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Note on the free sets and free subsemimodules in semimodules
Linear Algebra and its Applications 520 (2017) 125–133
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Note on the free sets and free subsemimodules in semimodules ✩ Qian-yu Shu, Xue-ping Wang ∗ College of Mathematics and Software Science, Sichuan Normal University, Chengdu, Sichuan 610066, People’s Republic of China
a r t i c l e
i n f o
Article history: Received 26 October 2016 Accepted 12 January 2017 Available online 17 January 2017 Submitted by R. Brualdi
a b s t r a c t Some conditions that a finite set in an L-semimodule is free are established. This gives an answer to an open problem raised by Tan (2016) in his work [7]. © 2017 Elsevier Inc. All rights reserved.
MSC: 15A03 16Y60 Keywords: Semiring Semimodule Free set Free basis
1. Introduction The study of semimodules over semirings has a long history. In 1966, Yusuf [9] introduced the concept of an inverse semimodule over a semiring and obtained some analogues to theorems in module theory for inverse semimodules (note that an inverse semimodule M is a semimodule in which the monoid (M, +) is an inverse semigroup). Since then, ✩
Supported by National Natural Science Foundation of China (No. 11401410).
* Corresponding author. Fax: +86 28 84761393. E-mail addresses: [email protected] (Q.-y. Shu), [email protected] (X.-p. Wang). http://dx.doi.org/10.1016/j.laa.2017.01.012 0024-3795/© 2017 Elsevier Inc. All rights reserved.
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Q.-y. Shu, X.-p. Wang / Linear Algebra and its Applications 520 (2017) 125–133
a number of works on semimodule theory were published (see e.g. [1–3,7]). We know that a semimodule structure is the one which arises naturally in the properties of sets of vectors with entries in a semiring. Thus, they turn out to be analogues for algebraic structure on semirings to the concept of a module for rings. However, we must be careful since not all properties are transferred in a straightforward way. For example, in general, a system of linearly independent vectors cannot be extended to a basis of a semimodule (see [2], Theorem 4). Some of facts known about free sets in modules have not yet been proved in semimodule, one of them is for a given finitely generated free semimodule over a commutative semiring, find a necessary and sufficient condition for a nonempty finite subset in this semimodule to be free (see [7], Open problem 3.1). In this note, we give conditions under which a nonempty finite subset in a free semimodule is free. So that we give an answer to Open problem 3.1 in [7]. 2. Definition and previous results In this section, we collect only some necessary notions for the presentations of the main result in the next section. Definition 2.1 (Golan [3]). A semiring L = L, +, ·, 0, 1 is an algebraic structure with the following properties: (i) (ii) (iii) (iv) (v)
(L, +, 0) is a commutative monoid, (L, ·, 1) is a monoid, r · (s + t) = r · s + r · t and (s + t) · r = s · r + t · r hold for all r, s, t ∈ L, 0 · r = r · 0 = 0 holds for all r ∈ L, 0 = 1.
A semiring L is commutative if r · r = r · r for all r, r ∈ L. Definition 2.2 (Golan [3]). Let L = L, +, ·, 0, 1 be a semiring. A left L-semimodule is a commutative monoid (M, +) with additive identity 0 for which we have a function L × M → M, denoted by (λ, a) → λa and called a scalar multiplication, which satisfies the following conditions for all λ, μ in L and a, b in M: (i) (ii) (iii) (iv) (v)
(λμ)a = λ(μa), λ(a + b) = λa + λb, (λ + μ)a = λa + μa, 1a = a, λ0 = 0 = 0a.
The definition of a right L-semimodule is analogous. In this paper, L-semimodules will always mean left L-semimodules. L-semimodules were studies in [2,5] under the name L-semilinear spaces.
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For convenience, we use N to denote the set of all positive integers. Let n = {1, · · · , n} for n in N. Then we can construct an L-semimodule as follows. Example 2.1 (Shu and Wang [5]). Let L = L, +, ·, 0, 1 be a semiring. For each n 1, let Vn (L) = {(a1 , a2 , · · · , an )T : ai ∈ L, i ∈ n}. Define x + y = (x1 + y1 , x2 + y2 , · · · , xn + yn )T , rx = (r · x1 , r · x2 , · · · , r · xn )T for all x = (x1 , x2 , · · · , xn )T , y = (y1 , y2 , · · · , yn )T ∈ Vn (L) and r ∈ L, where (x1 , x2 , · · · , xn )T denotes the transpose of (x1 , x2 , · · · , xn ). Then Vn (L) is an L-semimodule. Similarly, we can define the operations of addition and external multiplication on row vectors and obtain that V n (L) is also an L-semimodule, where V n (L) = {(a1 , a2 , · · · , an ) : ai ∈ L, i ∈ n}. In particular, V1 (L) = V 1 (L) = L is an L-semimodule. Let A be a nonempty subset of L-semimodule M. Then the intersection of all subsemimodules of M containing A is a subsemimodule of M, called the subsemimodule generated by A and denoted by Span(A). It is easy to verify that Span(A) = {
k
λi ai : λi ∈ L, ai ∈ A, i ∈ k, k ∈ N}.
i=1
k The expression i=1 λi ai is called a linear combination of the elements a1 , a2 , · · · , ak . If Span(A) = M, then A is called a generating set for M. An L-semimodule having a finite generating set is called finitely generated, otherwise, called infinitely generated. Definition 2.3 (Tan [7]). Let M be an L-semimodule. A nonempty subset A in M is called linearly independent if a ∈ / Span(A\{a}) for any a in A. If A is not linearly independent then it is called linearly dependent. The set A is called free if each element in M can be expressed as a linear combination of elements in A in at most one way. It is clear that any free set is linearly independent. Definition 2.4 (Tan [7]). Let M be an L-semimodule. A linearly independent generating set for M is called a basis for M and a free generating set for M is called a free basis for M. An L-semimodule having a free basis is called a free L-semimodule. Clearly, any finitely generated L-semimodule has at least a basis and any free basis is a basis.
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3. Free set and free L-semimodule In this section, we shall answer Open problem 3.1 raised by Tan in his work [7]. Definition 3.1 (Shu and Wang [6]). In a semiring L, for every element b, c ∈ L we define (b, c) ≡ 0 if and only if b = c. Otherwise, we define (b, c) ≡ 0. Similarly, we have the following definition. Definition 3.2. In an L-semimodule M, for every element b, c ∈ M we define (b, c) ≡ 0 if and only if b = c. Otherwise, we define (b, c) ≡ 0. Let M be an L-semimodule and a1 , a2 , · · · , am ∈ M. Consider the following system of equations: x1 a1 + x2 a2 + · · · + xm am ≡ 0,
(1)
where xi ∈ V 2 (L) is an unknown and ai = 0 for every i ∈ m. Let x0 = (x1 , x2 , · · · , xm )T be a solution of (1). If there exists j ∈ m such that xj ≡ 0, then we say that the equation (1) has a nontrivial solution. Otherwise, we say the equation (1) has only the trivial solution. Definition 3.3 (Golan [3]). An element a in a semiring L is said to be cancellable if and only if a + b = a + c implies b = c for every b, c ∈ L. We denote the set of all cancellable elements of L by K + (L). If K + (L) = L then we say that the semiring L is cancellative. Theorem 3.1. Let M be an L-semimodule and a1 , a2 , · · · , am ∈ M. Then {a1 , a2 , · · · , am } is free if and only if the equation (1) has only the trivial solution. Proof. Let ((k1 , k1 ), (k2 , k2 ), · · · , (km , km ))T be a solution of (1) with ki , ki ∈ L, i ∈ m. Then (k1 , k1 )a1 + (k2 , k2 )a2 + · · · + (km , km )am ≡ 0. From Definition 3.2, we have k1 a1 + k2 a2 + · · · + km am = k1 a1 + k2 a2 + · · · + km am .
Then ki = ki , i ∈ m since vectors a1 , a2 , · · · , am are free. Therefore, the equation (1) has only the trivial solution. Conversely, let k1 a1 + k2 a2 + · · · + km am = k1 a1 + k2 a2 + · · · + km am
with ki , ki ∈ L, i ∈ m. Then (k1 , k1 )a1 + (k2 , k2 )a2 + · · · + (km , km )am ≡ 0,
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i.e., ((k1 , k1 ), (k2 , k2 ), · · · , (km , km ))T is a solution of (1). Since the equation (1) has only the trivial solution, it is clear that ki = ki , i ∈ m, which means that {a1 , a2 , · · · , am } is free in M. 2
Definition 3.4 (Wang and Shu [8]). Let (a, b), (c, d) ∈ V 2 (L). Define that (a, b)(c, d) = (ac + bd, ad + bc). Let A = (aij ) ∈ Mn×m (L). Denote the column vectors of A by a1 , a2 , · · · , am . Consider the following system of equations in V 2 (L): ⎧ ⎪ ⎪ a11 x1 + · · · + a1m xm ≡ 0, ⎪ ⎨ a x + · · · + a x ≡ 0, 21 1 2m m ⎪··· ⎪ ⎪ ⎩ a x + · · · + a x ≡ 0, n1 1 nm m
(2)
which is equivalent to the matrix equation Ax ≡ 0 or x1 a1 + x2 a2 + · · · + xm am ≡ 0, where A is a known, xi ∈ V 2 (L) is an unknown and ai = 0 for every i ∈ m. 1 2 ··· n Let A ∈ Mn (L) and σ be the permutation . Then we define the j1 j2 · · · jn positive and negative determinants as |A|+ =
σ=even
a1,j1 a2,j2 · · · an,jn and |A|− =
a1,j1 a2,j2 · · · an,jn .
σ=odd
Moreover, we define the bideterminant det(A) of A is an ordered pair det(A) = (|A|+ , |A|− ). In what follows, denote D = {(a, b) ∈ V 2 (L) : a = b} by D ≡ 0. Let us recall the definition of the McCoy rank of a matrix as follows. Definition 3.5 (Shu and Wang [6]). Let A ∈ Mm×n (L). The McCoy rank of A is the largest t such that AnnihV 2 (L) (Ft (A)) ≡ 0, where Ft (A) is the set of the bideterminants of t × t submatrices of A and AnnihV 2 (L) (Ft (A)) = {r ∈ V 2 (L) : rFt (A) ≡ 0}.
Definition 3.6 (Poplin and Hartwig [4]). If n ∈ N then π(n) =
+ −
if n is even, if n is odd.
We define the (i, j)-minor matrix Aij to be the submatrix obtained by deleting the ith row and the jth column from a matrix A. We may subsequently define the positive and negative minors |Aij |+ and |Aij |− , which for convenience, we abbreviate to A+ ij and A− , respectively. ij
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Lemma 3.1 (Poplin and Hartwig [4]). Let p = q and A ∈ Mn (L). Then n
π(q+k)
apk Aqk
k=1
=
n
π(q+k−1)
apk Aqk
.
k=1
Lemma 3.2. In a commutative semiring L, let A = (aij ) ∈ Mn×m (L) with m ≤ n. If the McCoy rank of A is less than m, then Ax ≡ 0 has a nontrivial solution. Proof. Let the McCoy rank of A be t with t < m. Then AnnihV 2 (L) (Ft+1 (A)) ≡ 0, i.e., there is y ∈ AnnihV 2 (L) (Ft+1 (A)) such that 0 ≡ y. Thus by Definition 3.5, yFt+1 (A) ≡ 0. If t = 0, then yaij ≡ 0 for all i ∈ n, j ∈ m. Therefore, x0 = (y, y, · · · , y)T is a nontrivial solution of the matrix equation Ax ≡ 0. Next, suppose that t > 0. Then the product of y and a certain element of Ft (A) is nonzero since y ≡ 0. Such element of Ft (A) is the bideterminant of a t × t submatrix of A. By permuting rows and columns, we may assume that the submatrix is T with ⎛
a11 ⎜ ⎜ a21 T =⎜ ⎜ .. ⎝ . at1
a12 a22 .. . at2
··· ··· ···
⎞ a1t ⎟ a2t ⎟ .. ⎟ ⎟. . ⎠ att
Consider the submatrix ⎛ ⎜ ⎜ ⎜ T =⎜ ⎜ ⎜ ⎝
a11 a21 .. . at1
a12 a22 .. . at2
at+1,1
at+1,2
··· ··· ··· ···
a1t a2t .. . att at+1,t
⎞ a1,t+1 a2,t+1 ⎟ ⎟ ⎟ .. ⎟. . ⎟ ⎟ at,t+1 ⎠ at+1,t+1
For 1 ≤ k ≤ t + 1, let Tk denote the submatrix of T formed by deleting the row t + 1 and the column k. Consider the vector x = (x1 , x2 , · · · , xm )T with
xi =
y(0, 1)i+t+1 det(Ti ) (0, 0)
1 ≤ i ≤ t + 1, t + 2 ≤ i ≤ m.
Hence t+1 j=1
and
aij (0, 1)j+t+1 det(Tj ) ≡ 0 for 1 ≤ i ≤ t (by Lemma 3.1)
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aij (0, 1)j+t+1 det(Tj )y ≡ 0 for t + 1 ≤ i ≤ n
j=1
since y ∈ AnnihV 2 (L) (Ft+1 (A)) and for t + 1 ≤ i ≤ n. Therefore,
t+1 j=1
aij (0, 1)j+t+1 det(Tj ) is an element of Ft+1 (A)
⎞ ⎛ ⎞ ⎛ t+1 ( j=1 a1j (0, 1)j+t+1 det(Tj ))y 0 ⎟ ⎜ ⎟ ⎜ t+1 j+t+1 ⎟ ⎜0⎟ ⎜( a (0, 1) det(T ))y 2j j j=1 ⎟ ⎜ ⎟ ⎜ Ax = ⎜ ⎟ ≡ ⎜ . ⎟. .. ⎟ ⎝ .. ⎠ ⎜ ⎠ ⎝ . t+1 0 ( j=1 anj (0, 1)j+t+1 det(Tj ))y Finally, x ≡ 0 since xt+1 = y(0, 1)t+1+t+1 det(Tt+1 ) = y det(T ) ≡ 0. 2 Lemma 3.3 (Shu and Wang [6]). In a commutative and cancellative semiring L, let A = (aij ) ∈ Mn (L). Then the McCoy rank of A is less than n if and only if Ax ≡ 0 has a nontrivial solution. Theorem 3.2. In a commutative semiring L, let M be a free L-semimodule with a free basis {b1 , b2 , · · · , bn }. Let ai ∈ M, i ∈ m and (a1 , a2 , · · · , am ) = (b1 , b2 , · · · , bn )A, where A ∈ Mn×m (L) and m ≤ n. Then (1) If {a1 , a2 , · · · , am } is free, then the McCoy rank of A is equal to m. (2) If the semiring L is cancellative and the McCoy rank of A is equal to m, then {a1 , a2 , · · · , am } is free. Proof. Since (a1 , a2 , · · · , am ) = (b1 , b2 , · · · , bn )A, then from x1 a1 + x2 a2 + · · · + xm am ≡ 0 we have x1 (a11 b1 + a21 b2 + · · · + an1 bn ) + x2 (a12 b1 + a22 b2 + · · · + an2 bn ) + · · · + xm (a1m b1 + a2m b2 + · · · + anm bn ) ≡ 0.
(3)
This deduces that (a11 x1 + a12 x2 + · · · + a1m xm )b1 + (a21 x1 + a22 x2 + · · · + a2m xm )b2 + · · · + (an1 x1 + an2 x2 + · · · + anm xm )bn ≡ 0.
(4)
Therefore, from Theorem 3.1, we know that {a1 , a2 , · · · , am } is free if and only if the equation (4) has only the trivial solution, moreover, the equation (4) has only the trivial solution if and only if the equation
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Fig. 1. A commutative semiring which is not cancellative.
⎛
a11 ⎜ ⎜ a21 ⎜ . ⎜ . ⎝ . an1
a12 a22 .. . an2
··· ··· ···
⎞⎛ ⎞ ⎛ ⎞ x1 0 a1m ⎟⎜ ⎟ ⎜ ⎟ a2m ⎟ ⎜ x2 ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ .. ⎟ ⎟⎜ . ⎟ ≡ ⎜ . ⎟ . ⎠ ⎝ .. ⎠ ⎝ .. ⎠ 0 anm xm
(5)
has only the trivial solution, since {b1 , b2 , · · · , bn } is free. Thus {a1 , a2 , · · · , am } is free if and only if the equation (5) has only the trivial solution. (1) Suppose that {a1 , a2 , · · · , am } is free. Then the equation (5) has only the trivial solution. By Lemma 3.2, we know that the McCoy rank of A is equal to m. (2) If the McCoy rank of A is equal to m, then by Definition 3.5, there exists an m × m submatrix B in A such that the McCoy rank of B is m. Further, by Lemma 3.3, the equation Bx ≡ 0 has only the trivial solution, and so is the equation Ax ≡ 0. Consequently, {a1 , a2 , · · · , am } is free. 2 Remark 3.1. In general, the condition that the semiring L is cancellative in Theorem 3.2 (2) can not be deleted. For example, in the semiring as Fig. 1 (just replace the addition and the multiplication in Definition 2.1 by the join and meet in the lattice as Fig. 1, re1 0 ∈ M2 (L). Then the McCoy spectively), it is obvious that K + (L) = L. Let A = b 1 rank of A is equal to 2, but a(1, b)T + c(0, 1)T = a(1, b)T + 0(0, 1)T , i.e., {(1, b)T , (0, 1)T } is not free. Remark 3.2. Theorem 3.2 answered Open problem 3.1 raised by Tan in his work [7] over a commutative semiring. Acknowledgements The authors thank the reviewers for their valuable suggestions. References [1] P. Butkovič, Max-algebra: the linear algebra of combinatorics?, Linear Algebra Appl. 367 (2003) 313–335.
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[2] A. Di Nola, A. Lettieri, I. Perfilieva, V. Novák, Algebraic analysis of fuzzy systems, Fuzzy Sets and Systems 158 (2007) 1–22. [3] J.S. Golan, Semirings and Their Applications, Kluwer Academic Publishers, Dordrecht/Boston/London, 1999. [4] Phillip L. Poplin, Robert E. Hartwig, Determinantal identities over commutative semirings, Linear Algebra Appl. 387 (2004) 99–132. [5] Qian-yu Shu, Xue-ping Wang, Bases in semilinear spaces over zerosumfree semirings, Linear Algebra Appl. 435 (2011) 2681–2692. [6] Qian-yu Shu, Xue-ping Wang, The applications of the bideterminant of a matrix over commutative semirings, Linear Multilinear Algebra (2016), http://dx.doi.org/10.1080/03081087.2016.1242555. [7] Yi-jia Tan, Free sets and free subsemimodules in a semimodule, Linear Algebra Appl. 496 (2016) 527–548. [8] Xue-ping Wang, Qian-yu Shu, Bideterminant and rank of matrix, Soft Comput. 18 (2014) 729–742. [9] S.M. Yusuf, Inverse semimodules, J. Natur. Sci. Math. 6 (1966) 111–117.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Note on the free sets and free subsemimodules in semimodules ✩ Qian-yu Shu, Xue-ping Wang ∗ College of Mathematics and Software Science, Sichuan Normal University, Chengdu, Sichuan 610066, People’s Republic of China
a r t i c l e
i n f o
Article history: Received 26 October 2016 Accepted 12 January 2017 Available online 17 January 2017 Submitted by R. Brualdi
a b s t r a c t Some conditions that a finite set in an L-semimodule is free are established. This gives an answer to an open problem raised by Tan (2016) in his work [7]. © 2017 Elsevier Inc. All rights reserved.
MSC: 15A03 16Y60 Keywords: Semiring Semimodule Free set Free basis
1. Introduction The study of semimodules over semirings has a long history. In 1966, Yusuf [9] introduced the concept of an inverse semimodule over a semiring and obtained some analogues to theorems in module theory for inverse semimodules (note that an inverse semimodule M is a semimodule in which the monoid (M, +) is an inverse semigroup). Since then, ✩
Supported by National Natural Science Foundation of China (No. 11401410).
* Corresponding author. Fax: +86 28 84761393. E-mail addresses: [email protected] (Q.-y. Shu), [email protected] (X.-p. Wang). http://dx.doi.org/10.1016/j.laa.2017.01.012 0024-3795/© 2017 Elsevier Inc. All rights reserved.
126
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a number of works on semimodule theory were published (see e.g. [1–3,7]). We know that a semimodule structure is the one which arises naturally in the properties of sets of vectors with entries in a semiring. Thus, they turn out to be analogues for algebraic structure on semirings to the concept of a module for rings. However, we must be careful since not all properties are transferred in a straightforward way. For example, in general, a system of linearly independent vectors cannot be extended to a basis of a semimodule (see [2], Theorem 4). Some of facts known about free sets in modules have not yet been proved in semimodule, one of them is for a given finitely generated free semimodule over a commutative semiring, find a necessary and sufficient condition for a nonempty finite subset in this semimodule to be free (see [7], Open problem 3.1). In this note, we give conditions under which a nonempty finite subset in a free semimodule is free. So that we give an answer to Open problem 3.1 in [7]. 2. Definition and previous results In this section, we collect only some necessary notions for the presentations of the main result in the next section. Definition 2.1 (Golan [3]). A semiring L = L, +, ·, 0, 1 is an algebraic structure with the following properties: (i) (ii) (iii) (iv) (v)
(L, +, 0) is a commutative monoid, (L, ·, 1) is a monoid, r · (s + t) = r · s + r · t and (s + t) · r = s · r + t · r hold for all r, s, t ∈ L, 0 · r = r · 0 = 0 holds for all r ∈ L, 0 = 1.
A semiring L is commutative if r · r = r · r for all r, r ∈ L. Definition 2.2 (Golan [3]). Let L = L, +, ·, 0, 1 be a semiring. A left L-semimodule is a commutative monoid (M, +) with additive identity 0 for which we have a function L × M → M, denoted by (λ, a) → λa and called a scalar multiplication, which satisfies the following conditions for all λ, μ in L and a, b in M: (i) (ii) (iii) (iv) (v)
(λμ)a = λ(μa), λ(a + b) = λa + λb, (λ + μ)a = λa + μa, 1a = a, λ0 = 0 = 0a.
The definition of a right L-semimodule is analogous. In this paper, L-semimodules will always mean left L-semimodules. L-semimodules were studies in [2,5] under the name L-semilinear spaces.
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For convenience, we use N to denote the set of all positive integers. Let n = {1, · · · , n} for n in N. Then we can construct an L-semimodule as follows. Example 2.1 (Shu and Wang [5]). Let L = L, +, ·, 0, 1 be a semiring. For each n 1, let Vn (L) = {(a1 , a2 , · · · , an )T : ai ∈ L, i ∈ n}. Define x + y = (x1 + y1 , x2 + y2 , · · · , xn + yn )T , rx = (r · x1 , r · x2 , · · · , r · xn )T for all x = (x1 , x2 , · · · , xn )T , y = (y1 , y2 , · · · , yn )T ∈ Vn (L) and r ∈ L, where (x1 , x2 , · · · , xn )T denotes the transpose of (x1 , x2 , · · · , xn ). Then Vn (L) is an L-semimodule. Similarly, we can define the operations of addition and external multiplication on row vectors and obtain that V n (L) is also an L-semimodule, where V n (L) = {(a1 , a2 , · · · , an ) : ai ∈ L, i ∈ n}. In particular, V1 (L) = V 1 (L) = L is an L-semimodule. Let A be a nonempty subset of L-semimodule M. Then the intersection of all subsemimodules of M containing A is a subsemimodule of M, called the subsemimodule generated by A and denoted by Span(A). It is easy to verify that Span(A) = {
k
λi ai : λi ∈ L, ai ∈ A, i ∈ k, k ∈ N}.
i=1
k The expression i=1 λi ai is called a linear combination of the elements a1 , a2 , · · · , ak . If Span(A) = M, then A is called a generating set for M. An L-semimodule having a finite generating set is called finitely generated, otherwise, called infinitely generated. Definition 2.3 (Tan [7]). Let M be an L-semimodule. A nonempty subset A in M is called linearly independent if a ∈ / Span(A\{a}) for any a in A. If A is not linearly independent then it is called linearly dependent. The set A is called free if each element in M can be expressed as a linear combination of elements in A in at most one way. It is clear that any free set is linearly independent. Definition 2.4 (Tan [7]). Let M be an L-semimodule. A linearly independent generating set for M is called a basis for M and a free generating set for M is called a free basis for M. An L-semimodule having a free basis is called a free L-semimodule. Clearly, any finitely generated L-semimodule has at least a basis and any free basis is a basis.
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3. Free set and free L-semimodule In this section, we shall answer Open problem 3.1 raised by Tan in his work [7]. Definition 3.1 (Shu and Wang [6]). In a semiring L, for every element b, c ∈ L we define (b, c) ≡ 0 if and only if b = c. Otherwise, we define (b, c) ≡ 0. Similarly, we have the following definition. Definition 3.2. In an L-semimodule M, for every element b, c ∈ M we define (b, c) ≡ 0 if and only if b = c. Otherwise, we define (b, c) ≡ 0. Let M be an L-semimodule and a1 , a2 , · · · , am ∈ M. Consider the following system of equations: x1 a1 + x2 a2 + · · · + xm am ≡ 0,
(1)
where xi ∈ V 2 (L) is an unknown and ai = 0 for every i ∈ m. Let x0 = (x1 , x2 , · · · , xm )T be a solution of (1). If there exists j ∈ m such that xj ≡ 0, then we say that the equation (1) has a nontrivial solution. Otherwise, we say the equation (1) has only the trivial solution. Definition 3.3 (Golan [3]). An element a in a semiring L is said to be cancellable if and only if a + b = a + c implies b = c for every b, c ∈ L. We denote the set of all cancellable elements of L by K + (L). If K + (L) = L then we say that the semiring L is cancellative. Theorem 3.1. Let M be an L-semimodule and a1 , a2 , · · · , am ∈ M. Then {a1 , a2 , · · · , am } is free if and only if the equation (1) has only the trivial solution. Proof. Let ((k1 , k1 ), (k2 , k2 ), · · · , (km , km ))T be a solution of (1) with ki , ki ∈ L, i ∈ m. Then (k1 , k1 )a1 + (k2 , k2 )a2 + · · · + (km , km )am ≡ 0. From Definition 3.2, we have k1 a1 + k2 a2 + · · · + km am = k1 a1 + k2 a2 + · · · + km am .
Then ki = ki , i ∈ m since vectors a1 , a2 , · · · , am are free. Therefore, the equation (1) has only the trivial solution. Conversely, let k1 a1 + k2 a2 + · · · + km am = k1 a1 + k2 a2 + · · · + km am
with ki , ki ∈ L, i ∈ m. Then (k1 , k1 )a1 + (k2 , k2 )a2 + · · · + (km , km )am ≡ 0,
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i.e., ((k1 , k1 ), (k2 , k2 ), · · · , (km , km ))T is a solution of (1). Since the equation (1) has only the trivial solution, it is clear that ki = ki , i ∈ m, which means that {a1 , a2 , · · · , am } is free in M. 2
Definition 3.4 (Wang and Shu [8]). Let (a, b), (c, d) ∈ V 2 (L). Define that (a, b)(c, d) = (ac + bd, ad + bc). Let A = (aij ) ∈ Mn×m (L). Denote the column vectors of A by a1 , a2 , · · · , am . Consider the following system of equations in V 2 (L): ⎧ ⎪ ⎪ a11 x1 + · · · + a1m xm ≡ 0, ⎪ ⎨ a x + · · · + a x ≡ 0, 21 1 2m m ⎪··· ⎪ ⎪ ⎩ a x + · · · + a x ≡ 0, n1 1 nm m
(2)
which is equivalent to the matrix equation Ax ≡ 0 or x1 a1 + x2 a2 + · · · + xm am ≡ 0, where A is a known, xi ∈ V 2 (L) is an unknown and ai = 0 for every i ∈ m. 1 2 ··· n Let A ∈ Mn (L) and σ be the permutation . Then we define the j1 j2 · · · jn positive and negative determinants as |A|+ =
σ=even
a1,j1 a2,j2 · · · an,jn and |A|− =
a1,j1 a2,j2 · · · an,jn .
σ=odd
Moreover, we define the bideterminant det(A) of A is an ordered pair det(A) = (|A|+ , |A|− ). In what follows, denote D = {(a, b) ∈ V 2 (L) : a = b} by D ≡ 0. Let us recall the definition of the McCoy rank of a matrix as follows. Definition 3.5 (Shu and Wang [6]). Let A ∈ Mm×n (L). The McCoy rank of A is the largest t such that AnnihV 2 (L) (Ft (A)) ≡ 0, where Ft (A) is the set of the bideterminants of t × t submatrices of A and AnnihV 2 (L) (Ft (A)) = {r ∈ V 2 (L) : rFt (A) ≡ 0}.
Definition 3.6 (Poplin and Hartwig [4]). If n ∈ N then π(n) =
+ −
if n is even, if n is odd.
We define the (i, j)-minor matrix Aij to be the submatrix obtained by deleting the ith row and the jth column from a matrix A. We may subsequently define the positive and negative minors |Aij |+ and |Aij |− , which for convenience, we abbreviate to A+ ij and A− , respectively. ij
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Lemma 3.1 (Poplin and Hartwig [4]). Let p = q and A ∈ Mn (L). Then n
π(q+k)
apk Aqk
k=1
=
n
π(q+k−1)
apk Aqk
.
k=1
Lemma 3.2. In a commutative semiring L, let A = (aij ) ∈ Mn×m (L) with m ≤ n. If the McCoy rank of A is less than m, then Ax ≡ 0 has a nontrivial solution. Proof. Let the McCoy rank of A be t with t < m. Then AnnihV 2 (L) (Ft+1 (A)) ≡ 0, i.e., there is y ∈ AnnihV 2 (L) (Ft+1 (A)) such that 0 ≡ y. Thus by Definition 3.5, yFt+1 (A) ≡ 0. If t = 0, then yaij ≡ 0 for all i ∈ n, j ∈ m. Therefore, x0 = (y, y, · · · , y)T is a nontrivial solution of the matrix equation Ax ≡ 0. Next, suppose that t > 0. Then the product of y and a certain element of Ft (A) is nonzero since y ≡ 0. Such element of Ft (A) is the bideterminant of a t × t submatrix of A. By permuting rows and columns, we may assume that the submatrix is T with ⎛
a11 ⎜ ⎜ a21 T =⎜ ⎜ .. ⎝ . at1
a12 a22 .. . at2
··· ··· ···
⎞ a1t ⎟ a2t ⎟ .. ⎟ ⎟. . ⎠ att
Consider the submatrix ⎛ ⎜ ⎜ ⎜ T =⎜ ⎜ ⎜ ⎝
a11 a21 .. . at1
a12 a22 .. . at2
at+1,1
at+1,2
··· ··· ··· ···
a1t a2t .. . att at+1,t
⎞ a1,t+1 a2,t+1 ⎟ ⎟ ⎟ .. ⎟. . ⎟ ⎟ at,t+1 ⎠ at+1,t+1
For 1 ≤ k ≤ t + 1, let Tk denote the submatrix of T formed by deleting the row t + 1 and the column k. Consider the vector x = (x1 , x2 , · · · , xm )T with
xi =
y(0, 1)i+t+1 det(Ti ) (0, 0)
1 ≤ i ≤ t + 1, t + 2 ≤ i ≤ m.
Hence t+1 j=1
and
aij (0, 1)j+t+1 det(Tj ) ≡ 0 for 1 ≤ i ≤ t (by Lemma 3.1)
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aij (0, 1)j+t+1 det(Tj )y ≡ 0 for t + 1 ≤ i ≤ n
j=1
since y ∈ AnnihV 2 (L) (Ft+1 (A)) and for t + 1 ≤ i ≤ n. Therefore,
t+1 j=1
aij (0, 1)j+t+1 det(Tj ) is an element of Ft+1 (A)
⎞ ⎛ ⎞ ⎛ t+1 ( j=1 a1j (0, 1)j+t+1 det(Tj ))y 0 ⎟ ⎜ ⎟ ⎜ t+1 j+t+1 ⎟ ⎜0⎟ ⎜( a (0, 1) det(T ))y 2j j j=1 ⎟ ⎜ ⎟ ⎜ Ax = ⎜ ⎟ ≡ ⎜ . ⎟. .. ⎟ ⎝ .. ⎠ ⎜ ⎠ ⎝ . t+1 0 ( j=1 anj (0, 1)j+t+1 det(Tj ))y Finally, x ≡ 0 since xt+1 = y(0, 1)t+1+t+1 det(Tt+1 ) = y det(T ) ≡ 0. 2 Lemma 3.3 (Shu and Wang [6]). In a commutative and cancellative semiring L, let A = (aij ) ∈ Mn (L). Then the McCoy rank of A is less than n if and only if Ax ≡ 0 has a nontrivial solution. Theorem 3.2. In a commutative semiring L, let M be a free L-semimodule with a free basis {b1 , b2 , · · · , bn }. Let ai ∈ M, i ∈ m and (a1 , a2 , · · · , am ) = (b1 , b2 , · · · , bn )A, where A ∈ Mn×m (L) and m ≤ n. Then (1) If {a1 , a2 , · · · , am } is free, then the McCoy rank of A is equal to m. (2) If the semiring L is cancellative and the McCoy rank of A is equal to m, then {a1 , a2 , · · · , am } is free. Proof. Since (a1 , a2 , · · · , am ) = (b1 , b2 , · · · , bn )A, then from x1 a1 + x2 a2 + · · · + xm am ≡ 0 we have x1 (a11 b1 + a21 b2 + · · · + an1 bn ) + x2 (a12 b1 + a22 b2 + · · · + an2 bn ) + · · · + xm (a1m b1 + a2m b2 + · · · + anm bn ) ≡ 0.
(3)
This deduces that (a11 x1 + a12 x2 + · · · + a1m xm )b1 + (a21 x1 + a22 x2 + · · · + a2m xm )b2 + · · · + (an1 x1 + an2 x2 + · · · + anm xm )bn ≡ 0.
(4)
Therefore, from Theorem 3.1, we know that {a1 , a2 , · · · , am } is free if and only if the equation (4) has only the trivial solution, moreover, the equation (4) has only the trivial solution if and only if the equation
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Fig. 1. A commutative semiring which is not cancellative.
⎛
a11 ⎜ ⎜ a21 ⎜ . ⎜ . ⎝ . an1
a12 a22 .. . an2
··· ··· ···
⎞⎛ ⎞ ⎛ ⎞ x1 0 a1m ⎟⎜ ⎟ ⎜ ⎟ a2m ⎟ ⎜ x2 ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ .. ⎟ ⎟⎜ . ⎟ ≡ ⎜ . ⎟ . ⎠ ⎝ .. ⎠ ⎝ .. ⎠ 0 anm xm
(5)
has only the trivial solution, since {b1 , b2 , · · · , bn } is free. Thus {a1 , a2 , · · · , am } is free if and only if the equation (5) has only the trivial solution. (1) Suppose that {a1 , a2 , · · · , am } is free. Then the equation (5) has only the trivial solution. By Lemma 3.2, we know that the McCoy rank of A is equal to m. (2) If the McCoy rank of A is equal to m, then by Definition 3.5, there exists an m × m submatrix B in A such that the McCoy rank of B is m. Further, by Lemma 3.3, the equation Bx ≡ 0 has only the trivial solution, and so is the equation Ax ≡ 0. Consequently, {a1 , a2 , · · · , am } is free. 2 Remark 3.1. In general, the condition that the semiring L is cancellative in Theorem 3.2 (2) can not be deleted. For example, in the semiring as Fig. 1 (just replace the addition and the multiplication in Definition 2.1 by the join and meet in the lattice as Fig. 1, re1 0 ∈ M2 (L). Then the McCoy spectively), it is obvious that K + (L) = L. Let A = b 1 rank of A is equal to 2, but a(1, b)T + c(0, 1)T = a(1, b)T + 0(0, 1)T , i.e., {(1, b)T , (0, 1)T } is not free. Remark 3.2. Theorem 3.2 answered Open problem 3.1 raised by Tan in his work [7] over a commutative semiring. Acknowledgements The authors thank the reviewers for their valuable suggestions. References [1] P. Butkovič, Max-algebra: the linear algebra of combinatorics?, Linear Algebra Appl. 367 (2003) 313–335.
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[2] A. Di Nola, A. Lettieri, I. Perfilieva, V. Novák, Algebraic analysis of fuzzy systems, Fuzzy Sets and Systems 158 (2007) 1–22. [3] J.S. Golan, Semirings and Their Applications, Kluwer Academic Publishers, Dordrecht/Boston/London, 1999. [4] Phillip L. Poplin, Robert E. Hartwig, Determinantal identities over commutative semirings, Linear Algebra Appl. 387 (2004) 99–132. [5] Qian-yu Shu, Xue-ping Wang, Bases in semilinear spaces over zerosumfree semirings, Linear Algebra Appl. 435 (2011) 2681–2692. [6] Qian-yu Shu, Xue-ping Wang, The applications of the bideterminant of a matrix over commutative semirings, Linear Multilinear Algebra (2016), http://dx.doi.org/10.1080/03081087.2016.1242555. [7] Yi-jia Tan, Free sets and free subsemimodules in a semimodule, Linear Algebra Appl. 496 (2016) 527–548. [8] Xue-ping Wang, Qian-yu Shu, Bideterminant and rank of matrix, Soft Comput. 18 (2014) 729–742. [9] S.M. Yusuf, Inverse semimodules, J. Natur. Sci. Math. 6 (1966) 111–117.