Linear Algebra and its Applications

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1 Linear Algebra and its Applications 43 (009) Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: q-analogs of distance matrices of 3-hypertrees Sivaramakrishnan Sivasubramanian Department of Mathematics, Indian Institute of Technology, Bombay, India A R T I C L E I N F O A B S T R A C T Article history: Received 3 January 009 Accepted April 009 Available online 3 May 009 Submitted by R.A. Brualdi AMS classification: 5A09 5A5 5A4 Keywords: q-analog Distance matrix Determinant Pfaffian We consider the distance matrix of trees in 3-uniform hypergraphs (which we call 3-hypertrees). We give a formula for the inverse of a few q-analogs of distance matrices of 3-hypertrees T. Some results are analogs of results by Bapat et al. for graphs. We give an alternate proof of the result that the determinant of the distance matrix of a 3- hypertree T depends only on n, the number of vertices of T. Further, we give a Pfaffian identity for a principal submatrix of some (skewsymmetrized) distance matrices of 3-hypertrees when we fix an ordering of the vertices and assign signs appropriately. A result of Graham, Hoffman and Hosoya relates the determinant of the distance matrix of a graph and the determinants of its -connected blocks. When the graph has as blocks a fixed connected graph H which satisfy some conditions, we give a formula for the inverse of its distance matrix. This result generalises a result of Graham and Lovasz. When each block of G isafixedgraphg, we also give some corollaries about the sum of the entries of the inverse of the distance matrix of G and some of its analogs. 009 Elsevier Inc. All rights reserved.. Introduction Graham and Pollak [5] proved a very elegant theorem on the determinant of the distance matrix of atreeonn vertices. They showed that the determinant is ( ) n (n ) n. Thus the determinant value does not depend on the tree structure and only depends on n. Graham et al. [3] later proved a very attractive theorem about the determinant of the distance matrix of a digraph D as a function of the determinant of the -connected blocks (henceforth called blocks) of D. When each -connected block is a 3-clique (or a triangle graph), this implies results for the determinant of the distance matrix of trees in 3-uniform hypergraphs, which we call 3-hypertrees. We give an alternative proof of this result address: krishnan@math.iitb.ac.in Supported partially by Grant P07IR /$ - see front matter 009 Elsevier Inc. All rights reserved. doi:0.06/j.laa

2 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) for spanning trees in 3-hypertrees. We define spanning trees in r-uniform hypergraphs as follows. Let H = (V, E) be an r-uniform hypergraph (i.e. each e E is an r sized subset of V). Consider a bipartite graph B H associated with H as follows. The vertex set of B H is V E. For every e E, e ={v, v,..., v r }, there are r edges {e, v i } for i r in B H (see Fig. ). H is said to be an r-hypertree if the bipartite graph B H is a tree in the usual sense. It is known (see for example, Hirschman and Reiner [6]) that 3-hypertrees exist iff the number of vertices is odd. Let n = k +. Any such 3-hypertree on n vertices has k 3-hyperedges. See Fig. for an example of a 3-hypertree. We first reprove the following result which can be shown from the Theorem of Graham et al. [3]. Theorem. Let T be a spanning tree in a 3-uniform hypergraph on n = k + vertices. Then its distance matrix D has determinant k 3 k. We give several proofs of this result. The proof in Section is the simplest and seems redundant due to other proofs in this work, but we present it here because the other methods do not seem to give Corollary. We then consider two types of q-analogs of the above result. Both these analogs have been considered before for trees by Bapat et al. []. In the first analog, we replace entries i in the distance matrix by [i] q = + q + +q i where q is an indeterminate and where [0] q = 0. This matrix with polynomial entries is invertible and we find its inverse explicitly (see Section 3). We also give weighted generalisation of some of these results (see Section 5). The second analog is when we replace entry i in the distance matrix by q i (again q is an indeterminate) where q 0 =. The inverse of such a matrix is again given explicitly (see Section 4). We recall the symmetric distance matrix D G of a graph G. We skew-symmetrize D G by assigning positive signs to d i,j whenever i < j and negative signs otherwise. With this fixed sign pattern, it is possible that the order of the vertices (along the rows) might change the determinant value. For a fixed order π of the vertex set V ={,,..., n} of G, let the skew-symmetric distance matrix with rows and columns as in π be sd π. We recall that for a 3-hypertree n = k + is odd. Since the determinant of an odd skew-symmetric matrix is zero, we omit a leaf vertex v to get M π = v sdπ r v c v, which is sd π without the row and column corresponding to vertex v. Consider the (mixed) hypertree T {v}. This hypergraph gives a perfect matching PM on V {v} as follows. Since v is a leaf vertex, there is a unique 3-hyperedge e ={v, x, y }.Add{x, y } to PM and delete e. The resulting hypergraph st is a forest of 3-hypertrees, with each component being a 3-hypertree minus a vertex. Repeat this process till we get k (disjoint) edges e i ={x i, y i }. For the 3-hypergraph shown in Fig., the perfect matching obtained after removing vertex 9 is PM ={{3, 4}, {, 5}, {, 6}, {7, 8}}. We note that the perfect matching obtained by this procedure is unique. Let π be a permutation of the vertices in V {v} such that every edge 8 6 e 4 7 e 3 5 e 3 e 4 9 Fig.. The perfect matching on deleting vertex 9.

3 36 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) e 3 e Fig.. A 3-hypertree on 5 vertices. 5 of the perfect matching appears consecutively in π. Such permutations clearly exist and we call them nice-permutations. We show that if π is a nice permutation, then det(m π v ) = (see Section 6). Consider the case when all blocks of a graph G are identical, denoted H. For example, trees satisfy this condition where each block is an edge (i.e. is K ). We impose two conditions on H D H, the distance matrix of H is non-singular and secondly, the row-sum of H is identical for all vertices. When H satisfies these conditions, the inverse of the distance matrix of G can be explicitly found. This result (see Section 8) generalizes a result of Graham and Lovasz [4] where the inverse of the distance matrix of a tree is found. For such graphs G, the sum of the entries of the inverse of the distance matrix of G is found. This sum is shown to be independent of the tree structure on the blocks of G and only dependent on H and the number of copies of H in G. The analogous result for trees is implicit in the work of Graham and Lovasz. For the inverses of both the q-analog as well as the exponential distance matrix analog, the work of Bapat et al. [] implies a similar sum of entries result. We observe some such results in Section Hypertrees We concentrate on spanning trees of 3-uniform hypergraphs in this section. We call them 3- hypertrees. See Fig. for an example of a 3-hypertree. The tree T in this example has V ={,,...,5} and E ={{,, 3}, {3, 4, 5}}. Since vertices, are leaf vertices belonging to the same 3-hyperedge, we call them paired half-leaves. Similar to trees having at least two leaves, it is simple to note that any 3-hypertree has at least two paired half-leaves. The distance d i,j between vertices v i and v j in a 3-hypertree is defined as the number of 3-hyperedges on the unique path from v i to v j. Similar to graphs, we follow the convention that d i,i = 0. For the 3-hypertree of Fig., d, = as there is one 3-hyperedge (3) containing the vertices and. Similarly, d,4 = as two hyperedges 3, 345 are traversed on the (unique) path from to 4. The distance matrix of the 3-hypertree in Fig. is 0 0 D = Our proof of Theorem is similar to Yan and Yeh s proof [9] of Graham and Pollak s result, where the Condensation method for evaluating a determinant is used. Zeilberger [0] has given an elegant proof of this Theorem which is originally due to Reverend Dodgson []. We recall this Theorem below. Theorem (Condensation rule). Let A be a square matrix of order n >. Let A i,j be the submatrix obtained by deleting from A the ith row and the jth column. Let A be the submatrix obtained by deleting the first and last rows and the first and last columns. Then, det(a) det(a ) = det(a, ) det(a n,n ) det(a,n ) det(a n, ). Proof (Of Theorem ). We use induction on the number, n of vertices in the 3-hypertree. The base cases, when n = 3 and n = 5, being clearly true. Thus, we assume n 7. Any such 3-uniform hypertree has at least two pairs of half-leaves. Let v and v be a pair of half-leaves. Likewise, let v n and v n also be paired half-leaves.

4 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) Let D = (d i,j ) be the distance matrix. Let the ith row and the jth column of D be labelled r i and c j, respectively. Let the matrix obtained from D by omitting rows i, i and columns j, j be denoted D r i c j r i c j.lett = det(d) and denote det(d r c r c ) = det(d r n c n r n c n ) = t and det(d r c r c r n c n r n c n ) = s. By induction we know t = (k )3 k and s = (k )3 k 3. Lemma. With the above notation,. det(d r c ) = T+t.. det(d r c n ) = T 3t det(d r c r n c n ) = (T+3t). 9 ( )( ) ( ) ( ) 4. T+t t+s = (T+3t) t 9 T 3t 6. Proof of item. We note that c and c differ only in two entries (corresponding to their distance to each other). Let D be the matrix obtained from D by doing the elementary column operation c = c c. After this, c =[,, 0,...,0] t (where vector v t is the transpose of v) and as det(d) = det(d ),wegetdet(d) = det(d r c ) det(d r c ). Since r and r of D c differ only in one entry (i.e. D r,c = ), we get det(d r c ) = det(d r c r c ) + det(d r c ), from which we get det(d r c ) = det(d) + det(d r c r c ). Proof of item. Let (half-leaf) vertices and be connected to the remaining tree through the vertex x (i.e. the only 3-hyperedge containing vertices and is {,, x}) and similarly, let the half-leaves n and n be connected to the vertex y. Next, consider the matrix D obtained by the following column operations: c = c c x (c n c y ) and c = c c x (c n c y ). After these operations, c =[,, 0,...,0,,] t and c =[,, 0,...,0,,] t. Since det(d) = det(d ),weget det(d r c n ) = det(d) 3 det(d r c r c ) 6 = T 3t. 6 It is also simple to note that in the matrix D r c n, the columns corresponding to vertex and (ie c and c ) are such that c c =[, 0,...,0] t. Thus after performing this elementary column operation, we note that det(d r c n ) = det(d r c n r c ) and thus det(d r c r c n ) = T 3t. 6 Proof of item 3. Apply Dodgson s condensation rule to the matrix D to get det(d) det(d r c r n c n ) = det(d r c ) det(d r n c n ) det(d r c n ) det(d r n c ). By induction on the dimension of the matrix, det(d r c ) = det(d r n c n ) and this by the first part is ( T+t ). Similarly det(d r c n ) = det(d r n c ) = T 3t. Thus we get det(d r 6 c r n c n ) = (T+3t). 9 Proof of item 4. Apply Dodgson s rule on the matrix (D r c ) to get det(d r c ) det(d r c r c r n c n )=det(d r c r c ) det(d r c r n c n ) det(d r c r c n ) det(d r r c r n c ). Thus, T+t t+s = t (T+3t) ( ) T 3t. 9 6 This completes the proof of all parts of the Lemma. Using the inductive values of t and s, wegett = t or T = k3 k. The option T = t implies t = sand so implies that all the determinant values are identical, but alternate in sign. This is clearly seen to fail when k =,. The above proof gives us the following extra determinant value Corollary. With the above notation, det(d r c n ) = 3 k.

5 38 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) q-analogs of the distance matrix In this section, we consider the q-analog, D q of the distance matrix D of a 3-hypertree. Thus, we replace positive entries i in D by [i] q = + q + q + +q i where q is an indeterminate. Let D q be the q-analog of the n n matrix D. Lete be the n all-ones column vector and for a vertex u, deg(u) its degree, be the number of 3-hyperedges which contain u. Letd be the column vector of the degrees of the vertices of T. Define z = (d e) and let U = (e qz)(e qz) t. For a 3-hypergraph on n vertices, the adjacency matrix A is the n n 0/ matrix with A i,j = iff there is a 3-hyperedge containing vertices i and j.letdbethe n n diagonal matrix with D u,u = deg(u).let L = D A be the Laplacian matrix of the 3-hypertree. Let L = ql (q )I + q(q )diag(z).for example, for the 3-hypertree shown in Fig., itsq-analog of the distance matrix and the matrix L are 0 q + q + 0 q + q + D q = 0 q + q + 0 and q + q L = Theorem 3. With the above notation, e = n D q(e qz), () D q = (n )(q + ) U L. q + () Proof. We show the above by induction on n, the number of vertices of T. Eqs. () and () are both trivial for the 3-hypertree on three vertices. Thus let T be a 3-hypertree on n 5 vertices with vertices v n and v n being paired half-leaves in T connected through the vertex n to the remaining 3-hypertree. Let D q be the q-analog of the distance matrix of T. Lete be the all-ones n column vector. Let st ( s for smaller) be the smaller 3-hypertree obtained by deleting vertices v n and v n, and let sd q be the q-analog of its distance matrix. Let se be the (n ) all-ones column vector and let se k be the (n ) column vector with a one in precisely the kth coordinate for some k n and zero elsewhere. Let sz be the vector z for st (ie sz = sd se where sd is the degree sequence of st). Lastly, set v = se + q(sd q )se n. [ ] sdq V Clearly, D q = where V =[v v] is an (n ) dimensional matrix and L = V t [ ] 0. 0 L We first find (see Eq. (5)) an explicit matrix which we prove is Dq. We then check that this matrix can be expressed as given in Eq. (). To do this, we express the auxiliary matrices P, Q, S by their definitions to get Eqs. (5) (7). We then find the appropriate subblock matrices in Eq. () to get Eqs. (8) (0) and verify that these are identical. (3) (4)

6 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) Proof of Eq. (). By induction on n, with base case being simple. Thus, with base case being simple. we know (n 3) se = sd q (se q sz) and want to show that (n ) e = D q (e q z). It is simple to see that z t =[sz t + sen t,0,0]. Thus, sd q v v se q(sz + se n ) D q (e q z) = v t 0 v t 0 We calculate each of the three products separately. Firstly, sd q (se q sz) q sd q se n + (se + q sd q se n ), which is (n 3) se + se and thus (n ) se. For the next two products (which are identical), we note that se t sz = se t (sd se) which is 3(n 3)/ (n ) and hence se t sz = (n 5)/. (We have used the fact that the sum of the degrees of a 3-uniform hypergraph G = (V, E) is 3 E.) Thus se t (se q sz) = (n ) q (n 5)/. Thus the next product is =[se t + q se t n sd q] [se q (sz + se n )]+ = (n ) q(n 5) q + q se t n sd q(se q sz) q se t n sd q se n + = (n ) q(n 3) + q se t n (n 3) se q 0 = n. This completes the proof of Eq. (). Proof of Eq. (). Consider the (symmetric) matrix D α [ ] P Q D α = Q t, (5) S where P is an (n ) (n ) dimensional matrix, Q is an (n ) dimensional matrix and S is a matrix with values as shown below. We recall v from Eq. (3), and let N = V t (sd q ) V I. It is easy to see that the matrix N is invertible and we set M = N. (6) We note that V t (sd q ) VM = M + I. We also note that since M, N and sdq are symmetric, they are equal to their transpose. We set P = (sd q ) (sd q ) VMV t (sd q ), (7) Q = (sd q ) VM, (8) S = V t (sd q ) VM + L, (9) where in Eq. (9), we recall L from Eq. (4). With these values, it is simple to show that (sd q ) P + V Q t = I (n ) (n ), (0) (sd q ) Q + V S = 0, () V t P + L Q t = 0, () V t Q + L S = I. (3) Thus D α as given in Eq. (5) is the inverse of D q. Henceforth, instead of D α, we write Dq.Wenow expand the matrices P, Q, S according to its definition. Let a = v t (sd q ) v. Thus a = (se t + qse t n sd)(sd q) (se + qsd q se n ) = se t sd q se + qset n se + qset se n + q se t n sdse n = se t n q(n ) (se q sz) + q + 0 = + n 3 n 3 n. 3.

7 40 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) Hence (n )( + q) a =. n 3 If we let M = [ ] x y y x from Eq. (6), it is simple to note that x = (a ) a x + y = a. Further, VMV t = (x + y)vv t. We recall Eqs. (7) (9) and see that P = (sd q ) (sd q ) VMV t (sd q ) and y = (4) a and hence a su = (n 3)( + q) sl + (n 3) q (n )( + q) [ t su q(se q sz)sen + + q ] set n (se q sz) + q se (n 3) n se t n n 3 n 3 su = (n )( + q) sl + q (n 3)se n sen t q (n )( + q) q [ (se q sz)se t n (n )( + q) + set n (se q sz)]. (5) Similarly, Q = (sd q ) VM = (x + y)(sd q ) [v v] =[u u] where u = Likewise, (n )( + q) [se q sz]+ n 3 (n )( + q) q se n. (6) S = V t (sd q ) VM + L [ ][ ] x y = a + y x [ ] 0 = 0 where α = a = (n )+q(n ) (n )(+q) From Eq. (), we consider the matrix Since and α + = [ α α + α + α U (n )(+q) + +q(n ) (n )(+q). L +q. U = (e q z) (e t q z t ) (se q sz) q se n = [ ] (se t q szt) q se t n su q[(se q sz)se t + n se n (se t q sz t )] +4q se n sen t z z = z t, z t ], (7) where z t =[(se t q sz) q se t n ]. We note that diag(se n ) = se n se t n and L=qL (q ) I + q(q )diag(z) n I (n ) (n ) 0 n 0 n =q sen t (q ) 0 t sen t n 0 0 t n 0 diag(sz + se n ) 0 n 0 n +q(q ) 0 t n t n 0 0

8 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) sl + q se n sen t q se n q se n = q sen t q + q. q sen t q q+ ( The first (n ) (n ) block of Dq = U (n )(+q) ) L is +q su = (n )( + q) sl + q (n 3) q (n )( + q) diag(se n ) q (n )( + q) [(se q sz)set + n se n (se t q sz t )], (8) which is identical to P as in Eq. (5). Similarly, the next (n ) dimensional block of (D q ) is [b b] where b = q se n + + (es q sz) q se n, (9) q (n )( + q) which is identical to Q as given in Eq. (6). Likewise, the last block of (D q ) is (n )(+q) q+ +q (n )(+q) q +q, (0) (n )(+q) q +q (n )(+q) q+ +q which is identical to S as given in Eq. (7). This completes the proof. Corollary. Setting q = in the above, we see that D = 3(n ) U L, where D is the distance matrix 3 of a 3-hypertree T, U = (e z)(e z) t and L is as defined before the statement of Theorem 3. Remark. Theorem 3 can be generalized to trees in r-uniform hypergraphs, by changing all occurrences of into r and suitably changing L. The above proof can be modified for trees in r-hypergraphs when r > 3, but the proof gets rather cumbersome. As we will see in Section 7, these distance matrices are obtained when a graph G has as its blocks r-cliques. It would be nice to get a uniform proof of a generalization of Theorem 3 for all r. 4. Exponential distance matrices For a 3-hypertree T, with distances between vertices i and j given by d i,j, define the exponential distance matrix ED T = (e i,j ) as e i,j = iffi = j and e i,j = q d i,j where q is an indeterminate. We abuse notation and refer to the matrix as ED instead of ED T when the 3-hypertree T is clear from the context. We recall that A is the 0/ adjacency matrix of T with A i,j = iffi /= j and there is a 3-hyperedge containing i, j. We recall that deg(u) foravertexu, is the number of 3-hyperedges that u is in and let D be the diagonal matrix with d v,v = deg(v). We note that we are changing the definition of the matrix D (compared to the definition in Section 3) in this section. Theorem 4. For any 3-hypertree T and q /=,, the matrix ED is invertible and ED = I q q + q + A + q D. q () + q + Proof. We again induct on n and the base case when n = 3 is simple. Let T be a 3-hypertree on n vertices where vertices v n and v n are half-leaves connected by vertex v n in T. We recall that se k is the (n ) column vector with ainthekth position and zeroes

9 4 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) elsewhere. Denote st = T {v n, v n } and let sed, sa, sd be the corresponding matrices for st. Let V =[v v], where v t = (q d,n, q d,n,..., q d n,n ). () [ ] P Q We know that sed is invertible, and thus consider the n n matrix M =, where P, Q, S are of dimensions (n ) (n ), (n ) and, respectively, and defined as ( S = V t (sed) V [ ]) q, (3) q Q t = SV t (sed), (4) P = (sed) QV t (sed). (5) With the definition of P, Q and S as in Eqs. (5), (4) and (3), it is simple to check that M (if it exists) is the inverse of ED T. We now show that M exists and is as claimed in the statement of Eq. (). We first note that [ q sedse n = v. i.e. ] (sed) V =[u u], where u = q se n. Thus from Eq. (3), we see that S = q q(q ) q(q ) q, which implies that [ ] x y S =, (6) y x Q t S where x = x + y = q q +q+ and y = q. We also note that q +q+ q + q +. (7) Substituting in Q = (sed) VS, weget Q =[z z], (8) where z is an (n ) column vector with z = see that q q +q+ se n. Substituting for the value of P,we se t n 0 P = (sed) q q + q + se n se t. n (9) sa se n se n We note that A = sen t 0 and D = diag([d + sen,,]). With these observations and with matrices P, Q, S as in Eqs. (9), (8) and (6), we see that Eq. () is true, thus completing the proof. 5. Weighted distance matrices In this section, we look at weighted analogs of results of Section 3. Consider a 3-hypertree T = (V, E) with each 3-hyperedge e E assigned a weight w e. For two distinct vertices u, v V, if the path in T from u to v is e, e,..., e r (where u e, v e r ), define the distance d u,v as w + qw + q w 3 + q r w r. When u = v, define the distance d u,v = 0. Let D = (d u,v ) be the matrix of such weighted distances. D is not necessarily a symmetric matrix. As an example, consider the same 3-hypertree as in Fig. where w({,, 3}) = w and w({3, 4, 5}) = w. The matrix D for the 3-hypertree with these weights is

10 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) w w w + qw w + qw w 0 w w + qw w + qw D = w w 0 w w. w + qw w + qw w 0 w w + qw w + qw w w 0 Theorem 5. Let T = (V, E, w) be a 3-hypertree with a weight function w : E R (we refer to the value of this function on the 3-hyperedge e as w e rather than w(e)) and q-weighted distance matrix D. Let V =n where n = k +. Let W = e E w e. If q /=, then det(d) = (q + ) k (W) e E w e. Proof. We again proceed by induction on the number n = k + of vertices in T. It is simple to check the Theorem when n = 3. Let vertices n and n be paired half-leaves connected to the remaining tree through the vertex (n ). LetsT be the smaller 3-hypertree T {n, n } and let its weighted distance matrix be sd. Let the removed 3-hyperedge have weight w k.letw k = W w k. We assume that q /= and k i= w i /= 0. These assumptions do not cause any loss in generality as they can be removed by a continuity argument. We recall the vector v from Eq. (). Clearly, if p = sd se n + w k v and q = w k e + q(sd)se n, then sd p p D = q t 0 w k. (30) q t w k 0 We claim that for any 3-hypertree T, (e q z) t v = e t v qz t v = + q. We show this by induction on n. It is clear that the claim is true for the 3-hypertree on three vertices. As before let vertices v, v belong to a leaf block, connected in T through the vertex v 3. Suppose the statement is true for st = T {v, v }, where, we assume that the vector sv represents the exponential distance from vertex v n to v 3, v 4,..., v n. Thus we know (se q sz) t sv = + q. We rewrite this as n i=3 qd i,n ( + q) n i=3 qdeg st(i)q d i,n. Suppose d(v n, v 3 ) = d where deg st (i) is the degree of vertex i in st. In the above sum, we will have two extra terms corresponding to vertices v, v both of which are a distance d + fromv n. Thus, we add q d+ ( + q) and subtract q q d (on account of vertex v 3 ) and subtract q(q d+ + q d+ ) (on account of vertices v, v ). Since this contribution from T st is zero, the proof is complete. Similarly, we show that W k e t = (e t qz t )D. We again induct on n to show this. Clearly, (e t qz t ) =[se t ] q[sz t 0 0] =[se t q (sz + se n ) t ]. We look at sd sdse n + w k v sdse n + w k v [se t q(sz + se n ) t ] w k se t + qsen t sd 0 w k. w k se t + qsen t sd w k 0 The first block of this product is (se t qsz t )sd qsen t sd + w kse t + qsen t sd + w kse t + qsen t sd while the next two (identical) blocks are each (set qsz t )sdse n qsen t sdse n + (se t qsz t )w k v qsen t w kv + w k. Hence, the product is = [ W k se t + w k se t ] W k + w k W k + w k completing the proof of this statement. Next, we show that if b = p t (sd) p, then b = w k ( + q) + w k (+q).we W k make use of the following theorem (see Meyer s book [7, p. 475]).

11 44 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) [ ] A B Theorem 6. If M =, then det(m) = det(a) det(d CA B). C D If we define α = q t (sd) p, then, using the above theorem, we need to find the determinant of the matrix M = [ ] α w k α w k α α. We note that α = (w k se t + qse t n sd)(sd) (sdse n + w k v) = w k se t se n + w k set (sd) v + q se t n (sd)se n + q se t n w kv = w k + w (se t q sz t ) v k W k + q 0 + qw k = w k ( + q) + w k ( + q). W k Thus det(m) = α (w k α) = w k ( + q) w k+w k. Applying Theorem 6 to the matrix in Eq. W k (30), we get det(d) = det(sd) det(m) = (q + ) k (W k ) ( ) W ( + q) which completes the proof. w e e se w k W k Corollary 3. Let T be a 3-hypertree on n = k + vertices. When the weights on all the 3-hyperedges are, the weighted distance matrix is D q. Thus det(d q ) = k( + q) k, which is a q-analog of Theorem. 6. Pfaffian results 6.. Preliminaries on the Pfaffian We recall some preliminaries that we need about the Pfaffian of a skew-symmetric matrix. Any n n skew-symmetric matrix M = (m i,j ) has det(m) = 0 whenever n is odd and when n is even, det(m) = Pfaff(M) where Pfaff(M) is defined up to a ± sign. The Pfaffian can alternatively be defined as follows. Let n be even and PM(n) be the set of perfect matchings of [n]. Given an M PM(n), we call the matched elements as edges of M and denote the edge set as E(M). Then it is known (see [6]) that Pfaff(M) = ( ) cross(m) m i,j, (3) M PM(n) i<j E(M) where cross(m) = #{i < j < k < l :{i, k}, {j, l} E(M)} is the crossing number of M. We will use the following result (see Stembridge [8, Lemma.]) about the crossing number of perfect matchings. Lemma. Let M be a perfect matching of [n] in which vertices i, i + are not matched. Let s i Mbethe perfect matching obtained from M by exchanging the vertices i and i +. Then cross(m) and cross(s i M) differ by. 6.. Results Let T be any 3-hypertree on n = k + vertices and let v n be a half-leaf. From Section, we recall PM, the perfect matching obtained after deleting v n. We note that all pairs of half-leaves are edges of PM (since they are in only one 3-hyperedge, they cannot be parted). Let π be a nice-permutation of V {v n }.LetsD π = (d i,j ) be the skew-symmetric distance matrix with rows and columns of V {v n } ordered as they occur in π, i.e. if π = (π, π,..., π k ), then we can assume that {π i, π i } is an edge of the perfect matching for all i k. We recall that signs are assigned as d i,j = dist T (v i, v j ) if i < j and d i,j = d j,i, i.e. D is the distance matrix above the main diagonal and is skew-symmetric. We recall M π n = sdπ r n c n is the matrix obtained by deleting the v n th row and the v n th column from D π.

12 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) Theorem 7. With the notation as above, det(m π n ) =. Equivalenty, Pfaff(Mπ n ) =. Proof. We show the result for the Pfaffian. It is simple to check this for the unique 3-hypertree on 3 vertices. Consider any tree T on n 5 vertices, n = k +. Let π = (π, π,..., π k ) and consider the contribution arising from each perfect matching M of [k] when we use Definition 3. For some i, let vertices π i, π i be a pair of half-leaves of T {v n } and consider the total contribution to the Pfaffian from perfect matchings M where vertices π i and π i are not an edge of M, i.e {π i, m } and {π i, m } are in E(M) where m /= m. By swapping these two vertices in M,we get another matching M with {π i, m }, {π i, m } M. By Lemma, cross(m) = cross(m ) ±. We note that vertices π i and π i have the same distance to all vertices except themselves (because they are half-leaves). Thus in M π n, d π i,m = d πi,m and d πi,m = d πi,m. Because the signs change due to a switch, the total contribution from such perfect matchings M where {π i, π i } / E(M) is zero. Thus we only need to consider those perfect matchings M, where {π i, π i } E(M). We induct on the number of vertices and get the total contribution to Pfaff(M π ) n as d π i,π i multiplied by the Pfaffian of a smaller 3-hypertree T = T {π i, π i }. By induction, the Pfaffian of T r n c n is and since d πi,π i =, we get the total contribution from such perfect matchings is. The proof is complete. As the existence of nice-permutations is independent of the matrix entries, we assume that nicepermutations π exist for 3-hypertrees. More generally, let D q be the q-analog of the matrix D. From the above proof, we can infer the following two corollaries. Corollary 4. Let D q be the q-analog of the distance matrix of a 3-hypertree T and let v n be a half-leaf. Let π be a nice-permutation of V {v n }. Then, det(d π q v n) = and Pfaff(D π q v n) =. For a 3-hypertree T where each 3-hyperedge e, is given weight w e, define the q-weighted distance between vertices as follows. For two distinct vertices u, v V, letp = e, e,..., e k, where u e and v e k be the (unique) path between u and v in T. Define the q-weighted distance d q (u, v) to be w + qw + q w + +q k w k and the n nq-weighted distance matrix DW π = q (d q(u, v)) where π is a nice-permutation of the V {v n }. The proof of Theorem 7 also proves the following. Corollary 5. Let T = (V, E) be a 3-hypertree with weights w e on the 3-hyperedge e E. Let v n be a half-leaf vertex and let π be a nice-permutation of V {v n }. Then, Pfaff(DW π q v n) = e E w e and equivalently det(dw π q v n) = e E we. 7. Trees with identical graphs as blocks Consider the distance matrix D T for a 3-hypertree T.IfweaskforagraphG whose distance matrix D is identical to D T, then it is simple to see that corresponding to a 3-hypertrees T, the graph obtained by replacing each 3-hyperedge by K 3 (the complete graph on three vertices, or the triangle graph) and joining these triangles as in T gives a graph G which has the same distance matrix as T. Thus, we consider the case when G has k copies of a connected graph H as its blocks. We assume that Condition. D H is invertible and. the row-sum of D H is the same for all rows(this happens for example when H is vertex transitive). Since the row-sum of D H is independent of the rows, we denote it as rsum H. From the result of Graham, Hoffman and Hosoya it is clear that det(d) /= 0 iff det(d H ) /= 0. We are interested in D in this section. We recall Lemma 3, where for a tree T, D T is a scalar times a rank-one matrix plus a scalar

13 46 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) Fig. 3. A graph made from two copies of H = C5. times the Laplacian matrix of T. It is well known that the Laplacian matrix of any graph G has zero row-sum for all rows. Below, we give a similar expression for D when G is made of k copies of H. Suppose G has k copies of H connected in a treelike manner. Let T be this tree on the blocks of H. Let d be the vector of degrees of the vertices of G with respect to the underlying tree T. Thus, d v is equal to the number of blocks which contain the vertex v. For the graph in Fig. 3, the degree vector is d t =[,,,,,,,, ]. Let D be the distance matrix of G. Lete be the all ones column vector in G dimensions and let z = d e. Letu =[e ( H )z] and let U = u u t.foravertexx G, we write z x and u x for the xth components of the vectors z and u, respectively. Given tree T, it clearly has at least two leaf blocks which are terminal copies of H. Clearly, any such leaf block H i has exactly one connecting-vertex cv and we call the graph H i {cv} as a leaf-block. Theorem 8. With the notation as above, k rsum H e = D[e ( H )z], (3) (k rsum H H )D = U + R, (33) where R is a symmetric matrix with zero row-sum. Proof. We first show the proof of Eq. (3) by induction on k, the number of copies of H. When k =, the statement is trivially true. Let the statement be true when the graph has at most k copies of H and let G be a graph with k copies. As before, we denote the distance matrix of the smaller graph sg obtained after deleting a leaf-block of H as sd. We assume that sd satisfies Eq. (3) with vectors se and sz.letcv be the connecting vertex for the deleted leaf block and let se cv be the column vector with a in position cv and[ zero elsewhere. ] Let ee be the H dimensional column vector of all ones. sd N D(e ( H )z) = N t [ ] se ( H )(sz + se cv ) M ee where N is a sg ( H ) dimension matrix with the vth column being sd se cv + d(v, cv).se and M is the matrix D H restricted to the vertices in H {cv}. The first block of the product is (k ) rsum H se ( H )sdse cv + N ee. Since N ee = rsum H se + ( H )sdse cv, the first block has value (k )rsum H se + rsum H se = krsum H se. The second block of the product is N t (se ( H )sz) ( H )N t se cv + M ee. Clearly, M ee = (rsum H d v,cv ) ee, N t (se ( H )sz) = (k )rsum H ee + H d v,cv ee and ( H )N t se cv = ( H )d v,cv ee. Their sum is thus k rsum H ee. This completes the proof. We now prove Eq. (33). For x, y G,letcof x,y be the cofactor of D at the (x, y)th position. We show that for all vertices x G, the row-sum corresponding to vertex x (denoted rsum(x)) in the matrices (rsum H H k)d and U are identical. For a vertex x G in the matrix U, the corresponding row-sum is rsum(x) = u x u y = (e x ( H )z x ) y ( H )z y ) y G y G(e = G ( H ) G z x ( H ) y z y + ( H ) (k )z x

14 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) = G (k )( H ) ( H )z x [G ( H )(k )] = H ( ( H )z x ), while in the matrix (rsum H H k)d, rsum(x) = (k rsum H H ) y G cof x,y det(h), where we have used the following simple identities. x G z x = k (if G has k blocks) and G = k H (k ). The second identity implies G (k )( H ) = H. Thus, to show that the row-sums for each vertex x are identical, we need to show that k rsum H H cof x,y y G det(h) = H ( ( H )z x) which is the xth row of Eq. (3). Finally, as both D and U are symmetric matrices, R is also symmetric, completing the proof. 8. Sum of matrix entries For an n n matrix M = (m i,j ),letm = i,j m i,j be the sum of the entries of M. We recall the following lemma from Graham and Lovasz [4]. Lemma 3. Let T be a (usual) tree on n vertices with distance matrix D. Let L be its Laplacian matrix. Let e be the n all-ones column vector and let d be the column vector of degrees and let δ = e d. Then, D = (n ) δδt L. The above lemma implies the following corollary (implicit in [4,]). Corollary 6. Let D be the distance matrix of a tree T on n vertices. Then, D =. Similarly, (D n q ) = n q(n ). Thus, both D n and (Dq ) are independent of the tree T and only depend on n. Proof. We note that L = 0 and (δδ t ) = ( v V( d v )) which is 4. Thus D = n. Similarly, and L = (q )(n q(n the result for (Dq ) follows by observing that U = (n q(n )) (n )(+q) )). A statement similar to Corollary 6 can be made for 3-hypertrees as well. We note that by setting q = in Theorem 3, we get the following. Corollary 7. Let T be a 3-hypertree and let D be its distance matrix. Then D = 3 n n q(n 3). Hence, both D n and (D q ) are independent of T s structure and only depend on n. and (D q ) = When G is made of k copies of H satisfying conditions, then too, a similar statement can be made. With the notation of Section 7, we have the following. Corollary 8. Let D be the distance matrix of a graph G made from k copies of H satisfying Conditions. Then D = H and is independent of the tree structure on G s k rsum H blocks. U Proof. By Theorem 8, it suffices to determine k rsum. We finish the proof by observing that H H U = ( v G(e ( H )z)) = H. One result from both Corollaries 6 and 7 is generalized here.

15 48 S. Sivasubramanian / Linear Algebra and its Applications 43 (009) Similarly, for the exponential distance matrix of a 3-hypertree, we note the following corollary of Theorem 4. Corollary 9. Let T be a 3-hypertree on n = k + vertices and ED be its exponential distance matrix. Then, ED = n + 6q. Thus ED q+ depends only on n and not on the tree structure of T. Acknowledgments Some of the theorems in this work were in their conjecture form, tested using the computer package Sage. We thank the authors for generously releasing their software as an open-source package. We also thank a referee for pointing out several inaccuracies and for comments which enhanced the presentation. References [] R.B. Bapat, A.K. Lal, S. Pati, A q-analog of the distance matrix of a tree, Linear Algebra Appl. 46 (006) [] C.L. Dodgson, Condensation of determinants, Proc. Roy. Soc. Lond., 5 (866) [3] R.L. Graham, A.J. Hoffman, H. Hosoya, On the distance matrix of a directed graph, J. Graph Theory (977) [4] R.L. Graham, L. Lovasz, Distance matrix polynomials of trees, Adv. Math. 9 () (978) [5] R.L. Graham, H.O. Pollak, On the addressing problem for loop switching, Bell System Tech. J. 50 (97) [6] S. Hirschman, V. Reiner, Note on the Pfaffian Matrix-Tree Theorem, Graphs Combin. 0 () (004) [7] C.D. Meyer, Matrix Analysis and Applied Linear Algebra, SIAM, 000. [8] J.R. Stembridge, Non intersecting paths, Pfaffians and plane partitions, Adv. Math. 83 (990) [9] W. Yan, Y.-N. Yeh, A simple proof of Graham and Pollak s theorem, J. Combin. Theory, Ser. A 3 (5) (006) [0] D. Zeilberger, Dodgson s determinant-evaluation rule proved by two timing men and women, Electron. J. Combin. 4 () (997) R.