Linear Algebra and its Applications

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1 Lnear Algebra and ts Applcatons 438 (2013) Contents lsts avalable at ScVerse ScenceDrect Lnear Algebra and ts Applcatons journal homepage: wwwelsevercom/locate/laa A matrx sequence {Ɣ(A m )} m=1 mght converge even f the matrx A s not prmtve Woongbae Park a,borampark b,,1, Suh-Ryung Km c,2 a Department of Mathematcal Scence, Seoul Natonal Unversty, Seoul , South Korea b DIMACS, Rutgers Unversty, Pscataway, NJ 08854, Unted States c Department of Mathematcs Educaton, Seoul Natonal Unversty, Seoul , South Korea ARTICLE INFO Artcle hstory: Receved 29 May 2012 Accepted 15 October 2012 Avalableonlne1December2012 Submtted by RA Bruald AMS classfcaton: 05C20 05C50 Keywords: Irreducble Boolean (0, 1)-matrx Powers of Boolean (0, 1)-matrces Competton graph Graph sequence Powers of dgraphs ABSTRACT It s well-known that, for an rreducble Boolean (0, 1)-matrxA, the matrx sequence {A m } m=1 converges f and only f A s prmtve In ths paper, we ntroduce an operaton Ɣ on the set of Boolean (0, 1)- matrces such that a matrx sequence {Ɣ(A m )} m=1 mght converge even f the matrx A s not prmtve Gven a Boolean (0, 1)-matrx A, we defne a matrx Ɣ(A) so that the (, j)-entry of Ɣ(A) equals 0 f for = j, the nner product of the th row and jth row of A s 0 and equals 1 otherwse The am of ths paper s to study the convergence of {Ɣ(A m )} m=1 for a Boolean (0, 1)-matrx A whose dgraph has at most two strong components We show that {Ɣ(A m )} m=1 converges to a very specal type of matrx as m ncreases f A s an rreducble Boolean matrx Furthermore, we completely characterze a Boolean (0, 1)-matrxA whose dgraph has exactly two strongly connected components and for whch {Ɣ(A m )} m=1 converges, and fnd the lmt of {Ɣ(Am )} m=1 n terms of ts dgraph when t converges We derve these results n terms of the competton graph of the dgraph of A 2012 Elsever Inc All rghts reserved 1 Introducton The focus of ths paper s a problem about the convergence of a certan sequence of Boolean (0, 1)-matrces or equvalently the convergence of the m-step competton graphs of certan dgraphs Correspondng author E-mal addresses: boramp@dmacsrutgersedu, borampark22@gmalcom (B Park) 1 Ths work was supported by Natonal Research Foundaton of Korea Grant funded by the Korean Government (NRF C00004) 2 Ths work was supported by the Natonal Research Foundaton of Korea (NRF) Grant funded by the Korea Government (MEST) (No ) /$ - see front matter 2012 Elsever Inc All rghts reserved

2 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) Fg 1NotethatA, A 2, A 3 are all dstnct and A 4 = A Thus {A m } m=1 does not converge However Ɣ(Am ) = A for each postve nteger m Throughout the paper, we wll state defntons, facts, and theorems (where approprate) n terms of both Boolean (0, 1)-matrces and competton graphs For the two-element Boolean algebra B ={0, 1}, B n denotes the set of all n n matrces over B Under the Boolean operatons, we can defne matrx addton and multplcaton n B n Gvenamatrx A n B n, we defne a matrx Ɣ(A) = (γ j ) B n by 0f = j; γ j = 0f = j and the nner product of row and row j of A s 0; 1f = j and the nner product of row and row j of A s not 0 As a matter of fact, for a matrx A B n, Ɣ(A) s the adjacency matrx of the competton graph of the dgraph of A GvenamatrxA n B n, there exsts a unque dgraph whose adjacency matrx s A We call such a dgraph the dgraph of A and denote t by D(A) Gven a dgraph D, thecompetton graph C(D) of D has the same vertex set as D and has an edge between vertces u and v f and only f there exsts a common prey of u and v n DIf(u, v) s an arc of adgraphd, then we call v a prey of u (n D) and call u a predator of v (n D) A graph G s called the row graph of a matrx M f the rows of M are the vertces of G, and two vertces are adjacent n G f and only f ther correspondng rows have a nonzero entry n the same column of M Ths noton was studed by Greenberg et al [6] As noted n [6], the competton graph of a dgraph D s the row graph of ts adjacency matrx Thus t can easly be checked that the adjacency matrx of the competton graph of adgraphd s Ɣ(A) where A s the adjacency matrx of D The noton of competton graph s due to Cohen [5] and has arsen from ecology Competton graphs also have applcatons n codng, rado transmsson, and modelng of complex economc systems (See [13,14] for a summary of these applcatons) The greatest common dvsor of all lengths of drected cycles n a nontrval strongly connected dgraph D s called the ndex of mprmtvty of D A dgraph D s sad to be prmtve f D s strongly connected and has the ndex of mprmtvty 1 Let A be a matrx n B n IfD(A) s strongly connected, then we say A s rreducble We call the ndex of mprmtvty of D(A) the ndex of mprmtvty of A, when A s rreducble If D(A) s prmtve, then we say that A s prmtve Forundefnedtermsnthe followng, the reader s referred to [2] It s well-known that for an rreducble matrx A n B n, the matrx sequence {A m } m=1 converges f and only f A s prmtve Yet, a matrx sequence {Ɣ(A m )} m=1 mght converge even f the matrx A s not prmtve For example, the mth power of the matrx A gven n Fg 1 does not converge as m ncreases snce t s not prmtve However, the {Ɣ(A m )} m=1 converges to A snce Ɣ(A m ) = A for any postve nteger m In ths paper, we study the convergence of {Ɣ(A m )} m=1 for a matrx A B n whose dgraph has at most two strong components We show that {Ɣ(A m )} m=1 converges as m ncreases for any rreducble Boolean matrx A and ts lmt s a block dagonal matrx each of whose blocks conssts of all 1s up to conjugaton by smultaneous permutaton of rows and columns From now on, we call such a matrx J block dagonal (for short JBD) matrx (where J means a matrx wth all 1s) Furthermore, we completely characterze a matrx A B n whose dgraph has exactly two strongly connected components and for whch {Ɣ(A m )} m=1 converges, and fnd the lmt of {Ɣ(Am )} m=1 n terms of ts dgraph when t converges We derve these facts n terms of the competton graph of the dgraph of A

3 2308 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) Gven a dgraph D and a postve nteger m, avertexy s an m-step prey of a vertex x f and only f there exsts a drected walk from x to y of length m GvenadgraphD and a postve nteger m, the dgraph D m has the vertex set same as D and has an arc (u, v) f and only f v s an m-step prey of u It s well-known that a dgraph D s prmtve f and only f D m equals the dgraph whch has all possble arcs for any m N for some postve nteger N (we call the smallest such nteger N the exponent of D) Motvated by ths, we say that a graph sequence {G n } n=1 (resp dgraph sequence) converges f there exsts an nteger N such that G n s equal to G N for any n N In ths case, we call the graph G N the lmt of the graph sequence (resp dgraph sequence) Then the goals of ths paper proposed above are translated nto competton graph verson as follows We show that {C(D m )} m=1 converges to a graph wth only complete components as m ncreases f D s strongly connected, completely characterze a dgraph D wth exactly two strong components for whch {C(D m )} m=1 converges, and fnd the lmt of {C(D m )} m=1 when {C(Dm )} m=1 converges Gven a postve nteger m, them-step competton graph of a dgraph D, denotedbyc m (D), has thesamevertexsetasd and has an edge between vertces u and v f and only f there exsts an m- step common prey of u and v The noton of m-step competton graph s ntroduced by Cho et al [4] and one of the mportant varatons (see the survey artcles by Km [10] and Lundgren [12] forthe varatons whch have been defned and studed by many authors snce Cohen ntroduced the noton of competton) Snce ts ntroducton, t has been extensvely studed (see for example [1,3,7 9,11,16]) Cho and Km [3] showed that for any dgraph D and a postve nteger m, C m (D) = C(D m ) Thus the lmt of the graph sequence {C(D m )} m=1, f t exsts, s the same as that of the graph sequence {Cm (D)} m=1 Consequently studyng the graph sequence {C(D m )} m=1 s actually studyng the sequence of m-step competton graphs of D 2 {Ɣ(A m )} m=1 for an rreducble matrx A B n In ths secton, we study the convergence and the lmt graph of{ɣ(a m )} m=1 when A s an rreducble matrx n A B n As mentoned prevously, Ɣ(A m ) corresponds to the competton graph of D m where D s the dgraph of A for a postve nteger m We start wth the followng observaton Theorem 21 If a dgraph D s trval or each vertex of D has an out-neghbor, then {C(D m )} m=1 converges Proof The proposton mmedately holds for a trval dgraph Let D be a nontrval dgraph suchthat each vertex has an out-neghbor To show that E(C(D m )) E(C(D m+1 )) for any nteger m, takean edge uv n E(C(D m )) for some postve nteger m Then there exsts a vertex z n D such that (u, z) and (v, z) are arcs of D m for some vertex z InD, z s a common m-step prey of u and v By the assumpton on D,thereexstsavertexx such that (z, x) A(D)Thenx s a common (m + 1)-step prey of u and v n D, that s, x s a common prey of u and v n D m+1 Therefore u and v are adjacent n C(D m+1 )Thuswe have shown that E(C(D m )) E(C(D m+1 )) for any nteger m Snce the competton graph of a dgraph s defned to be smple, E(C(D m )) E(K n ) for each m where n = V(D) Therefore, t can easly be checked that there exsts an nteger N such that for any n N, C(D n ) = C(D N ), whch mples that {C(D m )} m=1 converges Theorem 21 s translated nto the matrx verson as follows: Corollary 22 If a Boolean (0, 1)-matrx A has order 1 or has at least one 1 n each row, then {Ɣ(A m )} m=1 converges It s known that f κ s the ndex of mprmtvty of a dgraph D, thend has an ordered partton {U 1, U 2,,U κ } of V(D) such that U κ+1 = U 1 and each arc of D ssues from U j and enters U j+1 for some j = 1, 2,, κ ThesetsU 1, U 2,,U κ are called the sets of mprmtvty of D

4 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) If D s a trval dgraph, then the ndex of mprmtvty of D s undefned Gven a strongly connected dgraph D,wedefneκ(D) as 1 f D s trval and as the ndex of mprmtvty of D otherwse In addton, f D s a trval dgraph, then we denote the vertex set by U 1 and call t the sets of mprmtvty of D If a dgraph D s strongly connected, then the graph sequence{c(d m )} m=1 converges by Theorem 21 Then t s natural to ask: What s the lmt of the sequence? In the rest of ths secton, we answer the queston by showng that the lmt of {C(D m )} m=1 s the unon of exactly κ(d) complete components Theorem 23 If D s strongly connected, then the lmt of {C(D m )} m=1 s the dsjont unon of complete graphs whose vertex sets are the sets of mprmtvty of D The above theorem mmedately mples the followng corollary Corollary 24 If A s an rreducble (0, 1)-Boolean matrx, then the lmt of {Ɣ(A m )} m=1 s transformable nto a block dagonal matrx by smultaneous permutatons of ther lnes n whch each block s n the form of J where J s a matrx wth all elements 1 and runs from 1 to the ndex of mprmtvty of A The followng s a well-known result related to the ndex of mprmtvty of a dgraph Theorem 25 [2, Theorem 345] Let D be a nontrval strongly connected dgraph of order n wth the ndex of mprmtvty κ, and A be the adjacency matrx of D Then there exsts a permutaton matrx P of order n such that A 1 O O PA l O A P T 2 O = O O A r where r = gcd(κ, l) and each of A 1,A 2,,A r s an rreducble matrx wth the ndex of mprmtvty l r If a matrx A n B n s prmtve, then there exsts a postve nteger m such that the mth power A m has only postve entres, and such smallest nteger m s called the exponent of A, whch s denoted by exp(a) Suppose that D s a nontrval strongly connected dgraph Let A be the adjacency matrx of D and κ(d) = κ ByTheorem25, there exsts a permutaton matrx P such that A 1 O O PA κ O A P T 2 O =, O O A κ (1) where A s prmtve for any 1 κ Let M = κ max{exp(a ) 1 κ} (2) For smplcty, let E = max{exp(a ) 1 κ}takeapostventegerssncepa sm P T = PA sκe P T = (PA κ P T ) se,

5 2310 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) A se 1 O O O A PA sm P T se = 2 O O O A se κ (3) by (1) Snce se exp(a ),anyentryofa se s postve for each 1 κ Therefore D(A se ) s a dgraph wth all possble arcs for each 1 κ, and so D(A sm ) s a dsjont unon of dgraphs wth all possble arcs, D(A se 1 ), D(AsE 2 ),, D(AsE κ ) Now we obtan the followng lemma: Lemma 26 Let D be a strongly connected dgraph Then there exsts a postve nteger M such that C(D sm ) has exactly κ(d) components all of whch are complete for each postve nteger s Proof Let κ = κ(d) IfD s trval, then κ = 1 and C(D m ) s trval wth the sets of mprmtvty of D as the vertex for any postve nteger m Thus the lemma mmedately holds for a trval dgraph Suppose that D s not trval and A s the adjacency matrx of D Thenby(3), for each postve nteger s, D sm s the dsjont unon of dgraphs wth all possble arcs for M defned n (2) Thus C(D sm ) s the dsjont unon of complete graphs whose vertex sets are V(D 1 ), V(D 2 ),, V(D κ ), respectvely As V(D 1 ), V(D 2 ),, V(D κ ) are the set of mprmtvty of D, we complete the proof Now we are ready to prove Theorem 23: Proof of Theorem 23 By Theorem 21, {C(D m )} m=1 converges By Lemma 26, there exsts a postve nteger M such that C(D sm ) has exactly κ(d) components all of whch are complete for each postve nteger s Thus a graph wth exactly κ(d) components all of whch are complete s the lmt of a subsequence {C(D sm )} s=1 of {C(Dm )} m=1 and must be the lmt of {C(Dm )} m=1 3 {Ɣ(A m )} m=1 for a matrx A B n whose dgraph has exactly two strong components In ths secton, we study the convergence of {Ɣ(A m )} m=1 for a matrx A B n such that PAP T = A 1 F O A 2 (4) for a permutaton matrx P of order n, amatrxf, rreducble matrces A 1 and A 2 Agan,wedoso by studyng the convergence of {C(D m )} m=1 for the dgraph D of A whch has exactly two strong components If a dgraph D has two strong components and ts underlyng graph s dsconnected, then{c(d m )} m=1 converges and the lmt s the dsjont unon of complete graphs as shown n Secton 2 Thus, throughout ths secton, we only consder a weakly connected dgraph whose underlyng graph s connected Throughout ths secton, for a dgraph D wth exactly two strong components, we denote the components by D 1 and D 2 so that there s no arc from a vertex n D 2 toavertexnd 1 For = 1, 2 and for the postve nteger M defned n (2), C(D sm ) has exactly κ(d ) components all of whch are complete for each postve nteger s We denote the sets of mprmtvty of D by U () 1, U() 2,,U() κ(d ) In the rest of ths secton, we characterze a dgraph D for whch {C(D m )} m=1 converges, and go further to present ts lmt when {C(D m )} m=1 converges If both D 1 and D 2 are trval, then t s clear that C(D m ) s an edgeless graph wth two vertces for any nteger m That s, f both D 1 and D 2 are trval then {C(D m )} m=1 converges and the lmt graph s an edgeless graph wth two vertces Thus, from now on, we only consder a dgraph wth exactly two strong components, at least one of whch s nontrval

6 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) We completely characterze a dgraph D wth two strong components for whch {C(D m )} m=1 converges For a dgraph D and a vertex v of D, N D (v) denotes the set of all n-neghbors of v Lemma 31 Let D be a weakly connected dgraph wth exactly two strong components D 1 and D 2 For any two vertces u U () j and v U () k, the length of a drected (u, v)-walk s congruent to k jmoduloκ(d ) Proof Snce there s no arc from a vertex n D 2 to a vertex n D 1,adrected(u, v)-walk belongs to D SnceD s strongly connected, we may apply one of the known propertes of a strongly connected dgraphwththendexofmprmtvtyκ(d ) to verfy the statement Lemma 32 Let D be a weakly connected dgraph wth exactly two strong components D 1 and D 2 Then there exsts an nteger M such that C(D m ) contans complete graphs whose vertex sets are U (1) 1,,U(1) κ(d 1 ), U (2) 1,,U(2) κ(d 2 ), respectvely, as subgraphs for m M Proof By Theorem 23, there exsts an nteger M such that C(D1 m) and C(Dm 2 ) equal dsjont unon of complete graphs whose vertex sets are U (1) 1,, U(1) κ(d 1 ), and complete graphs whose vertex sets are ) as subgraphs for any U (2) 1,, U(2) postve nteger m, the lemma holds κ(d 2 ), respectvely, for m MSnceC(Dm ) contans C(D m 1 ) and C(Dm 2 We also need the followng lemma: Lemma 33 [2, Lemma 343] Let D be a nontrval strongly connected dgraph, and U 1,U 2,,U κ(d) be the sets of mprmtvty of D Then there exsts a postve nteger N such that f x and y are vertces belongng respectvely to U and U j, then there are drected (x, y)-walks of every length j + tκ(d) wth t N Proposton 34 Let D be a weakly connected dgraph wth exactly two strong components D 1 and D 2 where D 1 s nontrval and D 2 s trval Then {C(D m )} m=1 converges f and only f for the vertex v of { D 2, U (1) N D (v) = } = 1 or κ(d1 ) Moreover, the lmt of {C(D m )} m=1 s a graph consstng of complete components f t converges Proof Snce D 1 s nontrval, by Lemma 33, there exsts a postve nteger N for whch the followng holds: If x and y are vertces belongng respectvely to U (1) and U (1) j, then there are drected (x, y)-walks of every length j + tκ(d 1 ) wth t N ( ) We show the f part of the proposton statement By Lemma 32, there exsts an nteger M such that C(D m ) contans the complete subgraphs whose vertex sets U (1) 1,, U(1) κ(d 1 ), U(2) 1, respectvely, as subgraphs for m M In addton, there s no edge jonng a vertex n D 1 and v n C(D m ) for any postve nteger m { snce v has no out-neghbor Suppose that U (1) N D (v) = { } = 1 Let U (1) N D (v) = } = { } Then for any drected walk from a vertex n U (1) to v, the term rght before v on the sequence belongs to U (1) and so t has length congruent to + 1moduloκ(D 1 )Snce + 1 j + 1 (mod κ(d 1 )) f j (mod κ(d 1 )), two vertces belongng to dstnct sets of mprmtvty cannot have an m-step common prey for any postve nteger m and so there s no edge jonng two vertces n dstnct sets of mprmtvty n C(D m ) for any postve nteger m Thus{C(D m )} m=1 converges to the dsjont unon of complete graphs whose vertex sets are U (1) 1,,U(1) κ(d 1 ), {v}

7 2312 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) { Now suppose that U (1) N D (v) = } = κ(d1 )Letu 1,,u κ(d1 ) be n-neghbors of v n U (1) 1,, U (1) κ(d 1 ), respectvely Take a vertex x of D 1Thenx U (1) j for some j {1, 2,,κ(D 1 )} and, by ( ), there are drected (x, u k )-walks of every length k j + tκ(d 1 ) wth t N for each k = 1,, κ(d 1 )Thusv s an m-step prey of x for each m Nκ(D 1 ) + 1 Snce x s arbtrarly chosen, C(D m ) s the unon of two complete graphs whose vertex sets are V(D 1 ) and {v} for each m Nκ(D 1 ) + 1 and so K V(D1 ) {v} s the lmt of {C(D m )} m=1 { We show the only f part Suppose that U (1) N D (v) = } = t for some t {2,,κ(D1 ) 1} Then there exsts j such that v has n-neghbors n U (1) j and U (1) j+r but no n-neghbor n U(1) j+1 where r {2, 3,,κ(D 1 ) 1}Takeavertexu n U (1) j and a vertex w n U (1) j+r Then, by ( ), v s a(tκ(d 1) + 1)- step common prey of u and w for each t NThusu and w are adjacent n C(D tκ(d 1)+1 ) for each t N However, u and w are not adjacent n C(D tκ(d 1)+2 ) for any postve nteger t N To show t, note that v s the only possble m-step common prey of u and w for any nteger m and so suppose that v s a (tκ(d 1 ) + 2)-step prey of u for some postve nteger t N We wll reach a contradcton By our assumpton, there s a drected (u, v)-walk of length tκ(d 1 ) + 2nD The vertex mmedately followed by v on the walk must belong to U (1) j+1, whch contradcts our assumpton that v does not have an n-neghbor n U (1) j+1 Thusv cannot be a (tκ(d 1) + 2)-steppreyofu for any postve nteger t N Therefore u and w are not adjacent n C(D tκ(d 1)+2 ) for any postve nteger t N Hencewe can conclude that {C(D m )} m=1 does not converge If D 2 s nontrval, then {C(D m )} m=1 converges by Theorem 21 Thus we have completely characterzed a dgraph D wth exactly two strong components for whch {C(D m )} m=1 converges Theorem 35 Let D be a weakly connected dgraph wth exactly two strong components D 1 and D 2 and wthout arc from D 2 to D 1 Then{C(D m )} m=1 converges f and only f D satsfes one of the followng: () D 2 s nontrval { () D 2 s trval and, for the vertex v of D 2, U (1) N D (v) = } = 1 or κ(d1 ) From the proof of Proposton 34 and Theorem 35, we obtan the followng: Corollary 36 LetAbeaBoolean(0, 1)-matrx n the followng form: A 1 F O A 2 where O s a zero matrx, F s a nonzero matrx, and A 1 and A 2 are rreducble Then {Ɣ(A m )} m=1 converges f and only f one of the followng holds: () A 2 has order at least 2 () A 2 has order 1 and there exsts a permutaton matrx P such that A 12 OO O O O A 23 O O O F PAP T = O O O O A κ(d1 ) 2,κ(D 1 ) 1 O O O O A κ(d1 ) 1,κ(D 1 ) O A 2

8 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) Fg 2 A bpartte graph B and an expanson G of B so that the rows contanng nonzero elements of F ntersect exactly one of A 12,A 23,,A κ(d1 ) 1,κ(D 1 ) or each of A 12,A 23,,A κ(d1 ) 1,κ(D 1 ) Moreover, the lmt of {Ɣ(A m )} m=1 s a JBD matrx We now examne the structure of {Ɣ(A m )} m=1 when t converges where A s a Boolean (0, 1)- matrxgvenn(4) where F s a nonzero matrx If A 2 s a trval matrx, then we presented the lmt of {Ɣ(A m )} m=1 n Corollary 36 Thus n the followng, we fnd the lmt when A 2 s nontrval To do so, we fnd the lmt of {C(D m )} m=1 where D s the dgraph of A and need the followng useful notons Gven a bpartte B = (X, Y), we construct a supergraph of B as follows We wrte each edge of B n the arc form (x, y) to make clear that x X and y YThenwereplaceeachvertexv wth a complete graph G v (of any sze) so that G v and G w are vertex-dsjont f v = w, and jon each vertex of G x and each vertex of G y whenever ether (x, y) s an edge of B or there exsts z Y such that (x, z) and (y, z) are edges of B We say that the resultng graph D s an expanson of B (See Fg 2 for an llustraton) Lemma 37 Let G be an expanson of some bpartte graph B = (X, Y) Then G has only complete components f and only f for each vertex x X, the degree of x s at most one n B Proof We show the f part by contradcton Suppose that there exst vertces x, y, z such that xy and xz are edges of G but y s not adjacent to z n G LetG u, G v, and G w be the complete graphs replacng vertces u, v, and w of B contanng x, y, z, respectvely By defnton, u, v, and w are dstnct Snce y and z are not adjacent whle x s adjacent to both y and z, t s true that u X Then, snce B s bpartte, v and w belong to YNow,bydefnton,u s adjacent to v and w and we reach a contradcton To show the only f part, suppose that G has only complete components and there exsts a vertex u X whch has two neghbors v, w n Y Then, by defnton, no vertex of G v s joned to any vertex of G w Takeavertexx G u,avertexy G v and a vertex z G w Then, by defnton, x s adjacent to y and z n G and so x, y, z belong to the same component Snce G has only complete components by our assumpton, y and z are adjacent n G, a contradcton Defnton 38 We take a weakly connected dgraph D wth exactly two strong components D 1 and D 2 where D 2 s nontrval Let I(D) = {(k, l) (x, y) A(D) for some x U (1) k, y U (2) l } Let B D = ( Z κ(d1 ), Z κ(d2 )) be the bpartte graph defned as follows If D1 s nontrval, then B D has an edge (, j) f and only f k p (mod κ(d 1 )) and j l + p (mod κ(d 2 )) for some (k, l) I(D) and some nteger pifd 1 s trval, then B D has an edge (, j) f and only f j l 1 (mod κ(d 2 )) for some (1, l) I(D), whch s obtaned by substtutng p = 1and k(d 1 ) = 1 n the nontrval case We note that when we consder an edge (x, y) of B D, the frst component and the second component are reduced modulo κ(d 1 ) and κ(d 2 ), respectvely Then followng s true Lemma 39 Let D be a weakly connected dgraph wth exactly two strong components D 1 and D 2 where D 2 s nontrval Then (, j) s an edge of B D f and only f there exsts a (u, v)-walk of length 2sκ(D 1 )κ(d 2 ) for some vertces u U (1) and v U (2) j and for some postve nteger s

9 2314 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) Proof For smplcty, we denote κ(d 1 )κ(d 2 ) by λ To show the only f part, suppose that (, j) s an edge of B D By defnton, for some nteger p and for some (k, l) I(D), k p (mod κ(d 1 )), j l + p (mod κ(d 2 )) We defne nonnegatve ntegers p 1 and p 2 as follows There exsts a postve nteger s such that both sλ p 1 and sλ + p are postve ntegers If D 1 s trval, then let p 1 = 0 and p 2 = 2sλ + p (note that p = 1) If D 1 s nontrval, then let p 1 = sλ p 1 and p 2 = sλ + p Snce (k, l) I(D), thereexstvertcesx U (1) such that (x, y) s an arc of D If k and y U (2) l D 1 s trval, then let u = x and then Q := u s a (u, x)-walk of length p 1 = 0 Suppose that D 1 s nontrval Snce D 1 s a nontrval strongly connected dgraph, any vertex n D 1 has an n-neghbor n D 1 and so there s a drected (u, x)-walk Q of length p 1 for some u V(D 1 )Sncex U (1) k,tstrue that u U (1) k p 1 However, k p 1 ( p 1) (sλ p 1) (mod κ(d 1 )) Therefore u U (1) Snce D 2 s nontrval and strongly connected D 2 has a drected (y, v)-walk R of length p 2 for some v V(D 2 ) Therefore y U (2) l+p 2 However, l + p 2 (j p) + (ksλ + p) j (mod κ(d 2 )), where k = 1fD 1 s nontrval and k = 2fD 1 s trval Thus v U (2) j Furthermore Q + (x, y) + R has the length p 1 + p = 2sλ To show the f part, suppose that there exsts a drected (u, v)-walk Q of length 2sλ for some vertces u U (1) and v U (2) j and for some postve nteger sthewalkq contans a unque arc (w, z) such that w V(D 1 ) and z V(D 2 ) snce there s no arc from a vertex n D 2 to a vertex n D 1 Then w U (1) r and z U (2) s for some r {1, 2,,κ(D 1 )} and some s {1, 2,,κ(D 2 )}Bydefnton, (r, s) I(D) Letl 1 and l 2 be the lengths of (u, w)-secton of Q and (z, v)-secton of Q, respectvely Then + l 1 r (mod κ(d 1 )), j l 2 s (mod κ(d 2 )) Snce l 1 + l = 2sλ, the frst congruence relaton s equvalent to + (2sλ l 2 1) r (mod κ(d 1 )) and so l 2 1 r (mod κ(d 1 )) Therefore r + l (mod κ(d 1 )), j s + l 2 (mod κ(d 2 )) By defnton, (, j) s an edge of B D Now we are ready to present the lmt of {Ɣ(A m )} m=1 f t exsts for a matrx A gvenn(4) when F s a nonzero matrx and A 2 has order at least 2, that s, D(A 2 ) s a nontrval strong component of D(A): Theorem 310 Let D be a weakly connected dgraph wth two strong components D 1 and D 2 such that no arc goes from D 2 to D 1,D 2 s nontrval, and {C(D m )} m=1 converges Then the lmt of {C(Dm )} m=1 s an expanson of the bpartte graph B D defned n Defnton 38 Proof Let G be the lmt of {C(D m )} m=1 ByTheorem32, complete graphs whose vertex sets are U(1) 1,, U (1) κ(d 1 ), U(2) 1,, U(2) κ(d 2 ), respectvely, are subgraphs of G

10 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) Snce there s no arc from a vertex n D 2 to a vertex n D 1, for any postve nteger m, anm-step common prey of two vertces of V(D 2 ) s n D 2 and so the unon of complete graphs whose vertex sets are U (2) 1,, U(2) κ(d 2 ), respectvely, s an nduced subgraph of G by Theorem 23 Therefore, to show that G s an expanson of B D, t s suffcent to prove the followng: () x U (1) and y U (2) j are adjacent n G f and only f (, j) s an edge of B D () For dstnct and j, x U (1) and y U (1) j edges of B D for some h Z κ(d2 ) are adjacent n G f and only f (, h) and (j, h) are For smplcty, we denote κ(d 1 )κ(d 2 ) by λ To show (), suppose that x U (1) and y U (2) j are adjacent n G Then there exst an nteger s and a2sλ-step common prey z of x and y n D, whch mples that there exsts an (x, z)-walk Q of length 2sλ n D On the other hand, snce z s a 2sλ-step prey of y, t s true that z V(D 2 ) Furthermore, snce 2sλ s a multple of κ(d 2 ), z U (2) j By Lemma 39, (, j) s an edge of B D Suppose that (, j) s an edge of B D Takevertcesx U (1) and y U (2) j By Lemma 39,thereexstsa (u, v)-walk Q of length 2sλ for some vertces u U (1) and v U (2) j and some postve nteger ssnce any drected (x, u)-walk belongs to D 1, the length of a drected (x, u)-walk s congruent to 0 modulo κ(d 1 ) Then there exsts a drected (x, u)-walk S of length s λ for some nteger s (If D 1 s trval then s = 0) Snce D 2 s nontrval and κ(d 2 ) dvdes λ, by Lemma 33, for some nteger N, thereexsta drected (v, v)-walk T of length Nλ, and a drected (y, v)-walk R of length (2s + s + N)λ SnceSQT s a drected (x, v)-walk of length (2s + s + N)λ, v s a (2s + s + N)λ-step common prey of x and y n D, and so x and y are adjacent n G Hence () holds Now we show that () holds Take dstnct and j n {1,,κ(D 1 )} To prove the f part, suppose that (, h) and (j, h) are edges of B D for some h Z κ(d2 ) By Lemma 39,thereexstadrected(u 1, v 1 )- walk S 1 of length 2s 1 λ and a drected (u 2, v 2 )-walk S 2 of length 2s 2 λ for some postve ntegers s 1 and s 2, and some vertces u 1 U (1), u 2 U (1) j, v 1, v 2 U (2) h Taketwovertcesx U(1) and y U (1) j Note that assumpton = j mples that D 1 s nontrval Therefore both D 1 and D 2 are nontrval, we may apply Lemma 33 to D 1 and D 2, respectvely, to have ntegers N 1 and N 2 satsfyng the followng Snce v 1, v 2 belong to the same set of mprmtvty, there exst a drected (v 1, v 2 )-walk T 1 of length (2s 2 +N 2 )λ and a drected (v 2, v 2 )-walk T 2 of length (2s 1 +N 2 )λthens 1 T 1 s a drected (u 1, v 2 )-walk of length (2s 1 + 2s 2 + N 2 )λ and S 2 T 2 s a drected (u 2, v 2 )-walk of length (2s 2 + 2s 1 + N 2 )λ Take any nteger t N 1 Then, snce x, u 1 belong to the same set of mprmtvty, there exsts a drected (x, u 1 )-walk Q 1 of length tλ For the same reason, there exsts a drected (y, u 2 )-walk Q 2 of length tλ NowQ 1 S 1 T 1 s a drected (x, v 2 )-walk of length (2s 1 + 2s 2 + N 2 + t)λ, and Q 2 S 2 T 2 s a drected (y, v 2 )-walk of length (2s 1 + 2s 2 + N 2 + t)λthusv 2 s a (2s 1 + 2s 2 + N 2 + t)λ-step common prey of x and y, and hence x and y are adjacent n C(D (2s 1+2s 2 +N 2 +t)λ )Sncet s arbtrarly chosen nteger greater than N 1, x and y are adjacent n G To show the only f part, suppose that x U (1) and y U (1) j are adjacent n G Then they have a 2sλstep common prey z n V(D 2 ) for some nteger s, that s, there exst a drected (x, z)-walk and a drected (y, z)-walk of length 2sλ Sncez V(D 2 ), t holds that z U (2) h for some h {1, 2,,κ(D 2 )} By Lemma 39, (, h) and (j, h) are edges of B D We can easly check that Theorem 310 s equvalent to the followng: Corollary 311 Let A B n be a matrx such that for a permutaton matrx P of order n, PAP T = A 1 F O A 2

11 2316 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) where O s a zero matrx, O A 12 O O O B 12 O O O O A 23 O O O B 23 O A 1 =, A 2 =, O O O A κ(d1 ) 1,κ(D 1 ) O O O B κ(d2 ) 1,κ(D 2 ) A κ(d1 )1 O O O B κ(d2 )1 O O O and A 2 has order at least two, and F s a nonzero matrx, F 11 F 12 F 13 F 1κ(D2 ) F 21 F 22 F 23 F 2κ(D2 ) F = F κ(d1 ) 1,1 F κ(d1 ) 1,2 F κ(d1 ) 1,3 F κ(d1 ) 1,κ(D 2 ) F κ(d1 )1 F κ(d1 )2 F κ(d1 )3 F κ(d1 )κ(d 2 ) Then {Ɣ(A m )} m=1 convergestoamatrxa such that where PA P T = C 1 F F T C 2 J C 12 C 13 C 1κ(D1 ) J O O O C 21 J C 23 C 2κ(D1 ) O J O O C 1 = ; C 2 = ; C κ(d1 ) 1,1 C κ(d1 ) 1,2 C κ(d1 ) 1,3 C κ(d1 ) 1,κ(D 1 ) OOO O C κ(d1 )1 C κ(d1 )2 C κ(d1 )3 J OOO J F = F 11 F 12 F 13 F 1κ(D 2 ) F 21 F 22 F 23 F 2κ(D 2 ) ; F κ(d 1 ) 1,1 F κ(d 1 ) 1,2 F κ(d 1 ) 1,3 F κ(d 1 ) 1,κ(D 2 ) F κ(d 1 )1 F κ(d 1 )2 F κ(d 1 )3 F κ(d 1 )κ(d 2 ) J represents a matrx of an approprate sze wth all the elements 1;C j = C j = JfF = k F jk = J for some k {1,,κ(D 2 )} and C j = C j = Ootherwse;F j = J f one of the followng holds: A 1 has order at least two and F k,l = O for some ntegers k, l satsfyng k + p + 1 (mod κ(d 2 )) and j l + p (mod κ(d 2 )) for some nteger p, A 1 has order one and F 1l = O for some nteger l such that j l 1 (mod κ(d 2 )),

12 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) and F j = Ootherwse Let A be a matrx gven n (4) where F s nonzero and D be the dgraph of A IfD 1 s trval, D 2 s nontrval and B D has at least two edges, then any expanson of B D cannot be the unon of complete subgraphs and so the lmt of {C(D m )} m=1 s not the unon of complete subgraphs by Theorem 310, that s, the lmt of {Ɣ(A m )} m=1 cannot be a JBD matrx If D 2 trval and {C(D m )} m=1 converges, then the lmt of {C(D m )} m=1 s always the unon of complete subgraphs by Proposton 34, that s, the lmt of {Ɣ(A m )} m=1 s a JBD matrx In the followng, we characterze a matrx A gven n (4) for whch the lmt graph of {Ɣ(A m )} m=1 s a JBD matrx when F s a nonzero matrx and both A 1 and A 2 are nontrval, that s, we characterze a dgraph D wth exactly two strong components both of whch are nontrval and for whch the lmt graph of {C(D m )} m=1 has only complete components Theorem 312 Let D be a weakly connected dgraph wth exactly two strong components D 1 and D 2 both of whch are nontrval and wthout arc from D 2 to D 1 Suppose that {C(D m )} m=1 converges to a graph G Then G s the unon of complete subgraphs f and only f κ(d 2 ) dvdes κ(d 1 ) and j j (mod κ(d 2 )) for any (, j), (, j ) I(D) Proof As we have shown n the proof of Theorem 310, G s an expanson of the bpartte graph B D defned n Defnton 38 For convenence, let B D = (X, Y) Suppose that κ(d 2 ) κ(d 1 ) and j j (mod κ(d 2 )) for any (, j), (, j ) I(D)Takeavertexa XIfa has no neghbor n B D,thentsdegree s zero Suppose that a has a neghbor n B D Let(a, b) and (a, c) are edges of B D Then by defnton, for some (, j), (, j ) I(D), for some ntegers l, l, a + l + 1 (mod κ(d 1 )), b j + l (mod κ(d 2 )), a + l + 1 (mod κ(d 1 )), c j + l (mod κ(d 2 )) Snce κ(d 2 ) κ(d 1 ), a + l l + 1 (mod κ(d 2 )), and so l l (mod κ(d 2 )) Therefore b c (j + l) (j + l ) (j j ) + (l l ) (j j ) + ( ) 0 (mod κ(d 2 )) Therefore, the vertex a has only one neghbor n B D Hence, by Lemma 37, G s the unon of complete subgraphs Now suppose that G s the unon of complete subgraphs Then, by Lemma 37, the degree of each vertex n X s at most one n B D Snce we have assumed that the underlyng graph of D s connected at the begnnng of ths secton, B D has an edge and so there exsts a vertex a X such that the degree of a s one Then (a, b) s an edge of B D and so for some (, j) I(D) and some nteger l, a + l + 1 (mod κ(d 1 )), b j + l (mod κ(d 2 )) Snce (, j) I(D),tstruethat( + l + κ(d 1 ) + 1, j + l + κ(d 1 )) s an edge of B D by the defnton of B D Snce + l + κ(d 1 ) + 1 a (mod κ(d 1 )),tholdsthat(a, j + l + κ(d 1 )) s an edge of B D Snce the degree of a s one, b s the unque neghbor of a n B D and so j + l + κ(d 1 ) b (mod κ(d 2 ))Thus j + l + κ(d 1 ) j + l (mod κ(d 2 )) and so κ(d 1 ) 0 (mod κ(d 2 )) Therefore κ(d 2 ) κ(d 1 ) Take (, j), (, j ) I(D) Wthout loss of generalty, we may assume that > Snce(, j) I(D), by the defnton of B D, ( + ( ) + 1, j + ( )) s an edge of B D and so s ( + 1, j + ) In addton, ( + 1, j ) s an edge of B D as (, j ) I(D) Therefore both ( + 1, j + ) and ( + 1, j ) are edges of B D Snce each vertex n X of B D has degree at most one, j + ( ) j (mod κ(d 2 )) Thus j j (mod κ(d 2 )) As a corollary of Theorem 312, we obtan the followng:

13 2318 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) Corollary 313 Let A B n be a matrx such that for a permutaton matrx P of order n, PAP T = A 1 F O A 2 where O A 12 O O O B 12 O O O O A 23 O O O B 23 O A 1 =, A 2 =, O O O A κ(d1 ) 1,κ(D 1 ) O O O B κ(d2 ) 1,κ(D 2 ) A κ(d1 )1 O O O B κ(d2 )1 O O O both A 1 and A 2 have order at least two, O s a zero matrx, and F s a nonzero matrx, F 11 F 12 F 13 F 1κ(D2 ) F 21 F 22 F 23 F 2κ(D2 ) F = F κ(d1 ) 1,1 F κ(d1 ) 1,2 F κ(d1 ) 1,3 F κ(d1 ) 1,κ(D 2 ) F κ(d1 )1 F κ(d1 )2 F κ(d1 )3 F κ(d1 )κ(d 2 ) Suppose that {Ɣ(A m )} m=1 convergestoamatrxa ThenA s a JBD matrx f and only f κ(d 2 ) dvdes κ(d 1 ) and j j (mod κ(d 2 )) whenever F,j = OandF,j = O 4 Concludng remarks In ths paper, we nvestgated the convergence and the lmt of the matrx sequence {Ɣ(A m )} m=1 for a matrx A n B n whose dgraph D has at most two strong components and, among such matrces, characterzed a matrx A for whch the lmt of {Ɣ(A m )} m=1 s a JBD matrx We would lke to see f our results can be generalzed for an arbtrary matrx n B n When a dgraph D has qute many strong components, vertces n the strong component whch has only outgong arcs n the condensaton of D have much more choces for prey and so the characterzaton of ts lmt, f t exsts, appears to be more dffcult We mentoned earler that studyng the matrx sequence {Ɣ(A m )} m=1 for a matrx A n B n s equvalent to studyng the graph sequence {C(D m )} m=1 and that {C(Dm )} m=1 s actually the sequence of m-step competton graphs of D In ths context, we propose to nvestgate the graph sequence obtaned by other varants of competton graph (see [4,9,15]) Acknowledgments We wsh to acknowledge the anonymous referee for nvaluable suggestons leadng to mprovements n the presentaton of the results References [1] E Belmont, A complete characterzaton of paths that are m-step competton graphs, Dscrete Appl Math 159 (2011)

14 W Park et al / Lnear Algebra and ts Applcatons 438 (2013) [2] RA Bruald, HJ Ryser, Combnatoral Matrx Theory, Cambrdge Unversty Press, Cambrdge, 1991 [3] HH Cho, HK Km, Competton ndces of strongly connected dgraphs, Bull Korean Math Soc 48 (2011) [4] HH Cho, S-R Km, Y Nam, The m-step competton graph of a dgraph, Dscrete Appl Math 105 (2000) [5] JE Cohen, Interval graphs and food webs: a fndng and a problem, RAND Corporaton Document PR, Santa Monca, Calforna, 1968 [6] HJ Greenberg, JR Lundgren, JS Maybee, Invertng graphs of rectangular matrces, Dscrete Appl Math 8 (1984) [7] GT Hellelod, Connected trangle-free m-step competton graphs, Dscrete Appl Math 145 (2005) [8] HK Km, Competton ndces of tournaments, Bull Korean Math Soc 45 (2008) [9] W Ho, The m-step, same-step, and any-step competton graphs, Dscrete Appl Math 152 (2005) [10] S-R Km, The competton number and ts varants, n: J Gmbel, JW Kennedy, LV Quntas (Eds), Quo Vads Graph Theory? Ann Dscrete Math, vol 55, North-Holland, Amsterdam, 1993, pp [11] B Park, JY Lee, S-R Km, The m-step competton graphs of doubly partal orders, Appl Math Lett 24 (2011) [12] JR Lundgren, Food webs, competton graphs, competton-common enemy graphs, and nche graphs, n: FS Roberts (Ed), Applcatons of Combnatorcs and Graph Theory n the Bologcal and Socal Scences, IMA Volumes n Mathematcs and ts Applcatons, vol 17, Sprnger-Verlag, New York, 1989, pp [13] A Raychaudhur, FS Roberts, Generalzed competton graphs and ther applcatons, Methods Oper Res 49 (1985) [14] FS Roberts, Competton graphs and phylogeny graphs, n: L Lovasz (Ed), Graph Theory and Combnatoral Bology, Bolya Math Stud, vol 7, J Bolya Math Soc, Budapest, 1999, pp [15] D Scott, The competton-common enemy graph of a dgraph, Dscrete Appl Math 17 (1987) [16] Y Zhao, GJ Chang, Note on the m-step competton numbers of paths and cycles, Dscrete Appl Math 157 (2009)