Bounds for Spectral Radius of Various Matrices Associated With Graphs

Transcription

1 45 5 Vol.45, No AVANCES IN MATHEMATICS (CHINA) Sep., 016 o: /sxjz b Bouns for Spectral Raus of Varous Matrces Assocate Wth Graphs CUI Shuyu 1, TIAN Guxan, (1. Xngzh College, Zhejang Normal Unversty, Jnhua, Zhejang, 31004, P. R. Chna;. College of Mathematcs, Physcs an Informaton Engneerng, Zhejang Normal Unversty, Jnhua, Zhejang, 31004, P. R. Chna) Abstract: In ths paper, usng a postve scale vector, we frst gve some bouns for the spectral raus of a nonnegatve rreucble matrx, whch mprove an generalze some nown results. Applyng these bouns to varous matrces assocate wth a graph G, we obtan some new upper an lower bouns for varous spectral ra of G wth parameter, whch theoretcally mprove an generalze some nown results. Keywors: nonnegatve matrx; spectral raus; ajacency matrx; sgnless Laplacan matrx; stance matrx; stance sgnless Laplacan matrx MR(010) Subject Classfcaton: 05C35; 05C50; 15A18 / CLC number: O157.5 ocument coe: A Artcle I: (016) Introucton We only conser smple graphs whch have no loops an multple eges. Let G = (V,E) be a smple graph wth vertex set V = {v 1,v,,v n } an ege set E. For any two vertces v,v j V, we wrte j f v an v j are ajacent. For any vertex v V, enote the egree of v by. The matrx A(G) = (a j ) s calle the ajacency matrx of G, where a j = 1 f j an 0 otherwse. Let (G) = ag( 1,,, n ) be the agonal matrx of vertex egrees. Then Q(G) = (G)+A(G) s calle the sgnless Laplacan matrx of G. For more revew about the sgnless Laplacan matrx of G, reaers may refer to [5 8] an the references theren. If G s connecte, then the stance matrx of G s the n n matrx (G) = ( j ), where j s the stance between two vertces v an v j n G. For any vertex v V, enote the transmsson of v by,.e., the sum of stances between v an other vertces of G. Let T R(G) = ag( 1,,, n ) be the transmsson agonal matrx. The stance sgnless Laplacan matrx of G s the n n matrx Q(G) = T R(G) + (G). For more revew about the stance sgnless Laplacan matrx of G, see [1]. Let A = (a j ) be an n n nonnegatve matrx. The spectral raus of A, enote by ρ(a), s the largest moulus of an egenvalue of A. Moreover, f A s symmetrc, then ρ(a) s just the largest egenvalue of A. For more revew about nonnegatve matrx, see [, 13]. Receve ate: Founaton tem: Ths wor was supporte n part by NSFC (No , No ) an the Natural Scence Founaton of Zhejang Provnce (No. LY15A010011). E-mal: gxtan@zjnu.cn

2 71 «45 In [3], Chen et al. obtane some upper bouns for the spectral raus of a nonnegatve rreucble matrx wth agonal entres 0 an characterze the equalty cases. Usng the same technque, uan an Zhou [9] got some upper an lower bouns for the spectral raus of a nonnegatve matrx an characterze the equalty cases whenever the matrx s rreucble n the general case. Applyng these bouns to the matrces assocate wth a graph as mentone above, they also obtane some upper an lower bouns on varous spectral ra of graphs. For example, () Let 1 n be the egree sequence of G. Then for 1 n, ρ(a(g)) 1+ ( +1) +4 1 =1 ( ) (see [9, 11]), (1) ρ(q(g)) ( 1 +1) +8 1 =1 ( ) (see [4, 9]); () () Let 1 n be the transmsson sequence of G an be the ameter of G. Then for 1 n, ρ((g)) + ( +) +4 1 =1 ( ) (see [3, 9]), (3) ρ(q(g)) ( 1 +) +8 1 =1 ( ) (see [9]), (4) an ρ((g)) n 1+ ( n +1) +4 n 1 =1 ( n ) (see [9]), (5) ρ(q(g)) 3 n 1+ ( n +1) +8 n 1 =1 ( n ) (see [9]). (6) In ths paper, usng a postve scale vector an ntersectons of ts components, we gve some upper an lower bouns for the spectral raus of a nonnegatve rreucble matrx an characterze the equalty cases. These bouns theoretcally mprove an generalze some nown results n [9]. Applyng these bouns to varous matrces assocate wth a graph, nclung the ajacency matrx, the sgnless Laplacan matrx, the stance matrx an the stance sgnless Laplacan matrx, we obtan some new upper an lower bouns on spectral ra of varous matrces of a graph, whch generalze an mprove some nown results as mentone above. 1 Upper an Lower Bouns for Spectral Raus of a Nonnegatve Matrx In [15], Tan et al. presente some new ncluson ntervals of matrx sngular values. These ntervals are base manly on the use of a postve scale vector an ntersectons of ts components. In ths secton, usng a smlar technque, we gve some upper an lower bouns for the spectral raus of a nonnegatve matrx. These bouns are use to obtan some upper an lower bouns on spectral ra of varous matrces of a graph n Secton.

3 5, : Bouns for Spectral Raus of Varous Matrces Assocate Wth Graphs 713 Theorem 1 Let A = (a j ) be an n n nonnegatve rreucble matrx an c = (c 1,c,, c n ) T be any vector wth postve components. For 1 n, tae M = 1 c n j=1 a j c j, M = max{a }, N = max j { aj c j Assume that M 1 M M n an N > 0. Then for 1 n, ρ(a) M +M N + (M M +N) +4N 1 =1 (M M ). (7) Moreover, the equalty n (7) hols f an only f M 1 = M = = M n or for some t, A satsfes the followng contons: () a = M for 1 t 1, () a lc l c = N for 1 n, 1 l t 1 wth l, () M t = M t+1 = = M n. Proof Let U = ag(c 1,c,,c n ). Obvously, A an B = U 1 AU have the same egenvalues. Thus, applyng [9, Theorem.1] to B, one may obtan the requre result. Corollary 1 [9] Let A = (a j ) be an n n nonnegatve rreucble matrx wth row sums r 1 r r n. Also let M = max 1 n {a } an N = max j {a j }. Assume that N > 0. Then for 1 n, ρ(a) r +M N + (r M +N) +4N 1 =1 (r r ). (8) Moreover, the equalty hols n (8) f an only f r 1 = r = = r n or for some t, A satsfes the followng contons: () a = M for 1 t 1, () a l = N for 1 n, 1 l t 1 wth l, () r t = r t+1 = = r n. Proof Tang c = (1,1,,1) T n Theorem 1, one may obtan the requre result. Corollary [3] Let A = (a j ) be an n n nonnegatve rreucble matrx wth agonal entres 0, ts row sums r 1 r r n. Also let N = max j {a j }. Assume that N > 0. Then for 1 n, ρ(a) r N + (r +N) +4N 1 =1 (r r ). (9) Moreover, the equalty n (9) hols f an only f r 1 = r = = r n or for some t, A satsfes the followng contons: result. () a l = N for 1 n, 1 l t 1 wth l, () r t = r t+1 = = r n. Proof Tang c = (1,1,,1) T an M = 0 n Theorem 1, one may obtan the requre c }.

4 714 «45 Remar 1 From Corollares 1 an, one may see that Theorem 1 mproves an generalzes some results n [3, 9]. On the other han, we always assume that M 1 M M n n Theorem 1 because the matrx B = U 1 AU n the proof of Theorem 1 may satsfy ths conton uner permutaton smlarty. Fnally, Chen et al. [3] clame that the equalty n (9) hols f an only f r 1 = r = = r n or for some n, A satsfes r 1 = r = = r 1 > r = = r n an a l = N for 1 n, 1 l 1 wth l (see [3, Theorem 1.9]). However, t seems to be naccurate. For example, let A = For = 3, applyng the nequalty (9), one gets ρ(a) In fact, by a rect calculaton, one has ρ(a) Hence, the equalty n (9) hols for the matrx A. But r 1 = 1, r = 11, r 3 = r 4 = r 5 = 10. Theorem Let A = (a j ) be an n n nonnegatve rreucble matrx an c = (c 1,c,,. c n ) T be any vector wth postve components. For 1 n, tae M = 1 c n j=1 a j c j, S = mn {a }, T = mn j { aj c j Assume that M 1 M M n. Then ρ(a) M n +S T + (M n S +T) +4T n 1 =1 (M M n ). (10) Moreover, the equalty n (10) hols f an only f ether M 1 = M = = M n, or T > 0 an for some t n, A satsfes the followng contons: () a = S for 1 t 1, () a lc l c = T for 1 n, 1 l t 1 wth l, () M t = M t+1 = = M n. Proof Let U = ag(c 1,c,,c n ). Obvously, A an B = U 1 AU have the same egenvalues. Thus, applyng [9, Theorem.] to B, one may obtan the requre result. Corollary 3 [9] Let A = (a j ) be an n n nonnegatve rreucble matrx wth row sums r 1 r r n. Also let M = mn 1 n {a } an N = mn j {a j }. Then ρ(a) r n +S T + (r n S +T) +4T n 1 =1 (r r n ). (11) Moreover, the equalty n (11) hols f an only f ether r 1 = r = = r n, or T > 0 an for some t n, A satsfes the followng contons: () a = S for 1 t 1, () a l = T for 1 n, 1 l t 1 wth l, c }.

5 5, : Bouns for Spectral Raus of Varous Matrces Assocate Wth Graphs 715 () r t = r t+1 = = r n. Proof Tang c = (1,1,,1) T n Theorem, one may obtan the requre result. Remar Corollary 3 mples that Theorem s a generalzaton an mprovement on [9, Theorem.]. Bouns for Spectral Ra of Varous Matrces Assocate Wth a Graph Let G = (V,E) be a smple graph wth vertex set V = {v 1,v,,v n } an ege set E. Let 1 n be a egree sequence of G. enote by m the average egree of a vertex v, whch s, j j. In [1, 14], the authors ntrouce the followng notaton: for any real number ( m) = j j, whch s calle the generalze average egree of v. Note that = ( 0 m) an m = ( 1 m). Moreover, they also presente some upper an lower bouns for the (Laplacan) spectral raus of G wth parameter. Next, usng a smlar metho, we also gve some bouns for spectral ra of varous matrces assocate wth G. Theorem 3 Let G be a connecte graph on n vertces wth generalzeaverageegree ( m) 1 ( m) ( m) n. Then for 1 n, ρ(a(g)) ( m) N + (( m) +N) +4N 1 =1 (( m) ( m) ), (1) where N = max j { j }. Moreover, the equalty n (1) hols f an only f ( m) 1 = ( m) = = ( m) n or for some t, G has one of the followng propertes: () If = 0, then G s a begree graph n whch 1 = = = t 1 = n 1 > t = t+1 = = n ; () If > 0, then G s a begree graph n whch ( m) 1 > ( m) = = ( m) n an 1 = n 1 > = = n. Proof Let c = ( 1,,, n) T n (7). Obvously, c s a postve vectoras G s connecte. Now apply Theorem 1 to A(G). Notce that M = ( m), M = 0, N = max j { j }. We realy get the requre upper boun (1). Agan from Theorem 1, the equalty n (1) hols f an only f ( m) 1 = ( m) = = ( m) n or for some t, A(G) = (a j ) satsfes the followng contons: (a) a l l = N for 1 n, 1 l t 1 wth l, (b) ( m) t = ( m) t+1 = = ( m) n. Conser the followng three cases: = 0, > 0 an < 0. If = 0, then the above (a) mples that 1 = = = t 1 = n 1. Snce ( 0 m) =, the above (b) becomes t = t+1 = = n. If > 0, then (a) mples 1 = n 1 an = 3 = = n as N 1 s a constant. Thus ( m) 1 ( m) = = ( m) n. Fnally, conser < 0. From (a) an N 1,

6 716 «45 one gets 1 = n 1 = = n, whch mples that G s a complete graph. Ths completes our proof. Remar 3 Tang = 0 n (1), we obtan the upper boun (1). Hence, the upper boun (1) mproves an generalzes some results n [9, 11]. Corollary 4 [10] Let G be a connecte graph on n vertces wth average egree m 1 m m n. Then for 1 n, ρ(a(g)) m N + (m +N) +4N 1 =1 (m m ), (13) where N = max j { j }. Moreover,the equalty n (13) hols f an only f m 1 = m = = m n,.e., G s pseuo-regular. Proof Tang = 1 n (1), we obtan the upper boun (13) as ( 1 m) = m. Moreover, the equalty n (13) hols f an only f m 1 = m = = m n or G s a begree graph n whch m 1 > m = = m n an 1 = n 1 > = = n. If G s a begree graph n whch m 1 > m = = m n an 1 = n 1 > = = n = δ, then m 1 = δ an m = δ 1+ n 1 δ. m 1 > m mples δ > n 1, whch s mpossble. Hence, m 1 = m = = m n. Theorem 4 Let G be a connecte graph on n vertceswth ( m) ( m) + ( m) n + n. Then for 1 n, ρ(q(g)) 1 ( ( m) + + N + (( m) + +N) +4N ) 1 =1 (( m) + ( m) ), where s the maxmum egree of G, N = max j { j }. Moreover, the equalty hols n (14) f an only f ( m) = ( m) + = = ( m) n + n or for some t, G has one of the followng propertes: () If = 0, then G s a begree graph n whch 1 = = = t 1 = n 1 > t = t+1 = = n ; ()If > 0,thenGsabegreegraphnwhch( m) > ( m) + = = ( m) n + n an 1 = n 1 > = = n. Proof Let c = ( 1,,, n) T. Obvously, c s a postve vector as G s connecte. Now apply Theorem 1 to Q(G). Notce that M = ( m) +, M =, N = max j { j }. Thus we obtan the requre upper boun (14). Agan from Theorem 1, the equalty n (14) hols f an only f ( m) = ( m) + = = ( m) n + n or for some t, Q(G) = (q j ) satsfes the followng contons: (a) q = for 1 t 1, (b) q l l = N for 1 n, 1 l t 1 wth l, (c) ( m) t + t = ( m) t+1 + t+1 = = ( m) n + n. Conser the followng three cases: = 0, > 0 an < 0. If = 0, then the above (b) mples 1 = = = t 1 = n 1. Snce ( 0 m) =, the above (c) becomes t = t+1 = = n. If > 0, then (b) shows that 1 = n 1 an = 3 = = n as N 1 s a (14)

7 5, : Bouns for Spectral Raus of Varous Matrces Assocate Wth Graphs 717 constant. Thus ( m) ( m) + = = ( m) n + n. Fnally, conser < 0. From (b) an N 1, one has 1 = n 1 = = n, whch mples that G s a complete graph. Ths completes our proof. Remar 4 Tang = 0 n (14), we may obtan the upper boun (). Hence, the upper boun (14) mproves an generalzes some results n [4, 9]. Corollary 5 Let G be a connecte graph on n vertces wth m m + m n + n. Then for 1 n, ρ(q(g)) m + + N + (m + +N) +4N 1 =1 (m + m ), (15) where s the maxmum egree of G, N = max j { j }. Moreover, the equalty n (15) hols f an only f m = m + = = m n + n or for some n, G s a begree graph n whch m > m + = = m n + n an 1 = n 1 > = = n. Proof Tang = 1 n (14), the requre result follows as ( 1 m) = m. Remar that the equalty n (15) hols for some n whenever G s a begree graph n whch m > m + = = m n + n an 1 = n 1 > = = n. For example, conser the followng graph G shown n Fg. 1. For =, applyng the nequalty (15), one gets ρ(q(g)) By a rect calculaton, one has ρ(q(g)) Hence, the equalty n (15) hols for the graph G shown n Fg. 1. In fact, 1 = 4 > = = 5 = an m = 6 > m + = = m = 5. Fgure 1 The shown graph G Let G = (V,E) be a smple graph wth vertex set V = {v 1,v,,v n } an ege set E. For any vertex v V, enote the transmsson of v by. enote by M the average transmsson of a vertex v, whch s, n j=1 jj. Now we ntrouce the followng notaton: for any real number n ( j=1 M) = jj, whch s calle the generalze average transmsson of v. Note that = ( 0 M) an M = ( 1 M). Theorem 5 Let G be a connecte graph on n vertces wth generalze average transmssons ( M) 1 ( M) ( M) n. Then for 1 n, ρ((g)) ( M) N + (( M) +N) +4N 1 =1 (( M) ( M) ), (16)

8 718 «45 where N = max 1,j n { j j }. Moreover, the equalty n (16) hols f an only f ( M) 1 = ( M) = = ( M) n or for some t, (G) = ( j ) satsfes the followng contons: () ll = N for 1 n, 1 l t 1 wth l, () ( M) t = ( M) t+1 = = ( M) n. Proof Let c = ( 1,,, n )T. Apply Theorem 1 to (G). Notce that M = ( M), M = 0, N = max 1,j n { j j }. We easly obtan the requre result. Corollary 6 [9] Let G be a connecte graph on n vertces wth transmssons 1 n. Also let be the ameter of G. Then for 1 n, the nequalty (3) hols. Moreover, the equalty n (3) hols f an only f 1 = = = n. Proof Tang = 0 n (16), the upper boun (3) follows. Agan from Theorem 5, the equalty n (3) hols f an only f 1 = = = n or for some t, (G) = ( j ) satsfes the followng contons: (a) l = N = for 1 n, 1 l t 1 wth l, (b) t = t+1 = = n. It follows from above (a) that = 1. Hence, G s a complete graph for the latter case. If we tae = 1 n (16), then the followng corollary s obtane: Corollary 7 Let G be a connecte graph on n vertces wth average transmssons M 1 M M n. Then for 1 n, ρ((g)) M N + (M +N) +4N 1 =1 (M M ), (17) where N = max 1,j n { jj }. Moreover, the equalty n (17) hols f an only f M 1 = M = = M n or for some t, (G) = ( j ) satsfes the followng contons: () l l = N for 1 n, 1 l t 1 wth l, () M t = M t+1 = = M n. Theorem 6 Let G be a connecte graph on n vertces wth generalze average transmssons ( M) 1 ( M) ( M) n. Then ρ((g)) ( M) n T + (( M) n +T) +4T n 1 =1 (( M) ( M) n ), (18) where T = mn 1 j n { j j }. Moreover, the equalty n (18) hols f an only f ( M) 1 = ( M) = = ( M) n or for some t n, (G) = ( j ) satsfes the followng contons: () l l = T for 1 n, 1 l t 1 wth l, () ( M) t = ( M) t+1 = = ( M) n. omtte. Proof Apply Theorem to (G). The rest proof s smlar to that of Theorem 5 an Remar 5 Tang = 0 n (18), we may obtan the boun (5). Hence, the lower boun (18) mproves an generalzes the result n [9]. If = 1 n (18), then ρ((g)) M n T + (M n +T) +4T n 1 =1 (M M n ),

9 5, : Bouns for Spectral Raus of Varous Matrces Assocate Wth Graphs 719 where T = mn 1 j n { jj }. Applyng Theorem 1 to the stance sgnless Laplacan matrx Q(G), we easly arrve at Theorem 7 Let G be a connecte graph on n vertces wth ( M) ( M) + ( M) n + n. Then for 1 n, ρ(q(g)) 1 ( ( M) + +M N + (( M) + M +N) +4N ) 1 =1 (( M) + ( M) ), where M = max 1 n { }, N = max 1 j n { j j }. Moreover, the equalty n (19) hols f an only f ( M) = ( M) + = = ( M) n + n or for some t, Q(G) = ( j ) satsfes the followng contons: () = M for 1 t 1, () l l = N for 1 n, 1 l t 1 wth l, () ( M) t + t = ( M) t+1 + t+1 = = ( M) n + n. If we apply Theorem to the stance sgnless Laplacan matrx Q(G), then the followng theorem s obtane. Theorem 8 Let G be a connecte graph on n vertces wth ( M) ( M) + ( M) n + n. Then ρ(q(g)) 1 ( ( M) n + n +S T + (( M) n + n S +T) +4T ) n 1 =1 (( M) + ( M) n n ), where S = mn 1 n { }, T = mn 1 j n { j j }. Moreover, the equalty n (0) hols f an only f ( M) = ( M) + = = ( M) n + n or for some t n, Q(G) = ( j ) satsfes the followng contons: () = S for 1 t 1, () l l = T for 1 n, 1 l t 1 wth l, () ( M) t + t = ( M) t+1 + t+1 = = ( M) n + n. Remar 6 If = 0 n (19), then the upper boun (4) s obtane. Hence, the upper boun (19) mproves an generalzes the result n [9]. If = 1 n (19), then ρ(q(g)) M + +M N + (M + M +N) +4N 1 =1 (M + M ), where M = max 1 n { }, N = max 1 j n { j j }. Smlarly, f = 0 n (0), then the lower boun (6) follows. Hence, the lower boun (0) mproves an generalzes the result n [9]. If = 1 n (0), then ρ(q(g)) M n + n +S T + (M n + n S +T) +4T n 1 =1 (M + M n n ), where S = mn 1 n { }, T = mn 1 j n { j j }. (19) (0)

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