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

3.4. Counting the number of spanning trees in graphs algebraically 61Now we want to show that if B is an (n − 1) × (n − 1) submatrix of M, thendet B = 0 if the corresponding n − 1 edges contain a cycle, and det B = ±1 if they forma spanning tree of G. If the edges corresponding to the columns contain a cycle C, thenthe columns sum to the zero vector when weighted with +1 or −1 determined by whetherthe directed edge is followed forward or backward when following the cycle. This columndependency implies det B = 0.For the other case, we use induction on n. For n = 1, by convention a 0 × 0 matrixhas determinant 1. Now suppose n > 1, and let T be the spanning tree whose edgesare the columns of B. Since T has at least two leaves, B contains a row correspondingto a leaf x of T . This row has only one nonzero entry in B. When computing thedeterminant by expanding along that row, the only submatrix B ′ given nonzero weightin the expansion corresponds to the spanning subtree of G − x obtained by deleting xand its incident edge from T . Since B ′ is an (n − 2) × (n − 2) submatrix of the incidencematrix for an orientation of G − x, the induction hypothesis implies that the determinantof B ′ is ±1, and multiplying it by ±1 gives the same result for B.Finally, we need to compute det L ∗ . Let M ∗ be the matrix obtained by deletingrow t of M, so L ∗ = M ∗ (M ∗ ) T . We may assume m ≥ n − 1, else both sides havedeterminant 0 and there are no spanning subtrees. The Binet-Cauchy formula expressesthe determinant of a product of matrices, not necessarily square, in terms of the determinantsof submatrices of the factors. In particular, if m ≥ p, A is a p × m matrix, andB is an m × p matrix, then det AB = ∑ S det A S det B S , where the summation runs overall S ⊆ [m] consisting of p indices, A S is the submatrix of A having the columns indexedby S, and B S is the submatrix of B having the rows indexed by S. When we apply theBinet-Cauchy formula to L ∗ = M ∗ (M ∗ ) T , the submatrix A S is an (n − 1) × (n − 1)submatrix of M as discussed before, and B S = A T S . Hence the summation counts1 = (±1) 2 for each set of n − 1 edges corresponding to a spanning tree and 0 for eachother set of n − 1 edges.□Example 3.4.4 Consider the graph G that shown in the following Figure 3.6, we shallcompute the number of spanning trees of this graph by the matrix-tree theorem.Figure 3.6: A graph G and its adjacency matrix, degree matrix and Laplacian matrixThe degree (diagonal) matrix D for this graph is the 8 × 8 matrix and the adjacencymatrix A is the 8 × 8 matrix as follows: