The lower and upper bounds on Perron root of nonnegative irreducible matrices

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1 Journal of Computatonal Appled Mathematcs 217 (2008) 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 , PR Chna b Department of Mathematcs, Schuan Unversty of Scence Engneerng, Zgong , 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) ] 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) /$ - see front matter 2007 Elsever BV All rghts reserved do:101016/cam

2 260 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 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)

3 t follows from (2) that G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 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 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

4 262 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 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)

5 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 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, 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) Note that ρ(a) 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]): A =

6 264 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 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) = , ẑ(t 0, β) = , t 0,z(t 0, α) = ρ(a) maxt 0, ẑ(t 0, β)= 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) = , ẑ(t 0, β) = , t 0,z(t 0, α) = ρ(a) maxt 0, ẑ(t 0, β)= So we have ρ(a) Notng that ρ(a) , 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]): A n = n 1 n n 2 n 1 n Let n = 20, then ρ(a 20 ) , 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) = , ẑ(t 0, β) = , t 0,z(t 0, α) = ρ(a) maxt 0, ẑ(t 0, β)= 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) = , ẑ(t 0, β) = , t 0,z(t 0, α) = ρ(a) maxt 0, ẑ(t 0, β)=

7 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) So we have ρ(a) 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 Example 4 Consder the nonnegatve matrx (see [5, Example 6] or [2, Example 3]): A =,

8 266 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 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) ρ(a) ρ(a) ρ(a) ρ(a) ρ(a) 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) ρ(a) ρ(a) ρ(a) 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, ρ(a) We see that the bounds mprove the bounds obtaned n [5, Example 6] Secondly, we get lower bound of ρ(a) when t 0 = = z(t 0, α) the upper bound of ρ(a) when t 0 = =ẑ(t 0, β), whch suggests that Theorem 31 provdes us wth good results Fnally, z(t 0, α) decreases strctly from to wth t 0 from 453 to 48 ẑ(t 0, β) decrease strctly from to 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 of ρ(a) when t 0 = =ẑ(t 0, β), whch suggest that the results n Lemmas 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 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 of ρ(a) when t 0 = =ẑ(t 0, β) holds, whch suggest that Theorem 31 provdes us wth a good method to choose t after α

9 G-X Huang et al / Journal of Computatonal Appled Mathematcs 217 (2008) 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,, ρ(a) α =11,, ρ(a) α =11,, ρ(a) α =13,, ρ(a) α =13,, ρ(a) α =13,, 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) [3] YZ Fan, Schur complement ts applcaton to symmetrc nonnegatve z-matrces, Lnear Algebra Appl 353 (2002) [4] LZ Lu, Perron complement Perron root, Lnear Algebra Appl 341 (2002) [5] LZ Lu, MK Ng, Locatons of Perron roots, Lnear Algebra Appl 392 (2004) [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) [8] M Neumann, Inverse of Perron complements of nverse M-matrces, Lnear Algebra Appl 313 (2000)