Edge–Laplacian of a tree

λ1x_j= d_jx_j−

∑

k:k∼ j

x_k. (4.2)

From (4.1) and (4.2) we get

λ1x_i= d_ix_i−

∑

k:k∼i,k∼ j

x_k−

∑

k:k∼i,k6∼ j

x_k (4.3)

and

λ1x_j= d_jx_j−

∑

k:k∼ j,k∼i

x_k−

∑

k:k∼ j,k6∼i

x_k. (4.4)

Subtracting (4.4) from (4.3),

λ1(x_i− x_j) = d_ix_i− d_jx_j−

∑

k:k∼i,k6∼ j

x_k+

∑

k:k∼ j,k6∼i

x_k

≤ d_ix_i− d_jx_j− (d_i− c(i, j))x_j+ (d_j− c(i, j))x_i

≤ (d_i+ d_j− c(i, j))(x_i− x_j). (4.5) If x_j= x_ifor all j ∼ i, then from (4.1) we see that λ1= 0, which is not possible if the graph has at least one edge. Therefore, there exists j such that i ∼ j and x_i> x_j. Now from (4.5) we see that

λ1≤ d_i+ d_j− c(i, j),

and the result follows. ut

Corollary 4.14. Let G be a graph with the vertex set V = {1, . . . , n}. Let L be the Laplacian of G with maximum eigenvalue λ1. Then

λ₁≤ max{d_i+ d_j: 1 ≤ i < j ≤ n, i ∼ j}.

4.5 Edge–Laplacian of a tree

Let G be a graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}. Assign an ori-entation to each edge, and let Q be the n × m incidence matrix. The matrix K = Q⁰Q has been termed the edge-Laplacian of G. The vertices and the edges of K are in-dexed by E(G). For i 6= j, the (i, j)-element of K is 0 if the edges e_iand e_jhave no vertex in common. If they do have a common vertex then the (i, j)-element of K is

−1 or 1 according as e_iand e_jfollow each other, or not, respectively. The diagonal entries of K are all equal to 2. Note that rank K = rank Q, and it follows that the edge-Laplacian of a tree is nonsingular. In the remainder of this section we consider the edge-Laplacian of a tree and obtain a combinatorial description of its inverse.

Let T be a tree with the vertex set {1, . . . , n} and the edge set {e1, . . . , e_n−1}.

Assign an orientation to each edge of T and let Q be the incidence matrix.

Lemma 4.15. Let H be the (n − 1) × n matrix defined as follows. The rows and the columns of H are indexed by the edges and the vertices of T, respectively. Set h_{i j}= 1 if edge e_iis directed away from vertex j, and h_{i j}= 0 otherwise. Then HQ = In−1. Proof. Fix i 6= j. Suppose edge e_jjoins vertices u, v and is directed from u to v. Then q_{u j}= 1, qv j= −1 and qk j= 0, k 6= u, k 6= v. Thus, the (i, j)-element of HQ equals

n

∑

k=1

h_ikq_{k j}= h_iu− h_iv.

Note that e_iis either directed away from both u and v or is directed towards both u and v. Therefore, h_iu= h_ivand hence the (i, j)-element of HQ is zero. If i = j then h_iu= 1 and hiv= 0 and then ∑ⁿ_k=1h_ikq_{k j}= h_iu− h_iv= 1. This completes the proof.

u t By Lemma 4.15 HQH = H and therefore H is a g-inverse of Q. It is well known that the class of g-inverses of Q is given by H + X (I − QH) + (I − HQ)Y, where X and Y are arbitrary. Since HQ = I by Lemma 4.15, the class of g-inverses of Q is given by H + X (I − QH), where X is arbitrary. We now determine the X that produces the Moore–Penrose inverse of Q.

By Lemma 2.2, rank HQ = rank Q = n − 1. Also rank (I − QH) = n − rank (QH) = 1. Therefore, rank X (I − QH) ≤ 1 and hence X (I − QH) = uv⁰ for some vectors u and v. Thus, we conclude that Q⁺= H + uv⁰for some vectors u and v, which we now proceed to determine.

For i = 1, . . . , n − 1, the graph T \ e_i has two components, both trees, one of which is closer to the tail of e_i, while the other is closer to the head of ei. We refer to these as the tail component and the head component of e_i, respectively. Let tibe the number of vertices in the tail component of e_i. Let t = (t1, . . . ,t_n−1)⁰. It is clear from the definition of H that H1 = t.

Considering Q⁺= H + uv⁰ and Q⁺1 = 0, we get H1 + (v⁰1)u = 0, and hence (v⁰1)u = −H1 = −t.

Also, I = Q⁺Q= HQ + uv⁰Q= I + uv⁰Q, and hence uv⁰Q= 0. Since Q⁺6= H, u and v are nonzero vectors. Hence, v⁰Q= 0 and v = α1 for some α. Therefore, t= −(v⁰1)u = −α1⁰1u = −αnu. It follows that

Q⁺= H + uv⁰

= H − t α n(α1⁰)

= H −1 nt1⁰.

Thus, we have obtained the formula for Q⁺given in the next result.

Theorem 4.16. The rows and the columns of Q⁺are indexed by the edges and the vertices of T, respectively. The (i, j)-element of Q⁺is−^t_nⁱ if j is in the head compo-nent of e_i, and it equals 1 −^t_nⁱ if j is in the tail component of e_i.

4.5 Edge–Laplacian of a tree 53 Example 4.17. Consider the following tree:

•1

satisfies HQ = I, while the Moore–Penrose inverse of Q is given by

Q⁺=1 Laplacian of the tree. We state the expression and omit the easy verification. We first introduce some notation. If i is a vertex and e_j an edge of T, then f (e_j, i) will denote the number of vertices in the component of T \ {e_j} that does not contain i.

Also, for vertices i, j and edge e_k, α(i, j, ek) will be −1 or 1 according as ekis on the (i, j)-path or otherwise, respectively.

Theorem 4.18. For i, j = 1, . . . , n, the (i, j)-element of L⁺is given by non-singular. It is easily seen by the Cauchy–Binet formula that det K = n. Since K⁻¹= K⁺= Q⁺(Q⁺)⁰, we may obtain an expression for K⁻¹using Theorem 4.16.

Again, we only state the expression. Extending our earlier notation, let us denote by

f(e_j, e_i) the number of vertices in the component of T \ {ej} that does not contain e_i.

Suppose edge e_ihas head u and tail v, while edge e_jhas head w and tail x. We say that e_iand e_jare similarly oriented if the path joining u and w contains precisely one of x or v. Otherwise, we say that e_iand e_jare oppositely oriented.

Theorem 4.19. For i, j = 1, . . . , n − 1, the (i, j)-element of K⁻¹is given by

±1

n(n − f (e_i, e_j))(n − f (e_j, e_i)),

where the sign is positive or negative according as e_iand e_jare similarly oriented or oppositely oriented, respectively.

Exercises

1. Let G be a graph with n vertices and let L be the Laplacian of G. Show that the number of spanning trees of G is given by ¹

n²det(L + J).

2. Let G × H be the Cartesian product of graphs G and H. Determine L(G × H) in terms of L(G) and L(H). Hence, determine the Laplacian eigenvalues of G × H in terms of those of G and H.

3. Let G be a graph with n vertices and m edges. Show that κ(G), the number of spanning trees of G, satisfies

κ (G) ≤1 n

 2m n− 1

n−1

.

4. Let G be a graph with vertex set V (G) = {1, . . . , n}. Let L be the Laplacian of G with maximum eigenvalue λ1. Show that λ1≤ n.

5. Let T be a tree with vertices {1, . . . , n} and edges {e₁, . . . , e_n−1}. Show that the edges of T can be oriented in such a way that the edge-Laplacian K becomes an entrywise nonnegative matrix.

6. Let A be an m × n matrix. Show that (A⁰)⁺= (A⁺)⁰and that (AA⁰)⁺= (A⁰)⁺A⁺.

Basic properties of the Laplacian are discussed in the books by Biggs and by Godsil and Royle quoted in Chapter 2. Other relevant references are as follows: Section 4.4: [4,1,3], Section 4.5: [2,5,6].

References and Further Reading

1. W.N. Anderson and T.D. Morley, Eigenvalues of the Laplacian of a graph, Linear and Multilinear Algebra,18(2):141–145 (1985).

2. R.B. Bapat, Moore–Penrose inverse of the incidence matrix of a tree, Linear and Multilinear Algebra,42:159–167 (1997).

4.5 Edge–Laplacian of a tree 55 3. K. ch. Das, An improved upper bound for Laplacian graph eigenvalues, Linear

Algebra Appl., 368:269–278 (2003).

4. R. Grone and R. Merris, The Laplacian spectrum of a graph, II SIAM J. Discrete Math.,7(2):221–229 (1994).

5. R. Merris, An edge version of the matrix-tree theorem and the Wiener index, Linear and Multilinear Algebra,25:291–296 (1989).

6. J.W. Moon, On the adjoint of a matrix associated with trees, Linear and Multi-linear Algebra,39:191–194 (1995).

Chapter 5

Cycles and Cuts

Let G be a graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}. Assign an ori-entation to each edge of G and let Q be the incidence matrix. The null space of Q is called the cycle subspace of G whereas the row space of Q is called the cut subspace of G. These definitions are justified as follows.

Consider a cycleC in G and choose an orientation of the cycle. Let x be the m×1 incidence vector of the cycle. We claim that Qx = 0, that is, x is in the null space of Q. The ith element of Qx is (Qx)_i= ∑^m_j=1q_{i j}x_j. If vertex i andC are disjoint, then clearly (Qx)_i= 0. Otherwise there must be precisely two edges ofC which are incident with i. Suppose e_pwith endpoints i, k and e_swith endpoints i, ` are inC . If e<sub>p</sub>has head i and tail k and if e<sub>s</sub>has head i and tail `, then we have q_ip= 1, qis= 1 and q_{i j}x_j= 0 for j 6= p, j 6= s. Also, xp= −x_s. It follows that (Qx)i= 0. The cases when e_pand e_s have other orientations are similar. Therefore, (Qx)_i= 0 for each i and hence x is in the null space of Q.

We now turn to cuts. Let V (G) = V₁∪ V₂be a partition of V (G) into nonempty disjoint subsets V₁and V₂. The set of edges with one endpoint in V1and the other endpoint in V₂is called a cut. Denote this cut byK . Given a cut K we define its incidence vector y as follows. The order of y is m × 1 and its components are indexed by E(G). If e_iis not inK , then yi= 0. If e_i∈K , then yi= 1 or −1 according as eiis directed from V1to V2, or from V2to V1, respectively.

Let u be a vector of order n × 1 defined as follows. The components of u are in-dexed by V (G). Set u_i= 1 or −1 according as i ∈ V1or i ∈ V2, respectively. Observe that y⁰=¹₂u⁰Qand hence y is in the row space of Q.