Discrete Dynamics in Nature and Society Volume 015, Article ID 96018, 6 pages http://dx.doi.org/10.1155/015/96018 Research Article On Factorizations of Upper Triangular Nonnegative Matrices of Order Three Yi-Zhi Chen Department of Mathematics, Huizhou University, Huizhou, Guangdong 516007, China Correspondence should be addressed to Yi-Zhi Chen; yizhichen1980@16.com Received February 015; Revised 18 March 015; Accepted 18 March 015 Academic Editor: Ivan Area Copyright 015 Yi-Zhi Chen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let T (N 0 ) denote the semigroup of upper triangular matrices with nonnegative integral-valued entries. In this paper, we investigate factorizations of upper triangular nonnegative matrices of order three. Firstly, we characterize the atoms of the subsemigroup S of the matrices in T (N 0 ) with nonzero determinant and give some formulas. As a consequence, problems 4a and 4c presented by Baeth et al. (011) are each half-answered for the case n=. And then, we consider some factorization cases of matrix A in S with ρ(a) = 1 and give formulas for the minimum factorization length of some special matrices in S. 1. Introduction and Preliminaries Upper triangular matrices are an important class of matrices, whichleadtoabroaderstudyofallintegral-valuedmatrices. There are many papers in the literature considering these matrices and similar topics. Factoring such matrices plays a vital role in the study of upper triangularmatrices (see [1]). The problem of factoring matrices was studied by Cohn [] as early as 196. Later, Jacobson and Wisner [, 4] and Ch uan and Chuan [5, 6] investigated these factorization problems in the context of semigroups of matrices. Motivated by these results, Baeth et al. [7] applied the concepts of contemporary factorization theory to semigroups of integral-valued matrices and calculated certain important invariants to give a sense of how unique or nonunique factorization is in each of these semigroups.in [7], Baeth et al. presented six open problems. In this paper, we will investigate factorizations of upper triangular nonnegative matrices of order three. Also, we will consider open problem 4 presented in [7]. Throughout this paper, N will denote the set of all positive integers and N 0 = N {0}. Also, T (N 0 ) will denote the semigroup of upper triangular matrices with nonnegative integral-valued entries. In the following, analogous with [7] or[8], some concepts and preliminaries are recalled. A semigroup is a pairing (S, ) where S is a set and is an associative binary operation on S. When the binary operation is clear from context and A, B S, we will simply write AB instead of A B.IfS contains an element I such that AI=IA= A for all A S,thenI is the identity of S. Let S be a semigroup with identity I.AnelementA Sis a unit of S if there exists an element B Ssuch that AB = BA = I.AnonunitA Sis called an atom of S if whenever A=BC for some elements B,C S,eitherB or C is a unit of S. The semigroup S is said to be atomic provided each nonunit element in S canbewrittenasaproductofatomsofs. Let S denote an atomic semigroup and let A be a nonunit element of S.Theset L (A) = {t :A=A 1 A A t with each A i an atom of S} (1) is called the set of lengths of A. We denote by L(A) = sup L(A) the longest (if finite) factorization length of A and l(a) = min L(A) the minimum factorizationlength ofa. The elasticity of A, denoted by ρ (A) = L (A) l (A), ()
Discrete Dynamics in Nature and Society givesacoarsemeasureofhowfarawaya is from having unique factorization. It is not hard to see that if A has a unique factorization A=A 1 A A t,whencel(a) = {t} and so l (A) =L(A) =t, ρ(a) = t =1. () t The elasticity of the semigroup S, denoted by ρ(s),isgivenby ρ (S) = sup {ρ (A) :A S}. (4) If L(A) = {t 1,t,...}with t i <t i+1 for each i, then the delta set of A is given by Δ (A) ={t i+1 t i :t i,t i+1 L (A)} (5) and Δ(S) = A S Δ(A). This paper will be divided into two sections. In Section, we will consider the semigroup S of upper triangular matrices with nonnegative entries and nonzero determinant. Firstly, we characterize the atoms of the subsemigroup S of the matrices in T (N 0 ) with nonzero determinant and give some formulas. As a result, problems 4a and 4c presented by Baeth et al. in [7] are each half-answered for the case n=. And then, we consider some factorization cases of matrix A in S with ρ(a) = 1 andgiveformulasfortheminimum factorization length of some special matrices in S.. Upper Triangular Nonnegative Matrices of Order Three In this section we consider the semigroup S of upper triangular matrices with nonnegative entries and nonzero determinant. In this case, I is the only unit of S. For each pair i, j {1,, }, lete i,j denote the matrix whose only nonzero entry is e ij =1. The following theorem gives some characterizations about the atoms of S. Theorem 1. Let S denotethesubsemigroupofthematricesin T (N 0 ) with nonzero determinant. The set of atoms of S consists of the matrices X 1 =( 0 1 0), X 1 =( 0 1 0) or X =( 0 1 1) and, for each prime p,thematrices p 0 0 Y 11 =( 0 1 0), Y =( 0 p 0) or Y =( 0 1 0). 0 0 p (6) (7) Proof. Suppose that X ij =X 1 X for some X 1,X S.Since det(x ij )=1, det(x 1 )=det(x )=1and we can write 1 b 1 b 1 X ij =X 1 X =( 0 1 b )( 0 1 c ), (8) 1 c 1 c 1 where k=1 b ikc kj =1and k=1 b hkc kl =0if h =iand l =j. As a result, either X 1 or X is the identity and hence X ij is an atom. Note that the proofs that Y ii (1 i )are atoms are similar; we only prove that Y is an atom in the following. Suppose now that p is prime and Y = Y 1 Y, Y 1,Y S. Since p is prime, either or 1 b 1 b 1 Y 1 Y =( 0 p b )( 0 1 c ) (9) 1 b 1 b 1 1 c 1 c 1 Y 1 Y =( 0 1 b )( 0 p c ). (10) 1 c 1 c 1 In either case, b ij =c ij =0for 1 i<j.consequently, either Y 1 or Y is the identity and hence Y ii is an atom of S. Finally, we will show that these are the only atoms of S. For any we can write A=( a 11 a 1 a 1 0 a a 0 0 a ) S, (11) A= [I +(a 1)E, ][I +a E, ][I +a 1 E 1, ] [I +(a 1)E, ][I +a 1 E 1, ][I +(a 11 1)E 1,1 ] 1 0 a 1 =( 0 1 0 )( 0 1 a )( 0 1 0 ) 0 0 a 1 a 1 0 a 1 ( 0 a 0)( 0 1 0). (1) Thus, the set of atoms of S consists of the matrices A=I +E i,j for each pair i and j with 1<i<j<or A=I +(p 1)E i,i for some prime p and 1 i. Recall that a unitriangular matrix is a matrix in T (N 0 ) whose diagonal elements are all 1s. Denote Σ(A) = Σ 1 i<j a ij.fromtheproofoftheorem 1,wecanimmediately obtain the following corollary.
Discrete Dynamics in Nature and Society Corollary. Let S denote the unitriangular matrices in T (N 0 ) and A S.ThenA is an atom if and only if Σ(A) = 1. Hereafter, for any given A T (N 0 ) with nonzero determinant, we let r(a) denote the number of (not necessarily distinct) prime factors of det(a). Proposition. Let S denote the subsemigroup of the matrices in T (N 0 ) with nonzero determinant. If A can be factored as A=A 1 A A t with each A i being an atom of S, thent= r(a) + k,where { { }} k= i:a { i ( 0 1 0),( 0 1 0),( 0 1 1). { }} { { }} (1) Proof. For each i, A i is an atom and thus det(a i ) is either 1 or prime. Since we have det (A) = det (A 1 ) det (A ) det (A t ), (14) {i:det (A i) is prime} =r(a). (15) If k = {i : det(a i )=1}, then the length of this factorization of A is (5) if a a a (a >0), a 1 =a 1 =0,andthen l (A) =r(a) +1, ρ(a) = r (A) +a r (A) +1. (0) Proof. (1) Suppose that A=A 1 A A t with each A i being an atom of S.ByProposition, t=r(a)+kwhere k= {i:a i {I +E 1,I +E 1,I +E }}. (1) Note that the numbers of factors I +E 1 (I +E 1,I +E, resp.) of A are not more than a 1 (a 1,a, resp.). Then we have and thus k a 1 +a 1 +a =Σ(A) () L (A) r(a) +Σ(A). () On the other hand, from the proof of Theorem 1, we know that A=A 1 A A A 4 A 5 A 6 =[I +(a 1)E, ][I +a E, ] [I +a 1 E 1, ][I +(a 1)E, ][I +a 1 E 1, ] (4) t= {i : det (A i) is prime} + {i : det (A i)=1} =r(a) +k. (16) and then, [I +(a 11 1)E 1,1 ], Lemma 4 (see [7, Theorem4.4]). Let S denote the subsemigroup of T n (N 0 ) of unitriangular matrices and let A S.Then L(A) = Σ(A). Theorem 5. Let S denotethesubsemigroupofthematricesin T (N 0 ) with nonzero determinant and A=( a 11 a 1 a 1 Then the following statements hold: 0 a a 0 0 a ) S. (17) (1) L(A) = r(a) + a 1 +a 1 +a ; () if a 1 =a 1 =a =0,thenl(A) = r(a) = L(A) and ρ(a) = 1; () if a 1 a 11 a (a 1 >0), a 1 =a =0,andthen l (A) =r(a) +1, ρ(a) = r (A) +a 1 r (A) +1 ; (18) (4) if a 1 a 11 a (a 1 >0), a 1 =a =0,andthen l (A) =r(a) +1, ρ(a) = r (A) +a 1 r (A) +1 ; (19) L (A) L(A 1 )+L(A )+L(A )+L(A 4 ) +L(A 5 )+L(A 6 ) =r(a 1 )+r(a 4 )+r(a 6 )+L(A )+L(A )+L(A 5 ) =r(a) +Σ 1 i<j n a ij =r(a) +Σ(A). (5) Thus, combining () and (5),wehaveL(A) = r(a) + Σ(A). () Suppose that A is a diagonal matrix; that is, a ij =0 for all i, j {1,, } with i<j,andwritet=r(a)+kas in Proposition.Ifk 1,thenA contains at least one factor of X 1 =I +E 1, X 1 =I +E 1,or X =I +E,and then there is at least one superdiagonal entry of A that is not 0. This contradicts with the fact that A is a diagonal matrix. Thus, t = r(a) = l(a) = L(A) and, in this case, ρ (A) = L (A) l (A) =1. (6) () Suppose that a 1 a 11 a (a 1 >0), a 1 =a =0. Denote l(a) = r(a) + k as in Proposition.Notice that r(a) isthenumberof(notnecessarilydistinct)primefactorsof det(a) and a 1 > 0.Thenk 1,andsol(A) r(a) + 1. Since a 1 >0, a 1 a 11 a, a 1 can be factored as a product of one factor of a 11 and one factor of a ;saya 1 =a 11 a where
4 Discrete Dynamics in Nature and Society a 11 =m 11 a 11 and a =n a forsomepositiveintegersm 11 and n. Factor A as A=A 1 A A 8 a 1 0 =( 0 n 0)( 0 1 0 ) 0 0 a a 1 m 1 ( 0 a 0)( 0 1 0) ( 0 a 0 ). (7) By the above factorization of A and (), it is not hard to see that l (A) l(a 1 )+l(a )+ +l(a 8 )=r(a) +1. (8) Thus, l(a) = r(a) + 1. In this case we immediately get ρ (A) = r (A) +Σ(A) r (A) +1 = r (A) +a 1 r (A) +1. (9) Hence, () holds. Similarly, we can show that (4) and (5) hold. Remark 6. Given A S, Theorem 5 provides a formula for L(A). Consequently, open problems 4a and 4c in [7] are each half-answered for the case n=. Theorem 7. Let S denote the subsemigroup of the matrices in T (N 0 ) with nonzero determinant and A=( a 11 a 1 a 1 If A satisfies one of the following conditions, 0 a a 0 0 a ) S. (0) Proof. (1) If a 1 =a 1 =a =0,byTheorem 5, l(a) = r(a) = L(A).Inthiscase,ρ(A) = 1. () Suppose that a 11 =a =a =1and a 1 =0.Ifa 1 = a =0, by (1), the conclusion holds. Now, assume that a 1 = a =0;thecasesofthata 1 =0, a =0or a 1 =0, a = 0 can be proved similarly. In this case, X 1 is not an atomic factor of A and X 1 and X are atomic factors of A. Note that X 1 can commute with X ;thenaonly has the following factorization: a 1 ( 0 1 1) a, (1) with ρ(a) = 1 and the minimum factorization length l(a) = a 1 +a. Analogous with the proof of (), we can show that (4) holds. () Suppose that a 11 =a =a =1and a 1 =0.Ifa 1 = a =0, by (1), the conclusion holds. Now, assume that a 1 = a =0;thecasesofthata 1 =0, a =0or a 1 =0, a = 0 can be proved similarly. In this case, X 1 is not an atomic factor of A and X 1 and X are atomic factors of A.Notethat X 1 X = X X 1 X 1 ; the factorizations of A must satisfy that all the X matrices are before X 1 matrices, and then A only has the following factorization: A=( 0 1 1) a ( 0 1 0) a 1, () with ρ(a) = 1 and the minimum factorization length l(a) = a 1 +a. (5) Suppose that a 1 = a = 1 and a 1 = 0.Inthis case, X 1 is not an atomic factor of A, andx 1 and X are atomic factors of A. Assume that A canbefactoredasa= A 1 A A t with each A i being an atom; then t=r(a)+k, where k= {i : A i {X 1,X }}. () By Theorem 5 (1), k a 1 +a +a 1 =. On the other hand, since a 1 =a =1, X 1, X are atomic factors of A, and we have k.thus,k=.inthiscase,a can factorize as follows: A= [I +(a 1)E, ][I +E, ][I +(a 1)E, ] (1) a 1 =a 1 =a =0; () a 11 =a =a =1, a 1 =0; () a 11 =a =a =1, a 1 =0; (4) a 11 =a =a =1, a =0; (5) a 1 =a =1, a 1 =0; (6) a 1 =a =1, a =0; (7) a 1 =a =1, a 1 =0, then ρ(a) = 1. [I +E 1, ][I +(a 11 1)E 1,1 ] =( 0 1 0 )( 0 1 1)( 0 a 0) 0 0 a a 1 ( 0 1 0). (4)
Discrete Dynamics in Nature and Society 5 So if we give the atomic factorizations of [I +(a 1)E, ], [I +(a 1)E, ],and[i +(a 11 1)E 1,1 ],thenwewillobtain the atomic factorization of A with ρ(a) = 1 and the minimum factorization length l(a) = r(a) +. Analogous with the proof of (5), we can show that (6) and (7) hold. Example 8. Consider 1 0 A=( 0 1 ) T (N 0 ). (5) Then by Theorem 7 (), we can obtain the atomic factorization of A with ρ(a) = 1 and the minimum factorization length l(a)=+=5as follows: Example 9. Consider ( 0 1 1). (6) 6 1 0 A=( 0 4 1) T (N 0 ). (7) 0 0 8 Then by Theorem 7 (5), we can obtain the atomic factorization of A with ρ(a) = 1 and the minimum factorization length l(a) = r(a) + = 9 as follows: A= ( 0 1 0) ( 0 1 1)( 0 0) 0 0 0 0 0 0 ( 0 1 1)( 0 1 0). (8) Theorem 10. Let S denote the unitriangular matrices in T (N 0 ) and Then we have 1 a 1 a 1 A=( 0 1 a ) S, a 1,a 1,a >0. (9) l (A) = { a 1 +a, if a 1 a 1 a, { a { 1 +a +a 1 a 1 a, if a 1 a 1 a. (40) Proof. It is a routine way to check that X 1 X 1 = X 1 X 1, X X 1 = X 1 X,andX 1 X = X X 1 X 1.Hencethe maximum length is achieved by putting all the X matrices before the X 1 matrices. In this case, we have the form X a Xa 1 1 Xa 1 1. The minimum length is achieved by having as fewaspossiblex 1 matrices, that is, by putting X matrices after X 1 matrices. If a 1 a 1 a,wecanrunoutofx 1 matrices, leaving a form X a 1 1 Xa X 1X a Xa 1 1,wherethese powers satisfy a 1 +1+a 1 =a 1, a +a =a,anda 1 a + a =a 1.And,inthiscase,wehavel(A) = a 1 +a.ifa 1 a 1 a,thenwecanputallthex matrices after all of the X 1 matrices, getting the form X a 1 1 Xa Xa 1 a 1 a 1.Thus,in this case, we have l(a) = a 1 +a +a 1 a 1 a. Example 11. Consider 1 A=( 0 1 ) T (N 0 ). (41) Then by Theorem 10, we can obtain a factorization of A with the minimum factorization length l(a) = a 1 +a =5as follows: A=( 0 1 1) ( 0 1 0) ( 0 1 0). (4) Example 1. Consider 1 6 A=( 0 1 ) T (N 0 ). (4) Then by Theorem 10, we can obtain a factorization of A with the minimum factorization length l(a) = a 1 +a =5as follows: Example 1. Consider ( 0 1 1). (44) 1 7 A=( 0 1 ) T (N 0 ). (45) Then by Theorem 10, we can obtain a factorization of A with the minimum factorization length l(a) = a 1 +a +a 1 a 1 a =6as follows: ( 0 1 1) ( 0 1 0). (46) Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper.
6 Discrete Dynamics in Nature and Society Acknowledgments The author would like to extend their sincere gratitude to the referee for his instructive advice and useful suggestions that contributed to this paper. This research was supported by grantsofthennsfofchina(nos.1140146,114611),thensf of Guangdong Province (nos. 014A0010087, 014A0010119), the NSF of Fujian Province (no. 014J01019), the Outstanding Young Innovative Talent Training Project in Guangdong Universities (no. 01LYM0086), and Science and Technology PlanProjectofHuizhouCity. References [1] V. Halava and T. Harju, On Markov s undecidability theorem for integer matrices, Semigroup Forum, vol. 75, no. 1, pp. 17 180, 007. [] P. M. Cohn, Noncommutative unique factorization domains, Transactions of the American Mathematical Society,vol.109,pp. 1 1, 196. [] B. Jacobson, Matrix number theory: an example of nonunique factorization, The American Mathematical Monthly,vol.7,no. 4, pp. 99 40, 1965. [4] B.JacobsonandR.J.Wisner, MatrixnumbertheoryI.Factorization of unimodular matrices, Publicationes Mathematicae Debrecen,vol.1,pp.67 7,1966. [5] J. C. Ch uan and W. F. Chuan, Factorizations in a semigroup of integral matrices, Linear and Multilinear Algebra,vol.18,no., pp.1,1985. [6] J. C. Ch uan and W. F. Chuan, Factorability of positive-integral matrices of prime determinants, Bulletin of the Institute of Mathematics. Academia Sinica,vol.14,no.1,pp.11 0,1986. [7] N. Baeth, V. Ponomarenko, D. Adams et al., Number theory of matrix semigroups, Linear Algebra and Its Applications, vol. 44, no., pp. 694 711, 011. [8] D. Adams, R. Ardila, D. Hannasch et al., Bifurcus semigroups and rings, Involve, vol., no., pp. 51 56, 009.
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