Properties for the Perron complement of three known subclasses of H-matrices

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1 Wang et al Journal of Inequalities and Applications 2015) 2015:9 DOI /s R E S E A R C H Open Access Properties for the Perron complement of three known subclasses of H-matrices Leilei Wang, Jianzhou Liu * and Shan Chu * Correspondence: liujz@xtueducn Department of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan , PR China Abstract In this paper, we analyze the diagonally dominant degree for the Perron complement upon several diagonally dominant cases by using the entries and spectral radius of the original matrix At the same time, we obtain closure properties for the Perron complement of several diagonally dominant matrices MSC: Primary 15A47; 15A48; secondary 65U05; 65J10 Keywords: diagonally dominant matrix; H-matrix; the Perron complement; nonnegative irreducible matrix; spectral radius 1 Introduction The Perron complement plays an important role in statistics, matrix theory, control theory, and so on see [1 4]) It was said in [2, 5]that the Perron vector or the stationary distribution vector) for a corresponding transition matrix of a certain Markov chain provides the long term probabilities of the chain being in a particular state Meyer [1] introduced the concept of the Perron complement, obtained the Perron complement of a nonnegative irreducible matrix is nonnegative irreducible, and first used the closure property of a nonnegative irreducible matrix to construct a divided and conquer algorithm to compute the Perron vector for a Markov chain Since then, a lot of work has been done on this topic The authors in [6 8] showed closure properties of the Perron complement for some special matrices When the Perron complement is primitive, we can design a related iterative algorithm to compute the Perron vector see [1])So farasweknow,ifthegivenmatrix has a sharper diagonally dominant degree, then the designed iterative algorithms has faster convergent rate than the ordinary ones see [9]) At the same time, if a given matrix has a sharper diagonally dominant degree, then we may discuss more properties about generalized nonlinear diagonal dominance in [10] Motivated by the useful applications, we will study the diagonal dominant degree for the Perron complement of several cases based on the nonnegative and irreducible nature, which may belong to inherent properties of the Perron complement Meanwhile, some closure properties for the Perron complement of three known subclasses of H-matrices are provided by the entries and spectral radius of the original matrix Let C n n R n n )denotethesetofalln n complex real) matrices, N = {1,2,,n},and G = {i, j):i, j N, i j}fora =a ij ) C n n,denote R i A)= a ij, S i A)= a ji, i N j N,j i j N,j i 2015 Wang et al; licensee Springer This is an Open Access article distributed under the terms of the Creative Commons Attribution License which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited

2 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 2 of 10 Choose N r A)= { i : a ii > R i A), i N }, N c A)= { j : a jj > S j A), j N } The comparison matrix μa)=μ ij )ofa =a ij ) C n n,definedby μ ij = { a ii, i = j, a ij, i j If A = μa) and the eigenvalues of A have positive real parts, then we call A a nonsingular) M-matrix We say that A is an H-matrix if μa)isanm-matrix For details and numerous conditions of M-matrices, please refer to [11] We denote by H n and M n the sets of all n n H-andM-matrices, respectively Recall that A =a ij ) C n n is row) diagonally dominant and denoted by A D n if a ii R i A) 11) for all i N holds A is said to be a strictly diagonally dominant matrix and denoted by A SD n if the representative inequality 11)isstrict A =a ij ) C n n is doubly diagonally dominant and denoted by A DD n if a ii a jj R i A)R j A) 12) for all i, j) G valid Moreover, A is said to be a strictly doubly diagonally dominant matrix and denoted by A SDD n if the representative inequality 12) is strict for all i, j) G If A SDD n, then there exists at most one index i 0 N such that a i0 i 0 R i0 A) 13) A =a ij ) C n n is γ -diagonally dominant and denoted by A D γ n if there exists γ [0, 1] such that a ii γ R i A)+1 γ )S i A), for all i N 14) If all the inequalities in 14)arestrict,thenA is called strict γ -diagonally dominant A =a ij ) C n n is product γ -diagonally dominant and denoted by A PD γ n if there exists γ [0, 1] such that a ii [ R i A) ] γ [ Si A) ] 1 γ, for all i N 15) If all the inequalities in 15)arestrict,thenA is said to be a strictly product γ -diagonally dominant matrix For A C n n,nonemptyindexsetsα, β N, wedenotebyα the cardinality of α Let Aα, β) denote the sub-matrix of A lying in the rows indexed by α and the columns indexed by β Aα, α) is abbreviated to Aα) If Aα) is nonsingular, then the Schur complement of Aα)inA is given by S A/Aα) ) = Aβ) Aβ, α) [ Aα) ] Aα, β) 16)

3 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 3 of 10 Let A be a nonnegative irreducible matrix of order n with spectral radius ρa), α, and β = N α Then the Perron complement of A with respect to Aα), which is denoted by PA/Aα)) or simply PA/α), is defined as Aβ)+Aβ, α) [ ρa)i Aα) ] Aα, β) 17) In addition, the extended Perron complement P t A/Aα)) or simply P t A/α)att is defined as Aβ)+Aβ, α) [ ti Aα) ] Aα, β) 18) 2 Properties for the Perron complement of diagonally dominant matrices In this section, we offer some properties for the Perron complement of diagonally dominant matrices by using the entries and spectral radius of the original matrix Lemma 1 see [11]) If A is an H-matrix, then [μa)] A Lemma 2 see [11]) If A =a ij ) SD n, or A SDD n, then μa) M n, ie, A H n Proposition 1 Let A =a ij ) R n n is nonnegative irreducible with spectral radius ρa), α = {i 1, i 2,,i k } N r A), β = N α = {j 1, j 2,,j l }, α < n, and denote x a jt i 1 a jt i k l B jt a i 1 j u μ[ρa)i Aα)], x >0 l a i k j u Then for ρa) 2a ii i α) and any j t β, B jt is an M-matrix of order k +1and det B jt >0 if x > k R iω A) a jt i ω ρa) a iω i ω 21) ω=1 Proof Denote B jt B b pq ) Equation 21) means that there exists ε >0suchthat x > k ω=1 ) R iω A) a jt i ω ρa) a iω i ω + ε 22) By ρa) 2a ii i α) andα N r A), we have 0 R iω A) ρa) a iωiω <11 ω k) Choose a positive matrix D = diagd 1, d 2,,d k+1 )andk = BD =k sv ), where { 1, v =1; d v = R iv A) ρa) a iv + ε, i v 2 v k +1 i) For s =1,22) follows from k ss v s k+1 k+1 k sv = k 11 k 1v = x v=2 v=2 a jt i v R iω A) ρa) a iω i ω + ε ) >0

4 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 4 of 10 ii) For s =2,3,,k +1,weobtain k ss k sv v s = [ ρa) a is i s ] ) R is A) ρa) a is i s + ε a is j u = R is A) k+1 v=2,v s a is j u [ + ε ρa) a is i s ) R iv A) a is i v ρa) a iv i v + ε k v=1,v s k+1 v=2,v s R iv A) a is i v ρa) a iv i v ] a is i v >0 Thus, K SD k+1 Further,B jt H k+1 Besides, observing that B jt = μb jt ), we have B jt M k+1 and det B jt >0 Proposition 2 Let A =a ij ) R n n be nonnegative irreducible with spectral radius ρa), α = {i 1, i 2,,i k } N, β = N α = {j 1, j 2,,j l }, α < n, PA/α) =a ts ), and ρa) 2a ii i α) i) If α N r A), then for all 1 t l,,,a jt i k ) [ ρa)i Aα) ] a i1 j s +,,a jt i k R jt A) ω jt 23) ii) If α N c A), then for all 1 t l, a js j t +a js i 1,,a js i k +,,a jt i k S jt A) ωj T t 24) Here ω jt = k a jt i u ρa) a i u i u R iu A), ωj T ρa) a iu i u t = k a iu j t ρa) a i u i u S iu A) ρa) a iu i u

5 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 5 of 10 Proof By α N r A), we have Aα) SD k Further,ρA) 2a ii i α) yieldsρa)i Aα)) SD k By Lemma 1 and Lemma 2,wehave { [ ]} [ ] μ ρa)i Aα) ρa)i Aα) Thus, by the definition of the Perron complement, we obtain,,a jt i k +,,a jt i k ) [ ρa)i Aα) ] a i1 j s +,,a jt i k ) [ ρa)i Aα) ] a i1 j s s=1 + ajt i 1,,a jt i k ){ μ [ ρa)i Aα) ]} a i1 j s s=1 l k = R jt A) a jt i u + a jt i 1,,a jt i k ){ μ [ ρa)i Aα) ]} s=1 a i 1 j s l s=1 a i k j s 1 = R jt A) ω jt det{μ[ρa)i Aα)]} k a j t i u ω jt a jt i 1 a jt i k l s=1 det a i 1 j s μ[ρa)i Aα)] l s=1 a i k j s R jt A) ω jt det B 1 det{μ[ρa)i Aα)]} On the basis of Proposition 1,wehavedet B 1 > 0, which implies 23) Moreover, 24)can be proved with a similar method to the above techniques Theorem 1 If A =a ij ) R n n is nonnegative irreducible with spectral radius ρa), α = {i 1, i 2,,i k } N r A), β = N α = {j 1, j 2,,j l }, α < n, PA/α)=a ts ), then for ρa) 2a ii i α) and 1 t l, ) a tt Rt PA/α) ajt j t R jt A)+ω jt a jt j t R jt A) 25) and a tt + Rt PA/α) ) ajt j t + R jt A) ω jt a jt j t + R jt A) 26)

6 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 6 of 10 Proof By α N r A) and Proposition 2,wehave a tt ) R t PA/α) = a tt a ts = a jt j t,,a jt i k,,a jt i k ) [ ρa)i Aα) ] a jt j t,,a jt i k,,a jt i k ) [ ρa)i Aα) ] a jt j t R jt A) ω jt ), a i1 j s a i1 j s which implies 25) Similarly, we can also verify 26) immediately When A SD n,25)andtheorem21of[1] show the following fact Corollary 1 If A =a ij ) R n n is nonnegative irreducible and strictly diagonally dominant with spectral radius ρa), α N, α < n, β = N α, then for ρa) 2a ii i α), PA/α)=Aβ)+Aβ, α) [ ρa)i Aα) ] Aα, β) is nonnegative irreducible and strictly diagonally dominant 3 Properties for the Perron complement of strictly γ -andproduct γ -diagonally dominant matrices In this section, we obtain several properties for the Perron complement of strictly γ -and product γ -diagonally dominant matrices through using the entries and spectral radius of the original matrix Lemma 3 see [12]) If a > b, c > b, b >0,and 0 r 1, then a r c 1 r a b) r c b) 1 r + b Theorem 2 If A =a ij ) R n n is nonnegative irreducible and with spectral radius ρa), N r A) N c A), α = {i 1, i 2,,i k } N r A) N c A)), α < n, β = N α = {j 1, j 2,,j l },

7 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 7 of 10 and PA/α)=a ts ), then for ρa) 2a ii i α), 1 t l, and 0 γ 1, a tt ) ) γ R t PA/α) 1 γ)st PA/α) a jt j t γ R jt A) 1 γ )S jt A)+γω jt +1 γ )ω T j t a jt j t γ R jt A) 1 γ )S jt A) and a tt ) ) + γ R t PA/α) +1 γ )St PA/α) a jt j t + γ R jt A)+1 γ )S jt A) γω jt 1 γ )ω T j t a jt j t + γ R jt A)+1 γ )S jt A) Proof In light of α N r A) N c A)) and Proposition 2,wehave a tt ) ) γ R t PA/α) 1 γ)st PA/α) = a tt γ a ts 1 γ) a st = a jt j t,,a jt i k γ,,a jt i k ) [ ρa)i Aα) ] a i1 j s 1 γ) a js j t +a js i 1,,a js i k a jt j t,,a jt i k γ,,a jt i k ) [ ρa)i Aα) ] a i1 j s 1 γ) a js j t +a js i 1,,a js i k = a jt j t γ,,a jt i k

8 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 8 of 10 γ,,a jt i k ) [ ρa)i Aα) ] a i1 j s 1 γ),,a jt i k 1 γ) a js j t +a js i 1,,a js i k a jt j t γ ) R jt A) ω jt 1 γ) Sjt A) ωj T ) t = a jt j t γ R jt A) 1 γ )S jt A)+γω jt +1 γ )ω T j t, which yields the first type of inequalities Similarly, the rest of the inequalities of Theorem 2 can be verified By Theorem 2 and Theorem 21 of [1], we get the following corollary Corollary 2 If A =a ij ) R n n is nonnegative irreducible and strictly γ -diagonally dominant with spectral radius ρa), N r A) N c A), α N r A) N c A), α < n, β = N α, then for ρa) 2a ii i α), PA/α)=Aβ)+Aβ, α) [ ρa)i Aα) ] Aα, β) is nonnegative irreducible and strictly γ -diagonally dominant, and so is P t A/α) Theorem 3 If A =a ij ) R n n is nonnegative irreducible with spectral radius ρa), N r A) N c A), α = {i 1, i 2,,i k } N r A) N c A), α < n, β = N α = {j 1, j 2,,j l }, PA/α)=a ts ), then for ρa) 2a ii i α), 1 t l, and 0 γ 1, and a γ ) 1 γ ) tt R t PA/α) S t PA/α) a jt j t [ ] γ [ R jt A) ω jt Sjt A) ωj T ] 1 γ ajt t j t R γ j t A)S 1 γ j t A) a γ ) 1 γ ) tt + R t PA/α) S t PA/α) a jt j t + [ ] γ [ R jt A) ω jt Sjt A) ωj T ] 1 γ ajt t j t + R γ j t A)S 1 γ j t A) Proof By the definition of the Perron complement, we obtain a tt [ Rt PA/α) )] γ [ St PA/α) )] 1 γ = [ a tt a ts ] γ [ a st ] 1 γ

9 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 9 of 10 = a jt j t,,a jt i k ) [ ρa)i Aα) ] a i1 j t,,a jt i k ) [ ρa)i Aα) ] a i1 j s γ a js j t +a js i 1,,a js i k ) [ ρa)i Aα) ] a i1 j t 1 γ a jt j t,,a jt i k ) [ ρa)i Aα) ] a i1 j t,,a jt i k ) [ ρa)i Aα) ] a i1 j s γ a js j t +a js i 1,,a js i k ) [ ρa)i Aα) ] a i1 j t 1 γ Denote h =,,a jt i k ) [ ρa)i Aα) ] a i1 j t In light of Lemma 3,weget a tt [ R t PA/α) )] γ [ St PA/α) )] 1 γ a jt j t h [ R jt A) ω jt h ] γ [ Sjt A) ω T j t h ] 1 γ a jt j t h [ R jt A) ω jt ) γ Sjt A) ω T j t ) h ] = a jt j t [ R jt A) ω jt ] γ [ Sjt A) ω T j t ] 1 γ Hence, we can get the first type of inequalities of Theorem 3 Similarly, we can immediately verify the other one By Theorem 3 and Theorem 21 of [1], we have the following corollary Corollary 3 If A =a ij ) R n n is nonnegative irreducible and strictly product γ -diagonally dominant with spectral radius ρa), N r A) N c A), α N r A) N c A), α < n, and

10 Wang et al Journal of Inequalities and Applications 2015) 2015:9 Page 10 of 10 β = N α, then for ρa) 2a ii i α), PA/α)=Aβ)+Aβ, α) [ ρa)i Aα) ] Aα, β) is nonnegative irreducible and strictly product γ -diagonally dominant, and so is P t A/α) Competing interests The authors declare that they have no competing interests Authors? contributions All authors contributed equally to the writing of this paper All authors read and approved the final manuscript Acknowledgements The author thanks the referee for the very helpful comments and suggestions This work was supported in part by National Natural Science Foundation of China ), National Natural Science Foundation of China ), National Natural Science Foundation for Youths of China ), the Key Project of Hunan Provincial Natural Science Foundation of China 2015JJ2134), the Key Project of Hunan Provincial Education Department of China 12A137) and the Hunan Provincial Innovation Foundation for Postgraduate CX2014B254) Received: 2 September 2014 Accepted: 13 December 2014 References 1 Meyer, CD: Uncoupling the Perron eigenvector problem Linear Algebra Appl 114/115, ) 2 Kirkland, SJ, Neumann, M, Xu, JH: A divide and conquer approach to computing the mean first passage matrix for Markov chains via Perron complement reductions Numer Linear Algebra Appl 8, ) 3 Lu, LZ: Perron complement and Perron root Linear Algebra Appl 341, ) 4 Yang, ZM: Some closer bounds of Perron root basing on generalized Perron complement J Comput Appl Math 235, ) 5 Neumann, M, Xu, JH: On the stability of the computation of the stationary probabilities of Markov chains using Perron complements Numer Linear Algebra Appl 10, ) 6 Neumann, M: Inverses of Perron complements of inverse M-matrices Linear Algebra Appl 313, ) 7 Fallat, SM, Neumann, M: On Perron complements of totally nonnegative matrices Linear Algebra Appl 327, ) 8 Zhou, SW, Huang, TZ: On Perron complements of inverse N 0 -matrices Linear Algebra Appl 434, ) 9 Liu, JZ, Huang, ZH, Zhu, L, Huang, ZJ: Theorems on Schur complement of block diagonally dominant matrices and their application in reducing the order for the solution of large scale linear systems Linear Algebra Appl 435, ) 10 Gan, TB, Huang, TZ, Gao, J: A note on generalized nonlinear diagonal dominance J Math Anal Appl 313, ) 11 Horn, RA, Johnson, CR: Topics in Matrix Analysis Cambridge University Press, New York 1991) 12 Liu, JZ, Huang, ZJ: The Schur complements of γ -diagonally and product γ -diagonally dominant matrix and their disc separation Linear Algebra Appl 432, )