Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 wwwelsevercom/locate/cam The lower upper bounds on Perron root of nonnegatve rreducble matrces Guang-Xn Huang a,, Feng Yn b,keguo a a College of Informaton Management, Chengdu Unversty of Technology, Chengdu 610059, PR Chna b Department of Mathematcs, Schuan Unversty of Scence Engneerng, Zgong 643000, PR Chna Receved 9 August 2006; receved n revsed form 26 March 2007 Abstract Let A be an n n nonnegatve rreducble matrx, let A[α] be the prncpal submatrx of A based on the nonempty ordered subset α of 1, 2,,n, defne the generalzed Perron complement of A[α] by P t (A/A[α]), e, P t (A/A[α]) = A[β]+A[β, α](ti A[α]) 1 A[α, β], t >ρ(a[α]) Ths paper gves the upper lower bounds on the Perron root of A An upper bound on Perron root s derved from the maxmum of the gven parameter t 0 the maxmum of the row sums of P t0 (A/A[α]), synchronously, a lower bound on Perron root s expressed by the mum of the gven parameter t 0 the mum of the row sums of P t0 (A/A[α]) It s also shown how to choose the parameter t after α to get tghter upper lower bounds of ρ(a) Several numercal examples are presented to show that our method compared wth the methods n [LZ Lu, MK Ng, Locatons of Perron roots, Lnear Algebra Appl 392 (2004) 103 117] s more effectve 2007 Elsever BV All rghts reserved MSC: 15A48; 05C50 Keywords: Nonnegatve rreducble matrx; Perron root; Lower upper bounds; Generalzed Perron complement 1 Introducton In ths paper the followng notatons are consdered used Let R n n, R n n Rn n > denote the sets of all n n real matrces, all n n real nonnegatve matrces all n n real postve matrces, respectvely For A, B R n n, we denote by A>B that each entry of the matrx A B s nonnegatve, A B has at least one postve entry For an arbtrary matrx A = (a ) R n n, let A T denote the transpose of A n r (A) = a k, = 1, 2,,n, k=1 r(a) = (r 1 (A), r 2 (A),,r n (A)) T, r (A) = r (A), r max (A) = max r (A) (1) 1 n 1 n Ths work was supported n part by the Foundaton of the Educaton Councl of Chongqng the Key Scence Fund of Chengdu Unversty of Technology Correspondng author E-mal address: huangx@cduteducn (G-X Huang) 0377-0427/$ - see front matter 2007 Elsever BV All rghts reserved do:101016/cam200706034
260 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 Let N =1, 2,,n Let α denote a nonempty ordered subset of N β = N\α, both consstng of strctly ncreasng ntegers We also denote the submatrx of the matrx A whose rows columns are detered by α β, respectvely, n A[α, β] The matrx A[α] s ust equal to the matrx A[α, α], the prncpal submatrx of A based on α For a nonnegatve rreducble matrx A R n n, a fundamental matrx problem s to locate the Perron root ρ(a) of A It s well known that for such a matrx A, the followng nequalty [1] holds: r (A) ρ(a) r max (A) (2) the equalty holds n one of the bounds f only f t holds n both For A R> n n, the bounds of ρ(a) were mproved by Brauer [6] Meyer [7] defned the Perron complement used t to compute the unque normalzed Perron egenvector of a nonnegatve rreducble A Neumann [8] used t to analyze the propertes of nverse M-matrces Fan [3] used t to derve the bounds of the Perron root of symmetrc rreducble nonnegatve matrces Z-matrces For a nonnegatve rreducble matrx A, n order to obtan the bounds on ρ(a), P t (A/A[α]) for t ρ(a) was frst defned by Neumann [8], followed by Lu [4] who defned used the generalzed Perron complement P t (A/A[α]) of A[α], whch s gven by P t (A/A[α]) = A[β]+A[β, α](ti A[α]) 1 A[α, β], t >ρ(a[α]) (3) It has been show n [4] that the use of the generalzed Perron complement of A[α] can gve tght bounds on ρ(a) Lu [5] has gven a new localzaton method that utlzes the relatonshp between the Perron root of a nonnegatve matrx the estmates of the row sums of ts generalzed Perron complement The man results n [5] can only obtan a tght upper bound or a tght lower bound of ρ(a), respectvely In ths paper, however, we am to solve the problems as follows It has always been supposed that matrx A R n n s rreducble wthout specal specfcaton Problem 1 How to obtan a tghter lower upper bounds of ρ(a) together by the estmates of the row sums of ts generalzed Perron complement? Problem 2 How to properly choose parameters α t after α to get an optmal lower bound an optmal upper bound of ρ(a), respectvely? Ths paper s organzed as follows In Secton 2, we wll gve the lower upper bounds on Perron root by the mum of the gven parameter t 0 the mum row sum of the row sums of P t (A/A[α]) the maxmum of the gven parameter t 0 the maxmum row sum of the row sums of P t (A/A[α]), respectvely Then n Secton 3, we wll properly choose the parameter t after α to get an optmal lower bound an optmal upper bound ρ(a), respectvely In Sectons 2 3, some numercal examples are also gven to show the applcaton of the correspondng results 2 The upper lower bounds on Perron root In ths secton, we wll show tghter lower upper bounds on Perron root by the mum of the gven parameter t 0 the mum row sum of the row sums of P t0 (A/A[α]) the maxmum of the gven parameter t 0 the maxmum row sum of the row sums of P t0 (A/A[α]), respectvely Three numercal examples are provded to llustrate the results For the generalzed Perron complement matrx P t (A/A[α]), the followng results are needed Lemma 21 (See [5, Theorem 5]) Assume that l u are found such that then Let l ρ(p t (A/A[α])) u, t > ρ(a[α]), (4) t,l ρ(a) maxt,u (5) z(t, α) = r (P t (A/A[α])), ẑ(t, α) = r max (P t (A/A[α])), (6)
t follows from (2) that G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 261 z(t, α) ρ(p t (A/A[α])) ẑ(t, α) By Lemma 21, we have t,z(t,α) ρ(a) maxt,ẑ(t, α) (7) Lemma 22 (See [5, Theorem 7]) If A 0 t 0 >r max (A[α]), then z(t 0, α) r (A[β]) + v 1 (t 0, α)r (A[β, α]), ẑ(t 0, α) max r (A[β]) + v 2 (t 0, α)r (A[β, α]), (8) where v 1 (t 0, α) = r (A[α, β]) t 0 r (A[α]) v 2 (t 0, α) = max A tghter lower bound of ρ(a) can be obtaned by Lemmas 21 22 r (A[α, β]) t 0 r (A[α]) (9) Corollary 21 Let A be an n n rreducble nonnegatve matrx wth n 3 r max (A)>r (A) If α (or β = N\α) t 0 are chosen, respectvely, such that max max r (A), r max (A[α]) < r (A), r (A[β, α])>0 (10) then max max r (A), r max (A[α]) <t 0 < r (A), (11) ρ(a) t 0,z(t 0, α) >r (A) (12) Proof Let α t 0 be chosen such that v 1 (t 0, α)>1, where v 1 (t 0, α) s defned n (9) Note that r max (A[α])<t 0 < r (A), α for any 1 α, so Therefore t 0 r (A[α]) r (A[α, β]) = t 0 r (A[α,N]) t 0 α r (A) < 0 0 <t 0 r max (A[α]) t 0 r (A[α])<r (A[α, β]) v 1 (t 0, α) = r (A[α, β]) t 0 r (A[α]) > 1, t follows from Lemma 22 (10) that z(t 0, α) [r (A[β])]+v 1 (t 0, α)r (A[β, α])> r (A) = r (A) (13) β Thus (7), (11) (13) gve (12) Ths completes the proof
262 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 Remark Corollary 21 was frst provded by Lu [5, Theorem 8]; however, t s obvous that the condton r (A[α, β]) > 0 s not necessary for t s mpled n nequalty (14) n [5], e, r (A[α, β])>0 s mpled n (10) The followng result gves a tghter upper bound of ρ(a) Lemma 23 Let A be an n n rreducble nonnegatve matrx wth n 3 r max (A)>r (A) If α (or β = N\α) t 0 are chosen, respectvely, such that max max r (A), r max (A[α]) < r (A), r (A[β, α])>0 (14) then max r (A)<t 0 <r max (A), (15) β ρ(a) maxt 0, ẑ(t 0, β) <r max (A) (16) Proof Let α t 0 be chosen such that 0 <v 2 (t 0, β)<1, where v 2 (t 0, β) = max Note that r (A[β, α]) t 0 r (A[β]) t 0 > max r (A), β t 0 r (A[β]) r (A[β, α]) = t 0 r (A[β,N]) t 0 max r (A) > 0, β from (14), t follows that So t 0 r (A[β])>r (A[β, α])>0 0 <v 2 (t 0, β) = max r (A[β, α]) t 0 r (A[β]) < 1, snce r (A[α, β])>0 s mpled n (14) By usng Lemma 22 (14) agan, we have ẑ(t 0, β) max [r (A[α])]+v 2 (t 0, β)r (A[α, β])<max r (A) = r max (A) (17) α Therefore, (16) s mpled by (7), (15) (17) Ths completes the proof By usng Corollary 21 Lemma 23, we have the followng result Theorem 21 Let A be an n n rreducble nonnegatve matrx wth n 3 r max (A)>r (A) If α (or β = N\α) t 0 are chosen, respectvely, such that max max r (A), r max (A[α]) < r (A), r (A[β, α])>0 (18) max max r (A), r max (A[α]) <t 0 < r max (A), r (A), (19)
G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 263 then r (A) < t 0,z(t 0, α) ρ(a) maxt 0, ẑ(t 0, β) <r max (A) (20) Proof When (18) (19) hold, t follows that (10), (11) (14), (15) hold, respectvely By usng Corollary 21 Lemma 23, we have that (20) holds Ths completes the proof It can be seen that (2) s mproved by (20) n Theorem 21 From the proof of Corollary 21, Lemma 23 Theorem 21, we have the followng result Corollary 22 Wth the condtons of Theorem 21, we have r max (A) < t 0, [r (A[β]) + v 1 (t 0, α)r (A[β, α])] ρ(a) α max t 0, max [r (A[α]) + v 2 (t 0, β)r (A[α, β])] r max (A), (21) α where v 1 (t 0, α) = r (A[α, β]) t 0 r (A[α]) v 2 (t 0, β) = max r (A[β, α]) t 0 r (A[β]) Next we wll consder the followng examples to llustrate the results of Theorem 21 Example 1 Consder the postve matrx (see [6] or [5]): ( 1 1 ) 2 A = 2 1 3, 2 3 5 we can compute that r(a) = (4, 6, 10) T,r max (A) = 10,r (A) = 4 Let α =3, β = N\α =1, 2, then r (A) = 10 > 6 = max r (A), r (A) = 10 > 5 = r max (A[α]), α max max r (A), r max (A[α]) = 6 <t 0 < 10 = r (A), r max (A) Accordng to Theorem 21, let t 0 = 7, then z(t 0, α) = 7, ẑ(t 0, β) = 78235, t 0,z(t 0, α)=7 ρ(a) maxt 0, ẑ(t 0, β)=78325 let t 0 = 75, then z(t 0, α) = 6, ẑ(t 0, β) = 75466, 75, 6=6 ρ(a) max75, 75466=75466 It follows that 7 ρ(a) 75466 Note that ρ(a) 75311 the upper lower bounds are better than those stated n [5, Examples 2 3] are better than those gven n [6, p 158] Example 2 Consder the followng 8 8 matrx (see [7] or [5]): 8 6 3 5 7 0 7 1 0 7 3 8 5 6 4 1 1 2 6 1 3 8 8 7 2 8 4 0 7 7 8 2 A = 2 4 6 2 5 7 6 5 4 1 0 4 8 4 8 2 3 1 6 6 4 5 5 0 0 1 1 6 7 0 3 4
264 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 We can compute that r max (A)=38,r (A)=22 r(a)=(37, 34, 36, 38, 37, 31, 30, 22) T Let α=1, 2, 3, 4, 5, β= N\α =6, 7, 8, then α r (A) = 34 > 31 = max β r (A), α r (A) = 34 > 29 = r max (A[α]) max max r (A), r max (A[α]) = 31 <t 0 < 34 = r (A), r max (A) Accordng to Theorem 21, f we let t 0 = 3101, then z(t 0, α) = 292713, ẑ(t 0, β) = 366773, t 0,z(t 0, α) = 292713 ρ(a) maxt 0, ẑ(t 0, β)=366773 Let α =1, 3, 4, 5, β = N\α =2, 6, 7, 8, then α r (A) = 36 > 34 = max β r (A), α r (A) = 36 > 16 = r max (A[α]), max max r (A), r max (A[α]) = 34 <t 0 < 36 = r (A), r max (A) Accordng to Theorem 21, f we let t 0 = 34575, then z(t 0, α) = 237308, ẑ(t 0, β) = 345760, t 0,z(t 0, α) = 237308 ρ(a) maxt 0, ẑ(t 0, β)=345760 So we have 292713 ρ(a) 345760 Notng that ρ(a) 332418, t can be seen that the upper lower bounds are better than the bounds gven n [5, Example 4] Example 3 Consder an n n postve matrx ([7] or [5]): 1 1 1 1 1 2 2 2 A n = 1 2 3 1 2 n 1 n 1 1 2 n 2 n 1 n Let n = 20, then ρ(a 20 ) 170404, r max (A) = 210,r (A) = 20 Let α =11,,20, β = N\α =1,,10, then α r (A) = 165 > 155 = max β r (A) α r (A) = 165 > 155 = r max (A[α]), max max r (A), r max (A[α]) = 155 <t 0 < 165 = r (A), r max (A) Accordng to Theorem 21, f we let t 0 = 15501, then z(t 0, α) = 446665, ẑ(t 0, β) = 1897912, t 0,z(t 0, α) = 446665 ρ(a) maxt 0, ẑ(t 0, β)=1897912 Let α =13,,20, β = N\α =1,,12, then α r (A) = 182 > 174 = max β r (A) α r (A) = 182 > 132 = r max (A[α]), max max r (A), r max (A[α]) = 174 <t 0 < 182 = r (A), r max (A) Accordng to Theorem 21, f we let t 0 = 1774, then z(t 0, α) = 232679, ẑ(t 0, β) = 1774019, t 0,z(t 0, α) = 232679 ρ(a) maxt 0, ẑ(t 0, β)=1774019
G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 265 So we have 446665 ρ(a) 1774019 Notng that when α =10,,20, β = N\α =1,,9, wehave r (A) = 155 > 144 = max r (A), α β but α r (A) = 155 < 165 = r max (A[α]), so ths do not satsfyng nequalty (18) n Theorem 21 From the prevous examples we can see that an approprate choce of α t 0 such that condtons (18) (19) hold n Theorem 21 makes us get a tghter upper bound a lower bound of Perron root 3 The optmal choce of t after α To get better bounds, we can see from Examples 1 3 that there stll exsts a problem as to how the parameters t after α should be chosen In ths secton, we wll dscuss ths problem Lemma 31 If A s a nonnegatve rreducble matrx, then (1) z(t, α) = r (P t (A/A[α])) s a strctly decreasng functon of t on (ρ(a[α], + )) (2) ẑ(t, β) = r max (P t (A/A[β])) s a strctly decreasng functon of t on (ρ(a[β], + )) Proof We wll only prove the frst part The second part can be proved smlarly Suppose t 2 >t 1 > ρ(a[α]), then t 2 I A[α] >t 1 I A[α], t follows by the characters of M matrces that (t 1 I A[α]) 1 >(t 2 I A[α]) 1 > 0, so we have that P t1 (A/A[α])>P t2 (A/A[α])>0, therefore r (P t2 (A/A[α]))<r (P t1 (A/A[α])) Ths completes the proof By Lemma 31, we have Lemma 32 Suppose A s a nonnegatve rreducble matrx, then (1) t,z(t,α) s a strctly ncreasng functon of t when ρ(a[α])<t z(t, α) s strctly decreasng functon of t when t z(t, α) (2) maxt,ẑ(t, β) s a strctly decreasng functon of t when ρ(a[β])<t ẑ(t, β) s a strctly ncreasng functon of t when t ẑ(t, β) Proof As for the frst part, when ρ(a[α])<t z(t, α), wehavet,z(t,α)=t a strctly ncreasng functon of t If t z(t, α), then t,z(t,α)=z(t, α) s a strctly decreasng functon of t by usng Lemma 31 We can prove the second part smlarly so t s omtted Ths completes the proof From Lemmas 31, 32 Theorem 21, we can easly have Theorem 31 Suppose A satsfes the condtons n Theorem 21, then the lower bound t,z(t,α) of ρ(a) s tghtest when t satsfes t = z(t, α) the upper bound maxt,ẑ(t, β) of ρ(a) s tghtest when t satsfes t =ẑ(t, β) Several examples are gven as follows to show the applcaton of Theorems 21 31 Example 4 Consder the nonnegatve matrx (see [5, Example 6] or [2, Example 3]): 2 5 1 0 0 0 1 2 A =, 1 4 1 2 1 1 0 1
266 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 Table 1 Estmates of the bounds on ρ(a) wth α =1, 3 t 0 z(t 0, α) ẑ(t 0, β) t 0,z(t 0, α) ρ(a) maxt 0, ẑ(t 0, β) 453 45445 49763 453 ρ(a) 49763 454 45327 49685 45327 ρ(a) 49685 455 45210 49608 45210 ρ(a) 49608 47 43582 48519 43582 ρ(a) 48519 479 42707 47921 42707 ρ(a) 47921 48 42614 47857 42614 ρ(a) 48 Table 2 Estmates of the bounds on ρ(a) wth α =3 t 0 z(t 0, α) ẑ(t 0, β) t 0,z(t 0, α) ρ(a) maxt 0, ẑ(t 0, β) 65 86667 81681 65 ρ(a) 81681 70 7 78235 70 ρ(a) 78235 75 6 75466 6 ρ(a) 75466 76 58462 74976 58 ρ(a) 76 ρ(a) 46182,r max (A) = 8,r (A) = 3 Let α =1, 3, β = N\α =2, 4, then α r (A) = 8 > 3 = max β r (A) α r (A) = 8 > 3 = r max (A[α]), choose dfferent t 0 satsfyng max max r (A), r max (A[α]) β = 3 <t 0 < 8 = r (A), r max (A), (22) α by computng we get dfferent values of z(t, α) ẑ(t, β) shown n Table 1 From Table 1 we can see the followng three facts Frstly, 45327 ρ(a) 47921 We see that the bounds mprove the bounds obtaned n [5, Example 6] Secondly, we get lower bound 45327 of ρ(a) when t 0 = 454 45327 = z(t 0, α) the upper bound 47921 of ρ(a) when t 0 = 479 47921 =ẑ(t 0, β), whch suggests that Theorem 31 provdes us wth good results Fnally, z(t 0, α) decreases strctly from 45445 to 42614 wth t 0 from 453 to 48 ẑ(t 0, β) decrease strctly from 49763 to 47857 wth t 0 from 453 to 48, whch supported the results n Lemma 31 Example 5 Consder the matrx n Example 1 Choose α =3, when t 0 s evaluated dfferently whch satsfes 6 <t 0 < 10 By Theorem 21 we have dfferent values of z(t, α) ẑ(t, β) shown n Table 2 Notng that 70 ρ(a) 75466, we get the lower bound 70 of ρ(a) when t 0 = 70 = z(t 0, α) the upper bound 75466 of ρ(a) when t 0 = 75 75466 =ẑ(t 0, β), whch suggest that the results n Lemmas 31 32 are perfect Remark 31 It s a pty that the parameter t n Theorem 31 satsfyng t = z(t, α) or t =ẑ(t, β) cannot always reach, respectvely, for the condtons n Theorem 21 However, we should choose t whch satsfes the condtons n Theorem 21 make t z(t, α) smallest so that we can get a much tghter lower bound of ρ(a) Smlarly, we should choose t whch satsfes the condtons n Theorem 21 makes t ẑ(t, β) smallest so that we can get a much tghter upper bound of ρ(a) Example 6 Consder the postve matrx n Example 3 Choose α=11,,20, 155 <t 0 < 165 α=13,,20, 174 <t 0 < 182 By Theorem 21 we have Table 3 From Table 3, we get the lower bound 44665 of ρ(a) when t 0 (=15501) makes t 0 z(t 0, α) (=110345) the smallest value n Table 3 satsfes the condtons n Theorem 23, the upper bound 1774019 of ρ(a) when t 0 = 1774 1774019 =ẑ(t 0, β) holds, whch suggest that Theorem 31 provdes us wth a good method to choose t after α
G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 259 267 267 Table 3 Estmates of the bounds on ρ(a) wth α =11,,20 α =13,,20 α t 0 z(t 0, α) ẑ(t 0, β) t 0,z(t 0, α) ρ(a) maxt 0, ẑ(t 0, β) α =11,,20 15501 44665 1897912 446665 ρ(a) 1897912 α =11,,20 156 443744 1894824 443744 ρ(a) 1894824 α =11,,20 160 37448 1882887 37448 ρ(a) 1882887 α =13,,20 176 235611 1779647 235611 ρ(a) 1779647 α =13,,20 1774 232679 1774019 232679 ρ(a) 1774019 α =13,,20 1775 232475 1773622 232475 ρ(a) 1775 Acknowledgement The authors are grateful to the anonymous referee for the constructve helpful comments Prof Lothar Rechel for all the communcaton References [1] A Berman, RJ Plemmons, Nonnegatve Matrces n the Mathematcs Scences, SIAM Press, Phladelpha, PA, 1994 [2] E Deutsch, Bounds for the Perron root of a nonnegatve rreducble parttoned matrx, Pacfc J Math 92 (1981) 51 56 [3] YZ Fan, Schur complement ts applcaton to symmetrc nonnegatve z-matrces, Lnear Algebra Appl 353 (2002) 289 307 [4] LZ Lu, Perron complement Perron root, Lnear Algebra Appl 341 (2002) 239 248 [5] LZ Lu, MK Ng, Locatons of Perron roots, Lnear Algebra Appl 392 (2004) 103 117 [6] M Marcus, H Mnc, A Survey of Matrx Theory Matrx Inequaltes, Dover Publcatons, New York, 1992 [7] CD Meyer, Uncouplng the Perron egenvector problem, Lnear Algebra Appl 114/115 (1989) 69 94 [8] M Neumann, Inverse of Perron complements of nverse M-matrces, Lnear Algebra Appl 313 (2000) 163 171