Network flows

is nonsingular. We prove the result by induction on k. The proof is easy for k ≤ 2.

Assume the result to be true for principal submatrices of order less than k. It will be sufficient to show that all the cofactors of B are nonnegative. The cofactor of a diagonal entry of B is the determinant of a principal submatrix of L, which is positive. We show that the cofactor of the (1, 2)-element of B is nonnegative, and the case of other cofactors will be similar. Partition B(1|2) as

B(1|2) = b₂₁ x⁰ y B(1, 2|1, 2)

 . Then

det B(1|2) = (det B(1, 2|1, 2))(b21− x⁰(B(1, 2|1, 2)⁻¹y). (9.1) By induction assumption, B(1, 2|1, 2)⁻¹≥ 0. Also x and y have all entries non-positive. Furthermore, det B(1, 2|1, 2) > 0 and b₂₁≤ 0. It follows from (9.1) that det B(1|2) ≤ 0. Thus, the cofactor of the (1, 2)-element of B is nonnegative and the

proof is complete. ut

We return to the proof of the fact that the resistance distance satisfies the triangle inequality. In order to prove r(i, j) + r( j, k) ≥ r(i, k), we must show that for any g-inverse H of L,

h_ii+ h_{j j}− 2h_{i j}+ h_{j j}+ h_kk− 2h_jk≥ h_ii+ h_kk− 2h_ik, and this is equivalent to

h_{j j}+ h_ik− h_{i j}− h_jk≥ 0. (9.2) Let B = L( j| j). By Lemma 9.1 B⁻¹≥ 0. We choose the following g-inverse of L: In L, replace entries in the jth row and column by zeros and replace L( j| j) by B⁻¹. Let the resulting matrix be H. It is easily verified that LHL = L, and hence H is a g-inverse of L. Note that h_{j j}= h_{i j}= h_jk= 0, while h_ik≥ 0 since B⁻¹≥ 0, as remarked earlier. Thus, (9.2) is proved and the resistance distance satisfies the triangle inequality.

We make the following observation in passing. Letting H be the g-inverse as defined above, we see that

r(i, j) = h_ii+ h_{j j}− h_{i j}− h_ji= h_ii=det L(i, j|i, j)

det L(i|i) . (9.3)

The corresponding result for a tree was noted in Corollary 4.10.

9.2 Network flows

If x is a vector of order n × 1 then the norm ||x|| is defined to be the usual Euclidean norm; ||x|| = (∑ⁿi=1x²_j)¹². We prove a preliminary result, which we will used.

Lemma 9.2. Let A be an n × m matrix and let b be a vector of order n × 1 in the column space of A. Let H be a g-inverse of A such that HA is symmetric. Then z= Hb is a solution of the equation Ax = b with minimum norm.

Proof. Note that Ax = b is consistent, as b is in the column space of A. Let y be a solution of Ax = b, so that Ay = b. We must show ||Hb|| ≤ ||y||, or that

||HAy|| ≤ ||y||. Squaring both sides of this inequality it will be sufficient to show that y⁰A⁰H⁰HAy≤ y⁰y. Now

y⁰A⁰H⁰HAy= y⁰(HA)⁰HAy= y⁰HAHAy= y⁰HAy,

since H satisfies AHA = A and A⁰H⁰= HA. Since HA is a symmetric, idempotent matrix, its eigenvalues are either 0 or 1, and hence I − HA is positive semidefinite.

It follows that y⁰(I − HA)y ≥ 0 and the result is proved. ut

Let G be a connected graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}.

We interpret the resistance distance between the two vertices i and j in terms of an

“optimal” flow from i to j. First we give some definitions. Let the edges of G be assigned an orientation and let Q be the vertex-edge incidence matrix. A unit flow from i to j is defined as a function f : E(G) → IR such that

Q









 f(e₁) f(e₂)

... f(e_m)











= e_{i j}. (9.4)

The interpretation of (9.4) is easy: At each vertex other than i, j, the incoming flow is equal to the outgoing flow; at i the outgoing flow is 1 whereas at j, the incoming flow is also 1. The norm of a unit flow f is defined to be

|| f || = ( m

∑

f(e_j)² )¹₂

.

Let L be the Laplacian matrix of G. As noted in the proof of Lemma 8.10, e_{i j} is in the column space of L, and hence in the column space of Q. Therefore, (9.4) is consistent. By Lemma 9.2, a solution of (9.4) with minimum norm is given by f₀= Q⁻e_{i j}, where Q⁻is a minimum norm g-inverse of Q, that is, satisfies QQ⁻Q= Q, and that Q⁻Qis symmetric. Since Q⁺ is a minimum norm g-inverse of Q, f₀= Q⁺e_{i j}is a solution of (9.4) with minimum norm. Then

|| f₀||²= e⁰_{i j}(Q⁺)^TQ⁺ei j= e⁰_{i j}L⁺ei j

since L⁺= (QQ^T)⁺= (Q^T)⁺Q⁺= (Q⁺)^TQ⁺ by well-known properties of the

9.2 Network flows 115 Moore–Penrose inverse. Thus, we have proved that r(i, j) = e⁰_{i j}L⁺e_{i j} is the mini-mum value of || f ||²where || f || is a unit flow from i to j.

We illustrate the interpretation to calculate r(u, v) in the following simple example.

Example 9.3. Consider the graph following:

•1

In the next result we show that resistance distance is dominated by classical dis-tance.

Theorem 9.4. Let G be a connected graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}, and let i, j ∈ V (G). Then

r(i, j) ≤ d(i, j). (9.5)

Proof. If i = j then r(i, j) = d(i, j) = 0. Assume that i 6= j. Choose and fix an i j-path P of length d(i, j). Orient each edge in P in the direction from i to j and assign an arbitrary orientation to the remaining edges of G. If g : E(G) → IR is defined as

g(e_k) = 1, if e_k∈P 0, otherwise

then g is a unit flow from i to j. Since r(i, j) is the minimum value of the squared

norm of a unit flow from i to j, we have

d(i, j) = ||g||²≥ r(i, j),

and the proof is complete. ut

It can be shown that equality holds in (9.5) if and only if there is a unique i j-path.

Before proving this statement we need a preliminary result.

Lemma 9.5. Let G be a connected graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}. Assume that each edge of G is oriented and let Q be the vertex-edge in-cidence matrix. Let y be a vector of order m× 1 such that Qy = 0. If e_kis a cut-edge of G, then y_k= 0.

Proof. Since e_k is a cut-edge, G \ {e_k} has two components, say G₁ and G₂. We assume that e_k is oriented in the direction from V (G₁) to V (G2). Let z be the in-cidence vector of V (G₁). Thus, z is a vector of order n × 1 with its components indexed by V (G). The component corresponding to a vertex is 1 if it is in V (G₁), and 0 otherwise. Then it can be verified that z⁰Qis a vector of order 1 × m with all the components 0 except that z_k= 1. It follows that z⁰Qy= y_k. Since Qy = 0, we

conclude that y_k= 0. ut

Theorem 9.6. Let G be a connected graph with V (G) = {1, . . . , n} and E(G) = {e₁, . . . , e_m}, and let i, j ∈ V (G). Then r(i, j) = d(i, j) if there is a unique i j-path. In particular, resistance distance and classical distance coincide for a tree.

Proof. Let g : E(G) → IR be the unit flow from i to j as defined in the proof of Theorem 9.4. Let g also denote the m × 1 vector with components g(e₁), . . . , g(e_m).

Then a general unit flow from i to j is given by g + y, where Qy = 0. If there is a unique i j-path, say P, then every edge on P must be a cut-edge and then, by Lemma 9.5, the components of y corresponding to the edges onP are zero. Thus, any unit flow from i to j coincides with g onP. Therefore, r(i, j), which equals the minimum value of the squared norm of a unit flow from i to j, must be ||g||²=

d(i, j). ut

The converse of Theorem 9.6 is true and the proof will be left as an exercise. The fact that resistance distance and classical distance coincide for a tree is also clear from Theorem 8.12.