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ONE-SIDED COMPLEMENTS AND SOLUTIONS
OF THE EQUATION
aXb
=
c
IN SEMIRINGS
SAM L. BLYUMIN and JONATHAN S. GOLAN
Received 22 July 2001
Given multiplicatively-regular elementsaandbin a semiringR, and given an elementcof R, we find a complete set of solutions to the equationaXb=c. This result is then extended to equations over matrix semirings.
2000 Mathematics Subject Classification: 16Y60.
1. Semirings. We follow the notation and terminology of [5], to which the reader is referred for all undefined notions and unproven assertions. LetRbe a semiring. An elementaismultiplicatively regular if and only if there exists an elementa−ofR, called ageneralized inverseofa, satisfyingaa−a=a. If such an element exists then the elementa×=a−aa− satisfies the conditionsaa×a=aanda×aa×=a×. We call the elementa×ofRaThierrin-Vagner inverseofa. The details are given in [5].
Ifais multiplicatively idempotent then it has a Thierrin-Vagner inverse and, indeed, we can choosea×=a. Thus we can always assume that 0×=0 and 1×=1. Ifahas a multiplicative inverse, we can choosea×=a−1. IfRis a semifield we see that every element is multiplicatively regular. This happens, for example, in such important and applicable semirings as the schedule algebra(R∪{−∞},max,+).
Regularity in fuzzy matrix rings is studied in [2]. For algorithms to calculate Moore-Penrose pseudoinverses of matrices over additively-idempotent semirings, which are special cases of Thierrin-Vagner inverses, refer to [7]. Also refer to [3] for calculation of generalized inverses for semirings of matrices over bounded distributive lattices.
We note too that ifa∈Ris multiplicatively regular then so isa×and so area×a andaa×, and indeed(a×a)×=a×aand(aa×)×=aa×. Moreover, both of these el-ements are multiplicatively idempotent. Thus we have two functions from the set of all multiplicatively-regular elements ofR to the setI×(R)of all multiplicatively-idempotent elements ofRgiven byλ:aa×aandρ:aaa× and these functions satisfyλ2=λandρ2=ρ. Moreover, for eacha∈Rwe have
aλ(a)=a=ρ(a)a,
λa×a×=a×=a×ρa×. (1.1)
We are interested in the following problem:given multiplicatively-regular elements
a,b∈R and given an elementc∈R,find a complete set of solutions to the equation
language theory. Also see [9]. They also appear in the consideration of fuzzy and semiring-valued relations [4] and fuzzy bilinear equations [8], and arise naturally in control theory with coefficients taken from the(max,+)algebra or from the semiring of fuzzy numbers. For certain noncommutative rings, such as rings of matrices or rings of operators over a linear space, they have an extensive literature, and the results there can often be extended to matrix semirings over semirings, for example.
Note that if there exists a solutionxto the equation
aXb=c, (1.2)
then
c=axb=ρ(a)(axb)λ(b)=ρ(a)cλ(b). (1.3)
Conversely, ifc∈Rsatisfiesρ(a)cλ(b)=c, thena×cb×is a solution for (1.2). Thus (1.2) has a nonempty set of solutions if and only ifc satisfies this condition. This allows us to rephrase our problem as follows:given multiplicatively regular elements
a,b∈Rand given an element c∈Rsatisfyingρ(a)cλ(b)=c,find a complete set of solutions of (1.2) inR.
Letabe an element of a semiringR. An elementa[r ]ofRis called aright complement ofaif and only ifaa[r ]=0 anda+a[r ]=1. An elementa[l]ofRis aleft complement ofaif and only ifa[l]a=0 anda[l]+a=1. Ifahas both a right complementa[r ]and a left complementa[l], then these must be equal. Indeed, we note that in this case
a[l]=a[l]a+a[r ]=a[l]a+a[l]a[r ]=a[l]a[r ]
=aa[r ]+a[l]a[r ]=a+a[l]a[r ]=a[r ]. (1.4)
Such an element is called acomplement ofa and is denoted bya⊥. Complements, when they exist, are necessarily unique.
Example1.1. Right and left complements need not be the same. For example, let Sbe the ring of all upper-triangular matrices over the ringZof integers, and letRbe the semiring ideal(S)consisting ofSand of all (two-sided) ideals ofS. The operations onRare the usual addition and multiplication of ideals. IfI=Z Z
0 0
andH=0Z
0Z
then it is easy to verify thatH=I[l]butH≠I[r ].
Complements of elements of a semiring are studied in [5, Chapter 5]; they play a very important role in the theory and applications of semirings. Since the inspiration for complements came from lattice theory, they were assumed to be two-sided. However, here we have to look at the notion of a one-sided complement.
Note that ifa∈Rhas a right complement thena∈I×(R)since
a=a1=aa+a[r ]=a2+aa[r ]=a2 (1.5)
and the same is, of course, true ifahas a left complement. Thus, if we denote the set of all elements ofR having a right (resp., left) complement by rcomp(R)(resp., lcomp(R)), and if we denote the set of all elements of R having a complement by comp(R), we see that
and if we denote the set of all elements of R having a one-sided complement by ocomp(R), that is, ocomp(R)=rcomp(R)∪lcomp(R), then we see that
ocomp(R)⊆I×(R). (1.7)
Also, we note that ifa∈rcomp(R)then any right complementa[r ]ofabelongs to lcomp(R)and, indeed,aitself is a left complement ofa[r ]. Similarly, ifa∈lcomp(R) then any left complement ofabelongs to rcomp(R). Thus we see that ocomp(R)is closed under taking left and right complements.
Note that ifγ:R→S is a morphism of semirings, thenγ(ocomp(R))⊆ocomp(S). Indeed, ifa∈Rhas a right complementa[r ]then 0
S=γ(0R)=γ(aa[r ])=γ(a)γ(a[r ]) and 1S=γ(1R)=γ(a+a[r ])=γ(a)+γ(a[r ])soγ(a[r ])is a right complement ofγ(a). Similarly, ifahas a left complementa[l]thenγ(a[l])is a left complement ofγ(a).
Assume thataandbare multiplicatively-regular elements ofRsuch thatλ(a)has a right complement λ(a)[r ] and thatρ(b)has a left complement ρ(b)[l]. Then we note thataλ(a)[r ]=ρ(a)aλ(a)[r ]=aλ(a)λ(a)[r ]=0 andρ(b)[l]b=ρ(b)[l]bλ(b)= ρ(b)[l]ρ(b)b=0.
Given an elementcofR, define a functionαc:R→Rby setting
αc:y→a×cb×+λ(a)yρ(b)[l]+λ(a)[r ]y. (1.8)
Then the foregoing discussion leads us to the following result.
Proposition1.2. Ifaandbare multiplicatively-regular elements of a semiringR
satisfying the condition thatλ(a)∈rcomp(R)andρ(b)∈lcomp(R), and ifc is an element ofRsatisfyingρ(a)cλ(b)=c, then a complete set of solutions of (1.2) is given by{αc(y)|y∈R}. Ifcdoes not satisfy this condition then (1.2) has no solutions inR.
Proof. Ifc does not satisfy the given condition then we have already seen that (1.2) has no solutions inR. Assume therefore that it does. From the hypothesis of the theorem we then see that
aαc(y)b=ρ(a)cλ(b)+ρ(a)ayρ(b)[l]b+aλ(a)[r ]yb =ρ(a)cλ(b)
=c,
(1.9)
soαc(y)is a solution to (1.2) for any y∈R. Moreover, we note that ifx∈R is a solution of (1.2) thenαc(x)=x. Indeed, ifaxb=cthen
αc(x)=a×cb×+λ(a)xρ(b)[l]+λ(a)[r ]x =λ(a)xρ(b)+λ(a)xρ(b)[l]+λ(a)[r ]x =λ(a)xρ(b)+ρ(b)[l]+λ(a)[r ]x =λ(a)x+λ(a)[r ]x
=λ(a)+λ(a)[r ]x =x
(1.10)
In particular, we have the following examples.
Example1.3. Suppose thatRis a semiring. Ifaandbare multiplicatively-regular elements ofRsatisfying the condition that bothλ(a)andρ(b)have additive inverses, then we can setλ(a)[r ]=1−λ(a) and ρ(b)[l]=1−ρ(b). In this case, both λ(a) andρ(b)in fact belong to comp(R). This surely happens ifRis a ring.
Example1.4. Suppose thatRis a Boolean algebra. Ifaandbare multiplicatively-regular elements ofR, we can setλ(a)[r ]=a andρ(b)[l]=ρ(b).
Example1.5. Following the terminology of [5], we say that a semiringRisplain
if and only ifa+b =b fora,b∈R implies that a=0. It is simple if and only if a+1=1 for alla∈R, and it isyoked if for each paira, bof elements ofR there exists an element c of R satisfyinga+c=b or b+c=a. By [5, Example 5.6] we see that every multiplicatively-idempotent element of a plain simple yoked semir-ing has a complement and so, for such semirsemir-ings,λ(a)[r ] and ρ(b)[l] exist for all multiplicatively-regular elementsaandbofR.
Among the most applicable families of semirings which are not rings are zerosum-freesemirings, namely semirings which satisfy the condition thata+b=0 when and only whena=b=0. Bounded distributive lattices are examples of such semirings, as are semirings of (two-sided) ideals of rings and information algebras in the sense of [6]. We make some remarks concerning the behavior of one-sided complements in such semirings.
Proposition1.6. IfRis a zerosumfree semiring and ifa∈rcomp(R)whileb∈ ocomp(R)thenaba[r ]=0.
Proof. Indeed, ifb is a one-sided complement ofbthen
aba[r ]+aba[r ]=ab+ba[r ]=aa[r ]=0, (1.11)
and soaba[r ]=0 sinceRis zerosumfree.
Similarly, ifa∈lcomp(R)whileb∈ocomp(R)thena[l]ba=0.
Proposition1.7. IfRis a zerosumfree semiring and ifa,b∈rcomp(R)thena+ a[r ]b∈rcomp(R).
Proof. Indeed, we note thata+a[r ]b+a[r ]b[r ]=a+a[r ](b+b[r ])=a+a[r ]= 1 while (a+a[r ]b)a[r ]b[r ] =a[r ]ba[r ]b[r ]. But we have already seen that a[r ] ∈ ocomp(R)so, byProposition 1.6,ba[r ]b[r ]=0. Thusa[r ]b[r ]is a right complement ofa+a[r ]b.
Similarly, we note that ifa,b∈lcomp(R)thena+ba[l]∈rcomp(R).
Proposition1.8. IfRis a zerosumfree semiring and ifa,b∈rcomp(R)thenab∈ rcomp(R). Moreover, ifrcomp(R)is closed under sums then every element ofrcomp(R)
is additively idempotent.
Now assume that rcomp(R) is closed under sums. Then, in particular, 1+1∈ rcomp(R)so, ifa∈rcomp(R)we see thata+a=a(1+1)∈rcomp(R). Letb be a right complement ofa+a. Thenab+ab=(a+a)b=0 and, by zerosumfreeness, we deduce thatab=0. Thereforea=a1=(a+a+b)=a2+a2=a+a, showing thata is additively idempotent.
Similarly, we note that ifa,b∈lcomp(R)thenab∈lcomp(R)and if lcomp(R)is closed under sums then each of its members is additively idempotent.
2. Semimodules over matrix semirings. IfRis a semiring then so is the setᏹn×n(R) of alln×nmatrices overR, with addition and multiplication defined in the standard manner. We denote the additive identity inᏹn×n(R)byOn×n and the multiplicative identity inᏹn×n(R)byIn×n. Moreover, ifkandnare positive integers then the set ᏹk×n(R)of allk×nmatrices overRis canonically a left semimodule overᏹk×k(R)and a right semimodule overᏹn×n(R). We denote the additive identity inᏹk×n(R)byOk×n. Furthermore, ifA∈ᏹk×n(R)andB∈ᏹn×k(R), then the productsAB∈ᏹk×k(R)and BA∈ᏹn×n(R)are defined in the usual manner. Ageneralized inverseofA∈ᏹk×n(R) is a matrix A−∈ ᏹn×k(R)satisfying AA−A= A. If such a generalized inverse ex-ists, thenAis multiplicatively regular. Again, ifAis multiplicatively regular then the
Thierrin-Vagner inverseofAis defined to beA×=A−AA−∈ᏹn×k(R)and this matrix satisfiesAA×A=AandA×AA×=A×. IfA∈ᏹk×n(R)is regular then, as before, we define the matricesλ(A)=A×A∈ᏹn×n(R)andρ(A)=AA×∈ᏹk×k(R).
Example 2.1. Consider the special case of A= a1
.. . ak
∈ᏹk×1(R). ThenA has a
generalized inverse A− =[b1,...,bk] if and only if the element e=ki=1biai of R satisfiesaie=aifor all 1≤i≤k.
GivenA∈ᏹk×n(R)andB∈ᏹn×k(R)having generalized inverses, and givenC∈ ᏹk×k(R), we then note, as above, that whenever there exists a matrixT ∈ᏹn×n(R) satisfyingAT B=Cwe have
C=AT B=AA×AT BB×B=AA×CB×B=ρ(A)Cλ(B). (2.1)
A matrixA∈ᏹk×n(R)isright regularly complementedif and only if it has a gener-alized inverseA−∈ᏹn×k(R)and there exists a multiplicatively-regular matrixA[r ]∈ ᏹn×n(R)satisfying the conditions AA[r ] =Ok×n and A×A+A[r ] =In×n. Similarly, B∈ᏹn×k(R)isleft regularly complementedif and only if it has a generalized inverse B−∈ᏹk×k(R)and there exists a multiplicatively-regular matrixB[l]ᏹn×n(R) satisfy-ing the conditionsB[l]B=O
n×kandBB×+B[l]=In×n.
Example2.2. Again, consider the special case of A= a1
.. . ak
∈ᏹk×1(R). ThenA
matrixA[l]=[d
ij]∈ᏹk×k(R)satisfying k
i=1biai=0 and
aibj+dij=
1 ifi=j,
0 ifi≠j. (2.2)
Suppose thatA∈ᏹk×n(R)and B∈ᏹn×k(R)are matrices having generalized in-verses and satisfying the condition thatAis right regularly complemented while B is left regularly complemented. Then each matrix C ∈ᏹk×k(R) defines a function αC:ᏹn×n(R)→ᏹn×n(R)by setting
αC:Y→A×CB×+λ(A)Y B[l]+λ(A)[r ]Y . (2.3)
We can now generalizeProposition 1.2as follows.
Proposition2.3. LetRbe a semiring. LetA∈ᏹk×n(R)andB∈ᏹn×k(R)be
ma-trices having generalized inverses and satisfying the condition thatAis right regularly complemented whileB is left regularly complemented. Furthermore, letC∈ᏹk×k(R)
be such that there exists a matrixT∈ᏹn×n(R)that satisfiesAT B=C. Then a complete
set of solutions of (1.2) is given by
αC(Y )|Y∈ᏹn×n(R). (2.4)
IfT does not satisfy this equation then (1.2) has no solutions inᏹn×n(R).
The proof is essentially the same as that ofProposition 1.2.
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Sam L. Blyumin: Lipetsk State Technical University,398055Lipetsk, Russia