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Invertible Matrices in Certain Commutative Subsemirings of Full Matrix Semirings

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Invertible Matrices in Certain Commutative Subsemirings of Full Matrix Semirings International Journal of Pure and Applied Mathematics Volume 106 No. 1 2016, 191-197 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu AP doi: 10.12732/ijpam.v106i1.14 ijpam.eu INVERTIBLE MATRICES IN CERTAIN COMMUTATIVE SUBSEMIRINGS OF FULL MATRIX SEMIRINGS N. Sirasuntorn1, R.I. Sararnrakskul2 § 1,2Department of Mathematics Faculty of Science Srinakharinwirot University 114 Sukhumvit 23, Wattana District, Bangkok 10110, THAILAND Abstract: Then every field is a semifield. For a semifield S, we let Dn(S) denote the set of all A ∈ Mn(S) of the form x1 0 ··· 0 y1 3 0 x2 ··· y2 0 ··············· 0 y2 ··· x2 0 y1 0 ··· 0 x1 where Mn(S) is the full n×n matrix semiring over S. Then Dn(S) is a maximal commutative subsemiring of the semiring Mn(S). If S is a field, it is known that A ∈ Dn(S) is invertible if and only if det A =6 0. In this paper, invertible matrices in Dn(S) where S is a semifield which is not a field are characterized. It is shown that if S is a semifield which is not a field, then A ∈ Dn(S) is an invertible matrix over S if and only if (xi = 0 if and only if yi =6 0). AMS Subject Classification: 16Y60, 20M17, 15A09 Key Words: semiring, semifield, full matrix semiring, invertible matrix 1. Introduction and Preliminaries A semiring S is an algebraic structure (S, +, ·) such that (S, +) and (S, ·) are semigroups and · is distributive over +. A semiring (S, +, ·) is called additively Received: Octobere 21, 2015 c 2016 Academic Publications, Ltd. Published: February 3, 2016 url: www.acadpubl.eu §Correspondence author 192 N. Sirasuntorn, R.I. Sararnrakskul [multiplicatively] commutative if x + y = y + x [x · y = y · x] for all x, y ∈ S. We say that (S, +, ·) is commutative if it is both addtitively commutative and multiplicatively commutative. An element 0 of S is called a zero of the semiring (S, +, ·) if x + 0 = x = 0+ x and x ·0 = 0 = 0·x for all x ∈ S and by an identity of (S+, ·) we mean an element 1 ∈ S such that x · 1 = 1 · x = x for all x ∈ S. Note that a zero and an identity of a semiring are unique. If a semiring (S, +, ·) has a zero 0 [an identity 1], we say that an element x ∈ S is additively [multiplicatively] invertible over S if there exists an element y ∈ S such that x + y = y + x = 0 [x · y = y · x = 1]. Note that such a y is unique and may be written as −x [x−1]. A commutative semiring (S, +, ·) with zero 0 and identity 1 is called a semifield if (S \{0}, ·) is a group. Then every field is a semifield. It is clearly seen that the following fact holds in any semifield. Proposition 1.1. If S is a semifield, then for all x, y ∈ S, xy = 0 implies that x = 0 or y = 0. Example 1.2 ([2]). Let R be the set of all real numbers, Q the set of + + + + rational numbers, R = {x ∈ R | x > 0}, R0 = R ∪{0}, Q = {x ∈ Q | x > 0} + + + + and Q0 = Q ∪ {0}. Then (R0 , +, ·) and (Q0 , +, ·) are semifields which are not fields. + + We note that both (R0 , +, ·) and (Q0 , +, ·) are semifields which are not fields. These semifields have the property that 0 is the only additively invertible element, that is, for x, y ∈ S, x + y = 0 implies x = y = 0. In fact, this property is generally true. Proposition 1.3 ([3]). If S is a semifield which is not a field, then 0 is the only additively invertible element of S. A maximal commutative subsemiring of a semiring S is defined naturally to be a maximal element of the set of all proper commutative subsemirings of S under inclusion. If S is a noncommutative ring, then a maximal commutative subsemiring of S is a maximal element of the set of all commutative subsemiring of S under inclusion. For a positive integer n and an additively commutative semiring S with zero, let Mn(S) be the set of all n × n matrices over S. Then under the usual addition and multiplication of matrices, Mn(S) is also an additively commu- tative semiring with zero and the n × n zero matrix over S is the zero of the matrix semiring Mn(S). For A ∈ Mn(S) and i, j ∈ {1, 2, . , n}, let Aij be the entry of A in the ith row and the jth column. In 2010, Sararnrakskul, Lertvijitsilp, Wassanawichit and Pianskool [1] prove INVERTIBLE MATRICES IN CERTAIN COMMUTATIVE... 193 that the ring Dn(R) of all A ∈ Mn(R) of the form x1 0 ··· 0 y1 0 x2 ··· y2 0 ··············· 0 y ··· x 0 2 2 y 0 ··· 0 x 1 1 is a maximal commutative subring of the ring Mn(R) where R is a commutative ring. Let S = (S, +, ·) be a commutative semiring with zero 0 and identity 1. An n × n matrix A over S is called invertible over S if there is an n × n matrix B over S such that AB = BA = In where In is the identity n × n matrix over S. Note that such a B is unique. The purpose of this paper is to show that when a square matrix in Dn(S) is invertible over S where S is a semifield. Moreover, Dn(S) is a maximal commutative subsemiring of the semiring Mn(S). 2. The Subsemiring Dn(S) of Mn(S) n From now on, let S be a semifield. Let n ∈ N \{1} and Λ = {1, 2,..., 2 }. Note 2.1. For A ∈ Mn(S),A ∈ Dn(S) if and only if (i) Aii = An−i+1,n−i+1 and Ai,n−i+1 = An−i+1,i for all i ∈ Λ and (ii) Aij = 0 for all i, j ∈ {1, 2, . , n} with j 6= i and j 6= n − i + 1. By the proof of Lemma 2.2 and Theorem 2.3 in [1], we have more generalized result for Dn(S) where S is a semifield as the following theorem. Theorem 2.2. The set Dn(S) is a maximal commutative subsemiring of the semiring Mn(S). It is well-known that a square matrix A over a field F is invertible if and only if det A 6= 0. Therefore if S is a field and A ∈ Dn(S) then A is invertible if and only if det A 6= 0. For this reason if S is a semifield which is not a field and A ∈ Dn(S) when A is invertible. So, we characterize invertible matrices in a commutative subsemiring of the semiring Mn(S) where S is a semifield which is not a field. Theorem 2.3. Let S be a semifield which is not a field. Then A ∈ Dn(S) is invertible if and only if every row and every column of A contains exactly one nonzero element, that is, for each i ∈ Λ,Aii = 0 if and only if Ai,n−i+1 6= 0. 194 N. Sirasuntorn, R.I. Sararnrakskul Proof. It is evident if n = 1. Assume that n > 1 and A ∈ Dn(S) is invertible. Let B ∈ Dn(S) such that AB = BA = In. Then we have the following equalities for i ∈ Λ, n 1 = (AB)ii = AikBki Xk=1 = AiiBii + Ai,n−i+1Bn−i+1,i (1) and n 0 = (AB)i,n−i+1 = AikBk,n−i+1 Xk=1 = AiiBi,n−i+1 + Ai,n−i+1Bn−i+1,n−i+1 (2) By (2) and Proposition 1.3 we obtain that AiiBi,n−i+1 = 0 and Ai,n−i+1Bn−i+1,n−i+1 = 0 (3) If Aii = 0, then by (1) Ai,n−i+1 6= 0. Assume that Ai,n−i+1 6= 0 and Aii 6= 0. By (3), Bi,n−i+1 = 0 and Bn−i+1,n−i+1 = 0. Since Bii = Bn−i+1,n−i+1 = 0 and Bn−i+1,i = Bi,n−i+1 = 0, by (1) we have AiiBii + Ai,n−i+1Bn−i+1,i = 0 a contradiction. So, we have if Ai,n−i+1 6= 0 then Aii = 0. Conversely, assume that for each i ∈ Λ,Aii = 0 if and only if Ai,n−i+1 6= 0. Define B ∈ Dn(S) by −1 Aij if Aij 6= 0, Bij = (0 if Aij = 0. We will show that AB = In. The following equalities for i ∈ Λ: n (AB)ii = AikBki Xk=1 = AiiBii + Ai,n−i+1Bn−i+1,i = AiiBii + Ai,n−i+1Bi,n−i+1 Ai,n−i Bi,n−i if Aii = 0 = +1 +1 (AiiBii if Aii 6= 0 = 1. INVERTIBLE MATRICES IN CERTAIN COMMUTATIVE... 195 and n (AB)i,n−i+1 = AikBk,n−i+1 Xk=1 = AiiBi,n−i+1 + Ai,n−i+1Bn−i+1,n−i+1 Ai,n−i Bn−i ,n−i if Aii = 0 = +1 +1 +1 (AiiBi,n−i+1 if Aii 6= 0 0 if Aii = 0 (∵ Bn−i ,n−i = Bii = Aii = 0) = +1 +1 (0 if Aii 6= 0 (∵ Bi,n−i+1 = Ai,n−i+1 = 0). Also if i, j ∈ {1, 2, . , n} are such that j 6= i and j 6= n − i + 1, then by Note 2.1, we have n (AB)ij = AikBkj Xk=1 = AiiBij + Ai,n−i+1Bn−i+1,j = Aii0 + Ai,n−i+10 = 0.
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