Resistance matrix and q-laplacian of a unicyclic graph

Transcription

1 Resistace matrix ad q-laplacia of a uicyclic graph R. B. Bapat Idia Statistical Istitute New Delhi, , Idia rbb@isid.ac.i Abstract: The resistace distace betwee two vertices of a graph ca be defied as the effective resistace betwee the two vertices, whe the graph is viewed as a electrical etwork with each edge carryig uit resistace. The cocept has several differet motivatios. The resistace matrix of a graph is a matrix with its (i, j)-etry beig the resistace distace betwee vertices i ad j. We obtai a explicit formula for the determiat of the resistace matrix of a uicyclic graph. Some properties of a q-aalogue of the Laplacia are also studied, with special attetio to the limitig behaviour as q approaches 1. A expressio for the iverse of the q-laplacia of a uicyclic graph is derived. 1 Itroductio ad Prelimiaries The classical defiitio of distace betwee two vertices i a graph is the legth of a shortest path betwee the two vertices. This defiitio is well-kow ad the cocept is widely studied. However, it does ot capture some features of distace, such as the degree of commuicatio betwee the vertices. For example, if there is a multitude of paths betwee two vertices, ituitively the two vertices should be thought of as havig shorter distace. The cocept of resistace distace, itroduced by Klei ad Radić [8], arises aturally from several differet cosideratios ad is also mathematically more attractive tha the classical distace. For more backgroud iformatio about resistace distace we refer to [2, 5, 8, 14]. It may be remarked that i the case of a tree, the cocepts of classical 1

2 distace ad resistace distace coicide. This perhaps explais some very attractive properties of distaces i trees which do ot carry over to arbitrary graphs. However, if oe uses the otio of resistace distace, the some of these attractive properties do have aalogs for arbitrary graphs. We cosider oly simple graphs, i.e., those with o loops or parallel edges. The results that we obtai ca easily be exteded to weighted graphs, though we cosider oly uweighted graphs for simplicity. Now we itroduce some otatio. Let G = (V, E) be a graph with vertices, labeled {1, 2,..., }. The Laplacia matrix L of G is defied as follows. For i j, the (i, j)-etry of L is zero, if vertices i ad j are ot adjacet, while it is 1, if i ad j are adjacet. The (i, i)-etry of L is defied to make the i-th row-sum equal to zero, i = 1, 2,...,. Thus L is a sigular matrix. For basic properties of the Laplacia matrix, see [1,10]. Let G be a coected graph with vertex set {1, 2,..., }. The resistace distace betwee two vertices ca be defied i a umber of differet, equivalet ways. We give two defiitios here. Let L be the Laplacia of G. Let i, j be vertices of G. The resistace distace r ij betwee i ad j is zero if i = j, ad if i j, the r ij = detl(i, j; i, j), detl(i, i) where L(i, i) is the submatrix obtaied by deletig row i, colum i, of L; while L(i, j; i, j) is the submatrix obtaied by deletig rows i, j ad colums i, j of L. As usual, det deotes determiat. Let χ(g) deote the complexity, that is, the umber of spaig trees of G. We remark that by the well-kow Matrix-Tree theorem [13], detl(i, i) equals χ(g) for i = 1, 2,...,. The secod defiitio is i terms of electrical etworks. Thik of G as a electrical etwork i which a uit resistace is placed alog each edge. Curret is allowed to eter the etwork oly at vertex i ad leave the etwork oly at vertex j. The the resistace distace betwee i ad j is the effective resistace betwee i ad j. If there is a uique (ij)-path i G, the it is clear from the secod defiitio that the resistace distace betwee i ad j equals the legth (that is, the 2

3 umber of edges) of the path. This explais why the resistace distace ad the classical distace coicide whe the graph is a tree. We ow state some kow results which will be required. The proofs are omitted ad ca be foud i [2]. Deote the vector of all oes by 1, the size of which will be clear from the cotext. The matrix of all oes will be deoted J. Let G be a coected graph with vertex set {1, 2,..., }, ad let L be the Laplacia of G. The L is sigular, has rak 1, ad ay vector i the ull space of L is a scalar multiple of 1. The matrix L+ 1 J is osigular. We set X = (L + 1 J ) 1. (1) The resistace matrix R of G is the matrix with its (i, j)-etry equal to 0 if i = j, ad r ij, the resistace distace betwee i ad j, otherwise. Let τ be a vector of order 1 with its compoets defied by τ i = 2 j:j i where j i deotes that j is adjacet to i. r ij, i = 1, 2,...,, (2) Theorem 1 Let G be a coected graph with vertex set {1, 2,..., }, let L be the Laplacia of G, let R be the resistace matrix of G, ad let X = (x ij ) ad τ be defied as i (1),(2). Let X be the diagoal matrix with x 11, x 22,..., x alog the diagoal. The the followig assertios are true: (i) r ij = x ii + x jj 2x ij, i, j = 1, 2,..., (ii) R = XJ + J X 2X (iii) τ = L X (iv) 1 τ = 2 (v) R is osigular ad R 1 = 1 2 L + 1 τ Rτ ττ (vi) detr = ( 1) τ Rτ χ(g) 3

4 For a graph G with vertices, D will deote the (classical) distace matrix of G; thus the (i, j)-etry of D is 0, if i = j ad it is the legth of the shortest path betwee i ad j, if i j. Whe the graph is a tree, oe gets the followig cosequece of Theorem 1. Assertios (iii) ad (iv) are well-kow results due to Graham ad Pollack [7] ad Graham ad Lovász [6] respectively. Theorem 2 Let G be a tree with vertex set {1, 2,..., }, let L be the Laplacia of G, let D be the distace matrix of G, ad let τ be defied as i (2). The the followig assertios are true: (i) τ i = 2 δ i, where δ i is the degree of vertex i, i = 1, 2,..., (ii) τ Dτ = 2( 1) (iii) D is osigular ad D 1 = 1 2 L + 1 2( 1) ττ (vi) detd = ( 1) 1 ( 1)2 2 2 Determiat of the resistace matrix of a uicyclic graph We ow cosider uicyclic graphs. Recall that a graph is uicyclic if it is coected ad has a uique cycle. The resistace distace betwee two vertices i a uicyclic graph G is particularly easy to determie. Clearly, if there is a uique path betwee vertices i ad j of G, the the resistace distace r ij betwee i ad j is the legth of the path betwee i ad j. Suppose a path betwee vertices i ad j meets the cycle i G i at least two vertices. Let u ad v be vertices o the cycle such that there is a uique path betwee i ad u, ad a uique path betwee j ad v (see Figure). Let the legth of the iu-path be a, that of the jv-path be b, ad suppose the two paths betwee u ad v have legths c ad d, where c + d = k, the legth of the cycle. By the iterpretatio of resistace distace as effective resistace i a electrical etwork, it follows that r ij is the sum of a, b ad the parallel sum of c ad d. Thus 4

5 i u v j r ij = a + b = a + b + cd c + d. (3) c d Sice the resistace matrix R of a uicyclic graph has simple structure, it is atural to seek a precise formula for the determiat of R, which we ow proceed to obtai. We first cosider the case of a cycle. Let G = C, the cycle o the vertices {1, 2,..., }, 3, (where, of course, i ad i+1 are adjacet, i = 1, 2,..., 1 ad 1 is adjacet to ). If i, j {1, 2,..., }, the there are two paths betwee vertices i ad j, of legths i j ad i j. If R deotes the resistace matrix of G, the i view of the precedig discussio, r ii = 0, i = 1, 2,..., ; r ij = r ji for all i ad j, ad if i < j. r ij = Let τ be defied as i (2). The (j i)( j + i) (4) τ i = 2 r i 1,i r i,i+1, i = 1, 2,..., ; where the subscripts are iterpreted modulo. It follows from (4) that r i 1,i = r i,i+1 = 1, i = 1, 2,...,, ad hece τ i = 2, i = 1, 2,...,. This last observatio also follows from the fact that 1 τ = 2 (see (iv), Theorem 1) ad by symmetry. Theorem 3 Let C be the cycle o the vertices {1, 2,..., }, 3, ad let R be the resistace matrix of C. The 5

6 (i) τ Rτ = 2(2 1) 3 (ii) detr = ( 1) ( 2 1) 3 2. Proof: (i). We first cosider the case whe is eve, say = 2m. It is clear by symmetry that each row-sum of R is the same. Usig the expressio (4) we see that the first row of R has sum r 11 + r r 1 = (r 12 + r r 1m ) + r 1,m+1 + (r 1,m+2 + r 1,m r 1,2m ) 1( 1) 2( 2) (m 1)( m + 1) = 2( ) m( m) + = 2 ( (m 1)) 2 ( (m 1) 2 ) + m( m) = 2 1) (m(m 2 = m 1 2 = (m 1)m(2m 1) m( m) ) + 6 (2m 2m 1 ) + m2 3 2m Thus each row-sum of R is 2 1. Sice τ 6 i = 2, i = 1, 2,...,, it follows that τ Rτ = i=1 j=1 r ij τ i τ j = = 2(2 1). 3 The proof whe is odd is similar ad we omit the details. (ii). By (vi), Theorem 1, detr = ( 1) τ Rτ. Sice τ χ(c ) Rτ = 2(2 1) 3 by the first part ad sice χ(c ) =, the result follows. We give a example. The resistace matrix of C 5 is

7 Accordig to Theorem 3, (ii), the determiat of the matrix is 23 (5 2 1) = This matrix is a symmetric Toeplitz matrix ad a formula for its determiat may be of idepedet iterest. The followig is the mai result of this sectio. Theorem 4 Let G be a uicyclic graph with vertices {1, 2,..., }, suppose the uique cycle C k of G has legth k, ad that it is formed by the vertices {1, 2,..., k}; k 3. Let R be the resistace matrix of G. The detr = ( 1) k 2k2 1 3k 2. (5) Proof: If k = 3, the G is C ad the result follows by Theorem 3, (ii). So let > k 3. We prove the result by iductio o. So suppose that the result is true for a graph with 1 vertices. Sice > k, the G has at least oe pedat vertex. We assume, without loss of geerality, that vertex is pedat ad that it is adjacet to vertex 1. Partitio R as R 1 z, z 0 where R 1 is of order 1 ad z is ( 1) 1. If 1 i, j 1, the a (ij)-path does ot pass through ad therefore R 1 is i fact the resistace matrix of G \ {}. Perform the followig row ad colum operatios o R. From row, subtract row 1 ad from colum, subtract colum 1. For ay 1 i 1, the resistace distace betwee i ad is 1 plus the resistace distace betwee i ad 1. Therefore after the row ad colum operatios, the resultig matrix, which has the same determiat as R, is R By Theorem 1, (v), R 1 is osigular. Usig the well-kow Schur formula for the determiat, detr = (detr 1 )( 2 1 R 1 11). (6) 7

8 Let L be the Laplacia of G \ {} ad let τ be defied for G \ {} as i (2). By Theorem 1, (v), R 1 1 = 1 2 L + 1 τ R 1 τ τ τ. (7) Sice L has row-sums zero, (7) ad Theorem 1, (iv) give 1 R 1 11 = It follows from (6) ad (8) that detr = (detr 1 )( 2 Agai, (9) ad Theorem 1, (vi) lead to By iductio assumptio, 4 τ R 1 τ. (8) 4 τ R 1 τ ) = 2detR 1 4 detr 1 τ R 1 τ. (9) detr = 2detR 1 4 ( 1) (10) k detr 1 = ( 1) k( 1) 2k2 1 3k 2. (11) Substitutig (11) i (10) we get ad (5) is proved. detr = ( 1) ( 3k( 1) 2k2 1 3k k ) = ( 1) k 2k2 1 3k 2 We coclude this sectio with the remark that a formula for the determiat of the (classical) distace matrix of a uicyclic graph has bee obtaied i [3]. 3 The q-laplacia Let G be a graph with vertices, labeled {1, 2,..., }. Let A be the adjacecy matrix of G. Thus A is a matrix with (ij)-etry equal to 1 if i ad j are adjacet, ad zero if i ad j are ot adjacet. Also, a ii = 0, i = 1, 2,...,. 8

9 Let be the diagoal matrix with its i-th diagoal etry equal to the degree of vertex i, i = 1, 2,...,. The L = A is the Laplacia of G. For a parameter q, the q-laplacia L q of G is defied as L q = I qa + q 2 ( I) = ql + (1 q 2 )I + q(q 1). (12) The matrix L q has bee called the geeralized Laplacia of G i [9]. The matrix was also itroduced i [4] for the case of a tree, i the cotext of a formula for the iverse of a q-aalogue of the distace matrix. The followig result has bee obtaied by Northshield [11]. Theorem 5 Let G be a graph with vertices ad m edges, ad let L q be the q-laplacia of G. If f(q) = detl q, the f (q) q=1 = 2(m )χ(g). Lemma 6 Let G be a graph with vertices ad m edges, ad let L q be the q-laplacia of G. Suppose G is coected ad ot uicyclic. The lim(1 q)l 1 q L = 0. (13) Proof: Let adjl q be the adjoit of L q, so that L 1 q Now lim (1 q)l 1 q = adjlq detl q. L = lim (1 q) (adjl q)l detl q. (14) Sice lim (1 q)(adjl q )L = 0 ad lim detl q L Hospital s rule ad the the limit i (14) equals = 0, we may apply ad Note that lim Furthermore, by Theorem 5, (1 q) d (adjl dq q)l (adjl q )L d (detl. (15) dq q) lim (1 q) d dq (adjl q)l = 0, (16) lim (adjl q)l = (adjl)l = 0. (17) 9

10 lim d dq (detl q) = 2(m )χ(g), (18) which is ozero, sice G is coected ad ot uicyclic. Substitutig (16), (17) ad (18) i (15) we get (13). We ow obtai a limitig property of the iverse of the q-laplacia, which is motivated by the expressio for resistace distace cotaied i Theorem 1,(i). Theorem 7 Let G be a graph with vertices ad m edges, let L q be the q- Laplacia of G ad let K q = L 1 q. Suppose G is coected ad ot uicyclic. The for i, j = 1, 2,...,, the resistace distace betwee i ad j. lim ((K q) ii + (K q ) jj 2(K q ) ij ) = r ij, (19) Proof: Sice L q L 1 q L = L, therefore, usig the defiitio of L q, After a routie simplificatio, (20) leads to (1 q 2 )L 1 q (I qa + q 2 q 2 I)L 1 q L = L. (20) L + (q 2 q) L 1 q L + qll 1 q L = L. (21) By Lemma 6 we see that the first two terms o the left had side i (21) approach 0 as q approaches 1. Therefore lim LL 1 q L = lim LK q L = L. (22) We ow itroduce some otatio. Fix vertices i j of G. Let e ij be the 1 vector with i-th coordiate 1, j-th coordiate 1 ad zeros elsewhere. Sice 1 e ij = 0, e ij is i the colum space of L ad hece there exists a vector w ij such that Lw ij = e ij. Let X be defied as i (1). The L1 = 0 implies (L + 1 J )1 = 1 ad hece X1 = 1. Therefore it easily follows that LXL = L. Hece e ijxe ij = w ijlxlw ij = w ijlw ij. (23) 10

11 Now lim ((K q) ii + (K q ) jj 2(K q ) ij ) = lim e ijk q e ij ad the proof is complete. = lim w ijlk q Lw ij = w ijlw ij by (22) = e ijxe ij = x ii + x jj 2x ij = r ij, by Theorem 1, (i), 4 Iverse of the q-laplacia of a uicyclic graph A walk without backtrackig is a walk i which ay two cosecutive edges are distict. Let G be a graph with the vertices {1, 2,..., }. For m 1, let A m be the matrix whose (i, j)-etry is the umber of walks i G of legth m with o backtrackig from i to j, i, j = 1,...,. We also set A 0 = I. Note that A 1 = A, the adjacecy matrix of G. The followig idetity has bee proved i [12],p.139. Lemma 8 Let G be a graph with the vertices {1, 2,..., }, ad let L q be the q-laplacia of G. The ( A m q m )L q = (1 q 2 )I. (24) m=0 For a tree or a uicyclic graph, matrices A m ca be described explicitly ad the Lemma 8 may be employed to get a expressio for L 1 q. First cosider the case of a tree. As usual, we deote the distace betwee vertices i ad j as r ij. Clearly, (A m ) ii = 0 for m 1 ad (A 0 ) ii = 1, i = 1, 2,...,. If i, j are distict vertices of the tree with d = r ij, the (A d ) ij = 1, whereas (A m ) ij = 0 for m d. It follows from (24) that (L 1 q ) ii = 1 ad (L 1 1 q2 q ) ij = qd, i, j = 1, 2,...,. (25) 1 q2 11

12 The matrix (q r ij ) has bee termed the expoetial distace matrix of a tree i [4], where its relatio with L 1 q is proved i a differet way. We retur to uicyclic graphs. For ay vertex i of a uicylic graph G, deote by α i the distace from α i to the cycle i G. Thus α i is the least distace from i to a vertex i the cycle. Theorem 9 Let G be a uicyclic graph with vertices {1, 2,..., } ad suppose the uique cycle C k of G has legth k, k 3. Let α i deote the distace of i from the cycle, i = 1, 2,...,. Let L q be the q-laplacia of G. Let d(i, j) deote the classical distace betwee i ad j. The for i, j = 1, 2,..., ; if a ij-path does ot meet the cycle ad if a ij-path meets the cycle. (1 q 2 )(L 1 q ) ij = q d(i,j) + 2qα i+α j +k 1 q k, (26) (1 q 2 )(Lq 1 ) ij = qd(i,j) + q 2(α i+α j )+k d(i,j), (27) 1 q k Proof: Let i, j be vertices of G ad suppose a ij-path does ot meet the cycle. There is a ij-path, which is also a ij-walk with o backtrackig, of legth d(i, j). There are two ij-walks with o backtrackig of legth α i +α j +k, two ij-walks with o backtrackig of legth α i + α j + 2k, ad, i geeral, two ij-walks with o backtrackig of legth α i + α j + sk, s 1. Therefore, by (24), it follows that (1 q 2 )(L 1 q ) ij = q d(i,j) + 2 q α i+α j +sk = q d(i,j) + 2qα i+α j +k, (28) s=1 1 q k ad (26) is proved. It may be remarked that whe i = j, (26) is applicable (whether or ot i is o the cycle) ad we have The proof of (27) is similar. (1 q 2 )(L 1 q ) ii = 1 + 2q2α i+k 1 q. (29) k We are ow i a positio to complete the case of uicyclic graphs which was left out i Theorem 7. Iterestigly, the coclusio is differet i case of uicyclic graphs. 12

13 Theorem 10 Let G be a uicyclic graph with vertices {1, 2,..., } ad suppose the uique cycle C k of G has legth k, k 3. Let α i deote the distace of i from the cycle, i = 1, 2,...,. Let L q be the q-laplacia of G ad let K q = L 1 q. The for i, j = 1, 2,...,, lim ((K q) ii + (K q ) jj 2(K q ) ij ) = r ij + (α i α j ) 2, (30) k where r ij is the resistace distace betwee i ad j. Proof: Let i, j be vertices of G ad suppose a ij-path does ot meet the cycle. By Theorem 9, (1 q 2 )((K q ) ii +(K q ) jj 2(K q ) ij ) = 2+ 2q2α i+k 1 q k + 2q2αj+k 1 q k 2(qd(i,j) + 2qαi+αj+k 1 q k ). The result follows by computig the limit of (K q ) ii + (K q ) jj 2(K q ) ij as q 1 by L Hospital s rule. We omit the details. The proof follows similarly whe a ij-path meets the cycle. 13

14 REFERENCES 1. R. B. Bapat, The Laplacia matrix of a graph, The Mathematics Studet, 65 (1996) R. B. Bapat, Resistace distace i graphs, The Mathematics Studet, 68 (1999) R. B. Bapat, S. Kirklad ad M. Neuma, O distace matrices ad Laplacias, Liear Algebra Appl., 401 (2005) R. B. Bapat, A. K. Lal ad Sukata Pati, A q-aalogue of the distace matrix of a tree, Liear Algebra Appl., 416 (2006) P. G. Doyle ad J. L. Sell, Radom Walks ad Electrical Networks, Math. Assoc. Am. Washigto, R. L. Graham ad L. Lovász, Distace matrix polyomials of trees, Adv. i Math., 29(1) (1978) R. L. Graham ad H. O. Pollack, O the addressig problem for loop switchig, Bell System Tech. J., 50 (1971) D. J. Klei ad M. Radić, Resistace distace, Joural of Mathematical Chemistry, 12 (1993) Ji Ho Kwak ad Iwao Sato, Zeta fuctios of lie, middle, total graphs of a graph ad their coverigs, Liear Algebra Appl., 418 (2006) R. Merris, Laplacia matrices of graphs: a survey, Liear Algebra Appl., 197,198 (1994) Sam Northshield, A ote o the zeta fuctio of a graph, Joural of Combiatorial Theory, Series B, 74 (1998) H. M. Stark ad A. A. Terras, Zeta fuctios of fiite graphs ad coverigs, Advaces i Math., 121 (1996)

15 13. D. B. West, Itroductio to Graph Theory, 2d Editio, Pretice-Hall Ic., Weju Xiao ad Iva Gutma, O resistace matrices, MATCH Commu. Math. Comput. Chem., 49 (2003)