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1 THE DISTANCE MATRIX OF A BIDIRECTED TREE R. B. BAPAT, A. K. LAL, AND SUKANTA PATI Abstract. A bidirected tree is a tree in which each edge is replaced by two arcs in either direction. Formulas are obtained for the determinant and the inverse of a bidirected tree, generalizing well-known formulas in the literature. Key words. Tree, Distance matrix, Laplacian matrix, Determinant, Block matrix. AMS subject classifications. 05C50, 5A5.. Introduction. We refer to 4, 8 for basic definitions and terminology in graph theory. A tree is a simple connected graph without any circuit. We consider trees in which each edge is replaced by two arcs in either direction. In this paper, such trees are called bidirected trees. We now introduce some notation. Let e,0 be the column vectors consisting of all ones and all zeros, respectively, of the appropriate order. Let J = ee t be the matrix of all ones. For a tree T on n vertices, let d i be the degree of the i-th vertex and let d = (d,d 2,...,d n ) t, δ = 2e d and z = d e. Note that δ + z = e. Let T be a tree on n vertices. The distance matrix of a tree T is a n n matrix D with D ij = k, if the path from the vertex i to the vertex j is of length k; and D ii = 0. The Laplacian matrix, L, of a tree T is defined by L = diag(d) A, where A is the adjacency matrix of T. The distance matrix of a tree is extensively investigated in the literature. The classical result concerns the determinant of the matrix D (see Graham and Pollak 7), which asserts that if T is any tree on n vertices then det(d) = ( ) n (n )2 n 2. Thus, det(d) is a function dependent only on n, the number of vertices of the tree. The formula for the inverse of the matrix D was obtained in a subsequent article by Graham and Lovász 6 who showed that D = (e z)(e z)t 2(n ) L. This result was 2 Received by the editors May 8, Accepted for publication April 0, Handling Editor: Bryan L. Shader. Stat-Math Unit, Indian Statistical Institute Delhi, 7-SJSS Marg, New Delhi , India (rbb@isid.ac.in). Indian Institute of Technology Kanpur, Kanpur , India (arlal@iitk.ac.in). Department of Mathematics, Indian Institute of Technology, Guwahati, India (sukanta pati@yahoo.co.in). 233

2 234 R. B. Bapat, A. K. Lal, and S. Pati extended to a weighted tree in. A q-analogue of the distance matrix was considered in 2. In this paper, we extend the result of Graham and Lovász by considering the distance matrix for a bidirected tree, denoted D = (D ij ). 2. Preliminaries. Let T be a tree on n vertices. Replace each undirected edge f i = {u,v} of T with two arcs (oppositely oriented edges) e i = (u,v) and e i = (v,u). Let u i > 0 and v i > 0 be the weights of the arcs e i and e i, respectively. We call the resulting graph a bidirected tree T with the underlying tree structure T. The distance D ij from i to j is defined as the sum of the weights of the arcs in the unique directed path from i to j. Thus if D ij = u i + v j, then D ji = v i + u j. Note i A that the diagonal entries of the matrix D are zero and in general the matrix D is not a symmetric matrix. We are interested in extending the definition of a Laplacian to the bidirected trees. The Laplacian matrix L = (L kl ) of a bidirected tree T with the underlying tree structure T is defined by j B i A j B 0 if {k,l} T L k,l = u i +v if f i = {k,l} T i u i+v i if k = l, f i k where e i k means that k is an endvertex of e i. Notice that, in view of the Gersgorin disc theorem, the matrix L is a positive semidefinite matrix. For the sake of convenience, we write w t = u t + v t. Then, the distance matrix D and the Laplacian matrix L of the bidirected tree T (shown in Figure 2.) are given by 4 u 4 v 4 u 2 v u 2 v 2 u 3 v 3 u v 5 3 Fig. 2.. A bidirected Tree on 6 vertices

3 Distance Matrix of a Bidirected Tree u u + u 2 u + u 3 + v 4 u + u 3 u + u 3 + u 5 v 0 u 2 u 3 + v 4 u 3 u 3 + u 5 D = v + v 2 v 2 0 v 2 + u 3 + v 4 v 2 + u 3 v 2 + u 3 + u 5 v + v 3 + u 4 v 3 + u 4 u 2 + v 3 + u 4 0 u 4 u 4 + u 5, v + v 3 v 3 u 2 + v 3 v 4 0 u 5 v + v 3 + v 5 v 3 + v 5 u 2 + v 3 + v 5 v 4 + v 5 v 5 0 and w w w w + w 2 + w 3 w 2 w L = w 2 w w 3 0 w 4. w w 4 w 3 + w 4 + w 5 w w 5 w 5 Observe that if u i = v i = for all i, then the matrices D and L reduce to the matrices D and 2L, respectively. We now introduce some further notation. Let T be a bidirected tree on n vertices. Let T be a spanning tree of T. Thus, T is obtained from T by choosing one arc and hence T has 2 n spanning trees. Let us denote the indegree and the outdegree of the vertex v in T by In T(v) and Out T(v), respectively. Consider the vectors z and z 2 defined by z (i) = ( ) n T In T(i) w( T) (2.) z 2 (i) = ( ) n T Out T(i) w( T), (2.2) where w( T) is the product of the arc weights of T. For example, the vectors z and z 2 for the bidirected tree T given in Figure 2. are z = u w 2 w 3 w 4 w 5 u 2 u 3 v + u u 3 v 2 + u u 2 v 3 + 2u v 2 v 3 + v v 2 v 3 w 4 w 5 v 2 w w 3 w 4 w 5 u 4 w w 2 w 3 w 5 w w 2 u 3 u 4 u 5 u 5 v 3 v 4 + 2u 3 u 4 v 5 + u 4 v 3 v 5 + u 3 v 4 v 5 v 5 w w 2 w 3 w 4

4 236 R. B. Bapat, A. K. Lal, and S. Pati and z 2 = v w 2 w 3 w 4 w 5 u u 2 u 3 + 2u 2 u 3 v + u 3 v v 2 + u 2 v v 3 u v 2 v 3 w 4 w 5 u 2 w w 3 w 4 w 5 v 4 w w 2 w 3 w 5 w w 2 u 4 u5v 3 + u 3 u 5 v 4 + 2u 5 v 3 v 4 u 3 u 4 v 5 + v 3 v 4 v 5 u 5 w w 2 w 3 w 4. Note that taking u i = v i = for all i, and putting k = In T (i), we see that ( ) n z (i) = T In T(i) = k 2 n k r=0 T In T (i)=r In T(i) k ( ) k = (r ) 2 n k = ( k2 k 2 k) 2 n k = 2 n 2 (k 2), r r=0 so that z = z 2 = ( ) n 2 n 2 (e z). Let T be a bidirected graph. Since each arc of a spanning tree T contributes to exactly one entry in In T, we have n In T(i) = n. Hence, i= n n z t e = z (i) = i= i=( ) n T In T(i) w( T) = ( ) n T w( T) n In i= T(i) = ( ) n = ( ) n i= A similar reasoning implies that n T w( T) w i. (2.3) n z t 2e = ( ) n i= For a bidirected tree T on n vertices we define w(t ) as w i. (2.4) w(t ) = T w( T) = n i= w i = ( ) n z t e = ( ) n z t 2e.

5 Distance Matrix of a Bidirected Tree 237 We use the convention that if T is a tree on a single vertex then z = e = z 2 and w(t) =. With this convention, for a bidirected forest F with the bidirected trees T, T 2,...,T k as components, the weight of F is defined as w(f) = k w(t i ). In the next section, we relate the matrices D and L and also obtain some properties of the matrix D with respect to minors. As corollaries, we obtain the results of Graham and Pollak 7) on det(d) and that of Graham and Lovasz 6 on D. i= 3. The main result. In this section, we extend certain results on distance matrices of trees to distance matrices of bidirected trees. Recall that a pendant vertex is a vertex of degree one. Denote by G v the graph obtained by deleting the vertex v and all arcs incident on it from G. By e k we denote the vector with only one nonzero entry which appears at the kth place. Given any tree T on vertices {,2,...,n} we may view it as a rooted tree and hence there is a relabeling of the vertices so that for each i > the vertex i is adjacent to only one vertex from {,...,i }. With such a labeling the vertex n is always a pendant vertex. Henceforth, unless stated otherwise, each bidirected tree will be assumed to have an underlying tree with such a labeling. Furthermore, for i < j, the weight of an arc e j = (i,j) will be assumed to be u j and the weight of the arc e j = (j,i) will be assumed to be v j. If T is a bidirected tree by T e j e j we denote the bidirected graph obtained by deleting the arcs (i,j) and (j,i) from T. We use the method of mathematical induction to prove our results. In the induction step, we start with a bidirected tree T on k + vertices, where the pendant vertex k + is adjacent to the vertex r. We use the definition of the distance matrix of the bidirected tree T = T {k + } to get the distance matrix of T. Putting D = D(T ), D = D(T ), L = L(T ), L = L(T ), we see that D = D u k e + De r v k e t + e t rd 0, L = L + e r e t r e t r e r. (3.) Furthermore, In T(k + ) w( T) ( ) k+ z (k + ) = T = w( T) (k+,r) T = w(t ) ( v k ).

6 238 R. B. Bapat, A. K. Lal, and S. Pati Also ( ) k+ z (r) = T = In T(r) w( T) (r,k+) T and for i k +,r, we have, In T(r) w( T) + = ( ) k z (r)u k + (k+,r) T z k+ (i) = ( ) In T T(i) w( T) = ( ) k+ In T(i) w( T) + ( ) k+ (r,k+) T = z (i)u k z (i)v k = z (i). In T(r) w( T) ( ) k z (r)v k + w(t )v k, (k+,r) T In T(i) w( T) Thus we have Similarly we have z = z 2 = wk z + ( ) k+ w(t ) v k e r ( ) k+ w(t ) ( v k ) wk z 2 + ( ) k+ w(t ) u k e r ( ) k+ w(t ) ( u k ). (3.2). (3.3) Note that these two equations provide an efficient way of computing the vectors z and z 2 for a bidirected tree. Combined with the next theorem they give an efficient way to compute D. We shall use our previous observations are in the proof of the next theorem. Theorem 3.. Let D be the distance matrix of a bidirected tree on n vertices where the pendant vertex n is adjacent to r. Then n det(d) = ( ) n u i v i w(t e i e i) (3.4) i= Dz = det(d)e, z t 2D = det(d)e t, and (3.5) D = L ( ) n z z t 2 det(d)w(t ). (3.6)

7 Distance Matrix of a Bidirected Tree 239 Proof. We prove the theorem by induction on the number of vertices of any bidirected tree. So, as the first step, let n = 2. In this case, the matrices D, L,z and z t 2 are respectively, D = 0 u v 0, L = w w w w u,z = v v, and z 2 = u As w(t e e ) =, det(d) = u v = ( ) 2 u v w(t e e ), D z = det(d) e and z t 2 D = det(d) e t. Thus (3.5) is true for n = 2. Also, for n = 2, the right hand side of (3.6) reduces to L z z t 2 det(d)w(t ) = = = w w w w w w w 0 v u 0 w = D u v u 2 w u v v 2 u v + u w v w v u w w Hence (3.6) holds for n = 2. We now assume that the equalities in (3.4), (3.5) and (3.6) are true for n = k. Let n = k + and T be a bidirected tree on k + vertices. Put T = T {k + }. To establish the first equality (3.5) we need to show that = ( ) k k u i v i w(t e i e i). As D is invertible, using (3.), the induction hypothesis and (2.3), we have i= = det(d) 0 (v k e t + e t rd)d (u k e + De r ) (3.7) = det(d) u k v k e t D e + v k e t e r + u k e t re + e t rde r = det(d) u k v k e t z det(d) + v k + u k = ( ) k u k v k w(t ) det(d) (3.8) k = ( ) k u k v k w(t ) + ( ) k u i v i w(t e i e i) i= = ( ) k k u k v k w(t e k e k) + u i v i w(t e i e i) k = ( ) k u i v i w(t e i e i). i= Hence the first equality holds for n = k +. i=.

8 240 R. B. Bapat, A. K. Lal, and S. Pati To prove the second equality we need to show that D z = e, z t 2 D = e t. Using the expressions given in (3.) and (3.2) we have D z D u = k e + De r wk z + ( ) k+ w(t ) v k e r v k e t + e t rd 0 ( ) k w(t ) v k The first block of the vector D z reduces to Dz + ( ) k u k v k w(t )e. Substituting det(d)e for Dz and using (3.8), The second block of the vector D z reduces to the first block of D z = e. (3.9) v k e t z e t rdz + ( ) k+ v k w(t ) ( v k e t e r + e t rde r ). Now using the equality e t rde r = 0, the equations (2.3), (3.4) and (3.8), we have the second block of D z =. (3.0) A similar reasoning gives that z t 2 D = e t. Hence the second equality is established for n = k +. We now prove that the matrix D is indeed given by (3.6). As 0, put W = 0 (v k e t + e t rd)d (u k e + De r ). From (3.7), it follows that W = det D. (3.). Since D D u = k e + De r v k e t + e t, it is straight- rd 0 Let D A A = 2 A 2 A 22 forward to see that A = D + D (u k e + De r )W (v k e t + e t rd)d, (3.2) A 2 = D (u k e + De r )W, (3.3) A 2 = W (v k e t + e t rd)d, (3.4) A 22 = W. (3.5) Using (3.) and the induction hypothesis, we have A = D + det D ( uk det(d D )( e + e r vk e t D + e t ) r) ( = D + det D = D + )( z t ) 2 v k det(d) + et r z u k det(d) + e r uk v k det(d) z z t 2 + ( u k z e t r + v k e r z t 2. ) + det(d)er e t r (3.6)

9 and A 2 = D (u k e + De r )W = det D Similarly Distance Matrix of a Bidirected Tree 24 uk D u k z + det(d)e r e + e r = det(d. ) (3.7) A 2 = v kz t 2 + det(d)e t r det(d. (3.8) ) We now determine the first and second blocks of the matrix L ( ) k+ z z t 2 w(t ). (3.9) Using Equations (3.), (3.2), (3.3), (3.7), (3.) and the induction hypothesis, the first block of (3.9) equals «( ) kz 2 L + eret r z t 2 + u k v k w(t ) 2 e re t r + u k w(t )z e t r + v k w(t )e rz t 2 + w(t ) = L + ( )k z z t 2 w(t ) e re t r = D + ( )k z z t 2 w(t ) + ( )k u k v k w(t )e r e t r e re t r + det(d ) det(d) + u kz e t r + v k e r z t 2 + det(d ) + det(d) det(d e r e t r ) + u kz e t r + v k e r z t 2 = D + u kv k z z t 2 det(d) + det(d) e re t r + u kz e t r + v k e r z t 2 and the second block of (3.9) equals (3.20) e r u k w(t )z ( ) k u k v k w(t ) 2 e r w(t ) = e r u kz = u kz + det(d)e r e r det(d ) + det(d) = (u k D e + e r )W. (3.2) Showing that A 2 is the (2,)-block of (3.9) is similar. The (2,2)-block of (3.9) is + ( )k u k v k w(t ) 2 w(t ) = + det(d ) + det(d) = W. Hence the third equality is established for n = k + and the proof is complete using induction.

10 242 R. B. Bapat, A. K. Lal, and S. Pati 4. Bidirected trees with two types of weights. Suppose T is a rooted tree with root r. Let u and v be two vertices of T. As we traverse the u-v path from u to v there exists a vertex, say w (which may be u itself), such that the path from u to v moves in the direction of r until it meets vertex w and then moves away from r. Let the lengths of the two paths u-w and w-v be l and l 2, respectively. Also, let x and y be two constants. We define the distance between u and v as D(u,v) = l y + l 2 x. (4.) Clearly, when x = y =, this reduces to the usual distance between u and v. We illustrate this with the following example Fig. 4.. A rooted tree Consider the tree given in Figure 4.. The distance matrix of the tree is as follows: 0 x x 2x 2x 2x 3x 3x y 0 x + y x x 2x + y 2x 2x y x + y 0 2x + y 2x + y x 3x + y 3x + y 2y y x + 2y 0 x + y 2x + 2y 2x + y 2x + y D =. 2y y x + 2y x + y 0 2x + 2y x x 2y x + 2y y 2x + 2y 2x + 2y 0 3x + 2y 3x + 2y 3y 2y x + 3y x + 2y y 2x + 3y 0 x + y 3y 2y x + 3y x + 2y y 2x + 3y x + y 0 Observe that if we apply a similar labeling to T as in the previous section and consider the bidirected tree T with the underlying tree structure T, and use the weights u i = x i, v i = y i, then the distance matrix D of the bidirected tree is nothing but the distance matrix D. Henceforth a rooted tree is assumed to have the root and the labeling as described earlier. Let u be a vertex of a rooted tree T. A vertex v is called a child of u if u and v are adjacent and u is on the v- path. Let us denote the number of children of u by ch(u). With the notations defined above, we have the following result. Corollary 4.. Let T be a rooted tree on n vertices and consider the distance

11 Distance Matrix of a Bidirected Tree 243 matrix D. Also, let z and z 2 be vectors of order n given by ( ) n( (ch(i) )y x ) (x + y) n 2, if i =, (z ) i = ( ) n y(x + y) n 2, if i is a pendant vertex, ( ) n (ch(i) )y(x + y) n 2, otherwise and ( ) n( (ch(i) )x y ) (x + y) n 2, if i = r, (z 2 ) i = ( ) n x(x + y) n 2, if i is a pendant vertex, ( ) n (ch(i) )x(x + y) n 2, otherwise. (4.2) (4.3) Then and where L is the usual Laplacian matrix. det(d) = ( ) n (n )xy(x + y) n 2, D = L x + y + z z t 2 (n )xy(x + y) 2n 3, Proof. Let T be the bidirected tree associated with T. As D is the same as D with u i = x and v i = y, the assertion about the determinant follows easily from (3.4). The vectors z, z 2 defined here are nothing but the vectors defined in (2.) and (2.2). In order to see this note that let T be a spanning tree of T and put k = ch(). ( ) n z () = T In T() w( T) = k (x + y) n k r=0 T In T ()=r In T() y r x k r = (x + y) n k k r=0 ( k )(r )y r x k r = (x + y) n k ky(x + y) k (x + y) k r = (x + y) n 2 (ch() )y x. If i is a pendant vertex, put k = ch(i) and observe that ( ) n z (i) = T In T(i) w( T) = x(x + y) n 2. If i is any other vertex, then put k = ch(i), and let p be the parent of i. We have In T(i) w( T) = In T(i) w( T)+ In T(i) w( T) ( ) n z (i) = T (i,p) T (p,i) T

12 244 R. B. Bapat, A. K. Lal, and S. Pati = (x + y) n 3 y (k )y x + kxy(x + y) n 3 = ch(i) y(x + y) n 2. The vector z 2 may be verified similarly. Now the assertion about inverse of D follows from (3.6). As a corollary, we obtain the result of Graham and Pollak 7 on det(d). Corollary 4.2. Let T be a tree on n vertices and let D be its distance matrix. Then det(d) = ( ) n (n )2 n 2. Proof. Let us denote by T the bidirected tree obtained from the given tree T. As observed earlier, the substitution of u i = v i = for i n, reduces the matrix D to the distance matrix D. Under this condition, we have w i = u i + v i = 2 and w(t e i e i ) = 2n 2 for i n. Therefore n det(d) = det(d) ui = X =v i = ( )n u iv iw(t e = i) ui =v i = ( )n (n )2 n 2. i= We now give a corollary to our result that gives a formula for D. This result was also obtained by Graham and Lovasz (see 6). Corollary 4.3. Let T be a tree on n vertices and let D be its distance matrix, L be its Laplacian matrix and let z and e be the vectors defined earlier. Then D = (e z)(e z)t 2(n ) L 2. Proof. Let us denote by T the bidirected tree obtained from the given tree T. Observe that under the condition, u i = v i =, the matrix D reduces to D, the matrix L reduces to L 2 and z = z 2 = ( ) n 2 2 n 2 (z e). So, we have D = D = L + z z t ( )n 2 ui=v i= det(d)w(t) = L n 4 (e z)(e z) t (n )2 n 2 2 n = L 2 Hence the required result follows. (e z)(e z)t +. 2(n ) ui=v i=

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