 35, 57, 52, 53, 92, 80, 81, 83, 91, 93].The Laplacian matrix of a graph and its eigenvalues can be used in several areas ofmathematical research and have a physical interpretation in various physical andchemical theories. The related matrix: the adjacency matrix of a graph and its eigenvalueswere extensively investigated in the past than the Laplacian matrix. From theLaplace spectrum of a graph, one can determine the number of spanning trees (whichwill be nonzero only if the graph is connected). Furthermore, by basic knowledge oflinear algebra, all eigenvalues of the Kirchhoff matrix L of a graph with n verticesare non-negative, and 0 is one of its eigenvalues because all its row vectors add up tothe 0 vector. So we can let λ 1 ≥ λ 2 ≥ ... ≥ λ n (= 0) denote all eigenvalues of L.Kel’mans and Chelnokov [64] have shown that the number of distinct spanning treesof a graph G is equal to the product of nonzero eigenvalues of the combinatorial Laplacian.For a graph G, the combinatorial Laplacian has non-negative eigenvalues exceptone, λ 1 ≥ λ 2 ≥ ... ≥ λ n (= 0). The number of spanning trees, can be related to theeigenvalues of L as follows: (A proof can be found in [11]. For completeness, we brieflydescribe the proof here). We now can state the following theorem and its proof.Theorem 3.4.6 (Kirchhoff’s Matrix Tree Theorem) For a given undirected connectedgraph G with n vertices, let λ 1 , . . . , λ n−1 be the non-zero eigenvalues of L(G). Then,the number of distinct spanning trees τ(G) is equal to:τ(G) = 1 n−1∏λ i .nEquivalently, τ(G) is equal to the absolute value of any cofactor of the Laplacian matrixof G.Proof: Suppose we consider the characteristic polynomial p(x) of the combinatorialLaplacian L.p(x) = det(L − xI).The coefficient of the linear term is exactlyn−1−i=1∏λ i .i=1On the other hand, the coefficient of the linear term of p(x) is -1 times the sum of thedeterminant of n principal submatrix of L obtained by deleting the i-th row and i-th