Incidence Matrix

Abstract

Let G be a graph with V (G) = {1, . . . , n} and E(G) = {e 1 , . . . , e m }. Suppose each edge of G is assigned an orientation, which is arbitrary but fixed. The (vertex-edge) incidence matrix of G, denoted by Q(G), is the n ×m matrix defined as follows. The rows and the columns of Q(G) are indexed by V (G) and E(G), respectively. The (i, j)-entry of Q(G) is 0 if vertex i and edge e j are not incident, and otherwise it is 1 or −1 according as e j originates or terminates at i, respectively. We often denote Q(G) simply by Q. Whenever we mention Q(G) it is assumed that the edges of G are oriented. Example 2.1 Consider the graph shown. Its incidence matrix is given by Q. Q = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ −1 1 −1 0 0 0 1 0 0 −1 0 0 0 −1 0 0 1 0 0 0 1 0 0 −1 0 0 0 1 −1 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ © Springer-Verlag London 2014 R.B. Bapat, Graphs and Matrices, Universitext, DOI 10.1007/978-1-4471-6569-9_2 13 14 2 Incidence Matrix 2.1 Rank For any graph G, the column sums of Q(G) are zero and hence the rows of Q(G) are linearly dependent. We now proceed to determine the rank of Q(G). Lemma 2.2 If G is a connected graph on n vertices, then rank Q(G) = n − 1. Proof Suppose x is a vector in the left null space of Q := Q(G), that is, x Q = 0. Then x i − x j = 0 whenever i ∼ j. It follows that x i = x j whenever there is an i j-path. Since G is connected, x must have all components equal. Thus, the left null space of Q is at most one-dimensional and therefore the rank of Q is at least n − 1. Also, as observed earlier, the rows of Q are linearly dependent and therefore rank Q ≤ n − 1. Hence, rank Q = n − 1. Theorem 2.3 If G is a graph on n vertices and has k connected components then rank Q(G) = n − k.