•1 e2 //
The fundamental cut associated with e3 corresponds to the partition V1 = {4, 6},V2= {1, 2, 3, 5} and its incidence vector is
The fundamental cut matrix B of G is an (n − 1) × m matrix defined as follows.
The rows of B are indexed by E(T ), while the columns are indexed by E(G). The ith row of B is the incidence vector of the fundamental cutKiassociated with ei, i = 1, . . . , n − 1. Since eiis the only edge of T that is inKi, i = 1, . . . , n − 1, B must be of the form [I, Bf] where Bf is of order (n − 1) × (m − n + 1).
The fundamental cycle matrix C of G is an (m − n + 1) × m matrix defined as fol-lows. The rows of C are indexed by E(Tc), while the columns are indexed by E(G).
The ith row of C is the incidence vector of the fundamental cycleCiassociated with ei, i = n, . . . , m. Since eiis the only edge of Tcthat is inCi, i = n, . . . , m, C must be of the form [Cf, I] where Cf is of order (m − n + 1) × (n − 1).
Lemma 5.4. Bf= −C0f.
Proof. Let Q be the incidence matrix of G. As seen earlier, each row vector of B is in the row space of Q. Also, the transpose of any row vector of C is in the null space of Q. It follows that BC0= 0. Therefore,
Example 5.5. Consider the graph G and the spanning tree T as in Example 5.3.
Let Q be the incidence matrix of G. There is a close relationship between Q, B and C, as we see next.
Theorem 5.6. Let Q1be the reduced incidence matrix obtained by deleting the last row of Q and suppose Q1is partitioned as Q1= [Q11, Q12], where Q11is of order (n − 1) × (n − 1). Then Bf = Q−111Q12and Cf = −Q012(Q011)−1.
Proof. The rank of Q11 equals n − 1, which is the rank of Q. Therefore, the rows of Q1form a basis for the row space of Q. Since each row of B is in the row space of Q, there exists a matrix Z such that B = ZQ1. In partitioned form, this equation reads
I Bf = Z Q11Q12 .
It follows that ZQ11= I and ZQ12= Bf. Thus, Z = Q−111 and Bf = Q−111Q12. The second part follows, since by Lemma 5.4, Cf = −B0f. ut
5.3 Minors
We continue to use the notation introduced earlier. We first consider minors of B and C containing all the rows.
Theorem 5.7. Let G be a connected graph with n vertices, m edges, and let B be the fundamental cut matrix of G with respect to the spanning tree T. Then the following assertions hold:
(i) a set of columns of B is a linearly independent set if and only if the correspond-ing edges of G induce an acyclic graph;
(ii) a set of n− 1 columns of B is a linearly independent set if and only if the corresponding edges form a spanning tree of G;
(iii) if X is a submatrix of B of order(n − 1) × (n − 1), then det X is either 0 or ±1;
(iv) det BB0equals the number of spanning trees of G.
Proof. Recall that the columns of B are indexed by E(G). Let Q be the incidence matrix of G. Let Q1be the reduced incidence matrix and let Q1= [Q11, Q12] as in
5.3 Minors 61 Theorem 5.6. By Theorem 5.6, B = Q−111Q1. Let Y be the submatrix of B formed by the columns j1, . . . , jk, and let R be the submatrix of Q1formed by the columns with the same indices. Then Y = Q−111R, and hence, rankY = rank R. In particular, the columns of Y are linearly independent if and only if the corresponding columns of R are linearly independent. By Lemma 2.5, the columns of R are linearly independent if and only if the corresponding edges of G form an acyclic graph. This proves (i).
Assertion (ii) follows easily from (i).
To prove (iii), note that det X is det Q−111 multiplied by the determinant of a sub-matrix of Q1of order (n − 1) × (n − 1). Since Q is totally unimodular (see Lemma 2.6), it follows that det X is either 0 or ±1.
To prove (iv), first observe that, by the Cauchy–Binet formula, det BB0 =
∑(det Z)2, where the summation is over all (n − 1) × (n − 1) submatrices Z of B.
By (ii), det Z is nonzero if and only if the corresponding edges form a spanning tree of G, and then by (iii), det Z must be ±1. Hence, det BB0equals the number of
spanning trees of G. ut
We now turn to the fundamental cycle matrix. Let C be the fundamental cycle matrix of G with respect to the spanning tree T. Recall that the columns of C are indexed by E(G).
Lemma 5.8. Columns j1, . . . , jk of C are linearly dependent if the subgraph of G induced by the corresponding edges contains a cut.
Proof. As usual, let Q be the incidence matrix of G. Suppose that the edges of G indexed by j1, . . . , jk contain a cut. Let u be the incidence vector of the cut. As observed earlier, u0is in the row space of Q and hence u0= z0Qfor some vector z.
Then Cu = CQ0z= 0, since CQ0= 0. Note that only the coordinates of u indexed by j1, . . . , jkcan possibly be nonzero. Thus, from Cu = 0 we conclude that the columns
j1, . . . , jkare linearly dependent. ut
If E(G) = E1∪ E2is a partition of the edge set of the connected graph G into disjoint subsets, and if E1does not contain a cut, then E2must induce a connected, spanning subgraph. We will use this observation.
Lemma 5.9. Let X be a submatrix of C of order (m − n + 1) × (m − n + 1). Then X is nonsingular if and only if the edges corresponding to the column indices of X form a cotree of G.
Proof. Let the columns of X be indexed by F ⊂ E(G). If X is nonsingular, then by Lemma 5.8, the subgraph induced by F contains a cut. Then Fcinduces a connected, spanning subgraph. Since |Fc| = n − 1, the subgraph must be a spanning tree of G.
Therefore, the edges in F form a cotree.
Conversely, suppose the edges in F form a cotree Sc, where S is a spanning tree of G. Let D be the fundamental cycle matrix with respect to S. Note that the columns of C, as well as D, are indexed by E(G), listed in the same order. Since the rows of C, as well as D, are linearly independent, and since their row spaces are the same, there exists a nonsingular matrix Z of order (m − n + 1) × (m − n + 1) such that C = ZD.
Therefore, an (m − n + 1) × (m − n + 1) submatrix of C is nonsingular if and only if the corresponding submatrix of D is nonsingular. The submatrix of D indexed by F is the identity matrix. Hence, the submatrix of C indexed by F is nonsingular. ut
We now prove the converse of Lemma 5.8.
Lemma 5.10. Let F ⊂ E(G) and suppose the columns of C indexed by F are linearly dependent. Then the subgraph of G induced by F contains a cut.
Proof. If the subgraph of G induced by F does not contain a cut, then the subgraph of G induced by Fcis spanning and connected. Therefore the subgraph induced by Fccontains a spanning tree S of G. By Lemma 5.9, the columns of C indexed by the edges in the cotree Scare linearly independent. These columns include all the columns indexed by F. Then the columns of F must also be linearly independent.
This is a contradiction and the result is proved. ut
Our next objective is to show that the fundamental cut matrix and the fundamen-tal cycle matrix are tofundamen-tally unimodular.
Lemma 5.11. Let G be a connected graph with n vertices, m edges, and let B be the fundamental cut matrix of G with respect to the spanning tree T. Then B is totally unimodular.
Proof. Consider a k × k submatrix D of B, and suppose D is indexed by E1⊂ E(T ) and E2⊂ E(G). If k = n − 1, then by Theorem 5.7, det D is either 0 or ±1. So, suppose k < n − 1. If det D = 0 then there is nothing to prove. So, suppose D is nonsingular. Then the columns of B indexed by E2are linearly independent, and by Theorem 5.7, the corresponding edges induce an acyclic subgraph of G. We may extend this subgraph to a spanning tree S, using only edges from T. The submatrix of B formed by the columns corresponding to the edges in S is a matrix of order (n − 1) × (n − 1), and it is nonsingular by Theorem 5.7. Thus, det S = ±1. We may expand det S using columns coming from the identity matrix and therefore det S =
± det D. Hence, det D = ±1. ut
Lemma 5.12. Let G be a connected graph with n vertices, m edges, and let C be the fundamental cycle matrix of G with respect to the spanning tree T. Then C is totally unimodular.
Proof. Recall that if B is the fundamental cut matrix with respect to the spanning tree T, then B = [I, Bf] and C = [−B0f, I]. Consider a submatrix F of C. If F is a submatrix of −B0f, then it follows by Lemma 5.11 that det F is either 0 or ±1. If F contains some part from the identity matrix, then we may expand det F along the columns coming from the identity matrix and again conclude that det F is either 0
or ±1. ut
We saw in Theorem 5.7 that det BB0 equals the number of spanning trees in G.
We now give an interpretation of the principal minors of BB0.
5.3 Minors 63 Theorem 5.13. Let G be a connected graph with n vertices, m edges, and let B be the fundamental cut matrix of G with respect to the spanning tree T. Let E ⊂ E(T ) and let BB0[E|E] be the submatrix of BB0with rows and columns indexed by E. Then det BB0[E|E] equals the number of ways of extending Ecto a spanning tree of G.
Proof. Let |E| = k. By the Cauchy–Binet formula, det BB0[E|E] =
∑
F⊂E(G),|F|=k
(det B[E|F])2, (5.1)
where B[E|F] denotes the submatrix of BB0 indexed by the rows in E and the columns in F. Note that since B[E(T )|E(T )] is the identity matrix, B[E|F] is non-singular if and only if B[E(T )|F ∪ Ec] is nonsingular, in which case by Lemma 5.11, det B[E|F] = ±1. Now B[E(T )|F ∪ Ec] is nonsingular if and only if the edges F ∪ Ec form a spanning tree of G, and hence the result follows by (5.1). ut Corollary 5.14. Let G be a connected graph with n vertices, m edges, and let B be the fundamental cut matrix of G with respect to the spanning tree T. Let ei∈ E(T ) and let BB0(ei|ei) be the submatrix of BB0 with row and column indexed by eideleted. Thendet BB0(ei|ei) equals the number of spanning forests of G with two components, such that the endpoints of eiare in different components.
Proof. By Theorem 5.13, det BB0(ei|ei) equals the number of ways of extending ei
to a spanning tree of G, which is precisely the number as asserted in the result. ut It may be mentioned that the theory of fundamental matrices may be developed for undirected graphs, resulting in 0 − 1 matrices. The treatment is similar, except the underlying field is that of integers modulo 2.
Exercises
1. Let G be a connected graph with n vertices, m edges, B the fundamental cut matrix, and C the fundamental cycle matrix of G. Show that the m × m matrix
B C
is nonsingular.
2. Let Ki be a cut in G with incidence vector xi, i = 1, . . . , n − 1. Suppose x1, . . . , xn−1 are linearly independent. Show that all nonzero (n − 1) × (n − 1) minors of the matrix X = [x1, . . . , xn−1] are equal.
3. Let G be a connected graph with n vertices, m edges, and let C be the fundamental cycle matrix of G with respect to the spanning tree T. Let E ⊂ E(T )c. Show that det CC0[E|E] equals the number of ways of extending Ecto a cotree of G.
4. Let G be a connected graph with n vertices, m edges, and let B be the fundamental cut matrix of G with respect to the spanning tree T. Let T1be a subtree of T. Show that det BB0[E(T1)|E(T1)] equals det L(V (T1)|V (T1)), where L is the Laplacian matrix of G.
5. Let G be a connected planar graph and let G∗be its dual. Let T be a spanning tree of G and let T∗be its dual spanning tree of G∗. Show that the fundamental cut matrix of G with respect to T equals the fundamental cycle matrix of G∗with respect to T∗.
The material in this chapter is generally covered in most basic texts, but the level and the depth to which it is covered may vary. We list below only two selected references: Deo [1] is recommended for an elementary treatment, while Recski [2], Chapter 1, is more advanced. The statements and the proofs of several results in Section 5.3 have not appeared in the literature in the present form.
References and Further Reading
1. N. Deo, Graph Theory with Applications to Engineering and Computer Science, Prentice-Hall, Inc., New Jersey, 1974.
2. A. Recski, Matroid Theory and Its Applications in Electric Network Theory and in Statics,Algorithms and Combinatorics, 6. Springer-Verlag, Berlin, 1989.
Chapter 6
Regular Graphs
A graph is said to be regular if all its vertices have the same degree. If the degree of each vertex of G is k, then G is said to be k-regular. Examples of regular graphs include cycles, complete graphs and complete bipartite graphs with bipartite sets of the same cardinality.