交換半環(huán)上矩陣的秩和坡上矩陣的可逆條件

交換半環(huán)上矩陣的秩和坡上矩陣的可逆條件

1標(biāo)準(zhǔn)質(zhì)值introviphingkim-a、columbiac（a）和schrgreecerandom（a）是甲殼類動物ma復(fù)合體的2-elfour，a.mo，kim和ro湖州的a.ma育種中的最小傅里葉變換。AninclineSisaspecialtypeofcommutativesemiring,inwhichs+1=1holdsforalls∈S.InclinesareintroducedandstudiedearliestbyCao,KimandRoush.TheinvertibleconditionsformatricesoverdistributivelatticesareconsideredbySkornyakovandZhaoindependentlybydifferentmethods.Heretheinvertibleconditionsformatricesoverinclinesarestudied.Byintroducingthepermanentwegivethenecessaryandsufficientconditionsforinvertibilityofmatrices,whichisageneralizationoftheresultsinand.2風(fēng)bhenabhenaAsemiringisanalgebraicstructure(S,+,·)suchthat(S,+)isanAbelianmonoid(identity0),(S,·)isamonoid(identity1),multiplication·distributesoveraddition“+”fromeitherside,s·0=0·s=0foralls∈Sand0≠1.UsuallySdenotesboththesemiringandthesetandabdenotesa·bfora,b∈S.Ifthemonoid(S,·)isAbelian,Sissaidtobecommutative.AsemiringSissaidtobezero-sum-freeiffora,b∈S,a+b=0impliesthata=b=0.Ler(S,+,·)beasemiringand≤beapartialorderrelationonsetS.Ifthefollowingconditionsaresatisfiedforalla,b,c∈S:(1)Ifa≤bthena+c≤b+c;(2)Ifa≤bandc≥0thenac≤bcandca≤cb.wecallSpartially-orderedsemiring(withthepartialorderrelation≤).Apartially-orderedsemiringSissaidpositiveifs≥0foralls∈S.Itisclearthatapositivepartially-orderedsemiringiszero-sum-free.Inthesequelofthissection,Swilldenoteanycommutativesemiring.Thesetofm×nmatriceswithentriesinSisdenotedbyMm,n(S).WesetMn(S)=Mn,n(S),Vn(S)=M1,n(S)andamemberofVn(S)iscalledarowvector.TheentriesofthematricesA,B,C,etc.willbedenotedbyaij,bij,cij,etc.Them×nzeromatrixOm,nandthen×nidentitymatrixInaredefinedasifSwereafield.Addition,multiplicationbyscalars,andtheproductofmatricesarealsodefinedasifSwereafield.IfSispartially-orderedwecandefinethepartialorderrelationonMm,n(S):A≤Bifandonlyifaij≤bij.Definition1AnonemptysubsetWofVn(S)issaidtobeasubspaceifforalls∈S,u,v∈W,su∈Wandu+v∈W.AlinearcombinationofelementsofasetTofrowvectorsisafinitesum∑siuiwheresi∈Sandui∈T.ThesetofalllinearcombinationsofelementsofTiscalledthespanofT,denotedby〈T〉.If〈T〉=WthenTiscalledaspanningsetforW.Westipulate〈?〉={0}.AsubspaceanditsspanningsetforthesetofcolumnvectorsVn(S)=Mn,1(S)aredefinedsimilarly.Definition2Therow(column)spaceofamatrixAoverSisthesubspacespannedbyitsrow(column)vectors.Therow(column)spaceofamatrixAisdenotedbyR(A)(C(A)).Therow(column)rankρr(A)(ρc(A))ofamatrixAistheleastcardinalityofallspanningsetsfortherow(column)space.Definition3TheScheinrankρs(A)ofanonzerom×nmatrixAoverSistheleastintegerksuchthatA=BCforsomem×kandk×nmatricesBandCoverS.Forthezeromatrixρs(Om,n)=0.Fromthedefinitionsthefollowingareobtained:Proposition1LetA∈Mm,n(S)andB∈Mn,k(S).Then(1)ρs(A)≤ρr(A)≤m,ρs(A)≤ρc(A)≤n.(2)ρs(AT)=ρs(A),ρr(AT)=ρc(A),whereATdenotesthetransposeofA.(3)ρs(AB)≤min{ρs(A),ρs(B)}.Definition4Anm×nmatrixAoverSisrank-fullifρs(A)=min{m,n}.AisregularifthereexistsamatrixBsuchthatABA=A.Ann×nmatrixAoverSisinvertibleifAB=BA=Inforsomen×nmatrixBoverS.Lemma1()ForanymatrixAoveracommutativesemiringS,ρr(A)=min{ρs(X):XA=A}andρc(A)=min{ρs(Y):AY=A}.3nmaelitys...............................................................Theorem1IfAisaregularmatrixoveracommutativesemiringSthenρs(A)=ρr(A)=ρc(A).ProofThereexistsamatrixBsuchthatABA=AduetotheregularityofA.Thusρr(A)≤ρs(AB)≤ρs(A)byLemma1andProposition1(3).Thereforeρr(A)=ρs(A)byProposition1(1).ApplyigtheresulttotheregularmatrixATwehaveρr(AT)=ρs(AT).Thereforeρc(A)=ρs(A)byProposition1(2).Theorem2LetSbeazero-sum-freecommutativesemiringandA∈Mn(S).If∏ni=1ni=1aii≠0andaijaji=0(i≠j)thenAisrank-full.ProofSupposeρs(A)=s<n.ThusA=BCforsomen×sands×nmatricesBandC.Bythehypothesisaijaji=(∑sk=1sk=1bikckj)(∑sk=1sk=1bjkcki)=0(i≠j).Sobikckibjkckj=0fori≠jandk=1,2,…,ssinceSiszero-sum-free.Nowweconsider∏ni=1aii=∏ni=1∑sk=1bikcki=∑1≤j1,j2,?,jn≤s∏ni=1bijicjii.∏ni=1aii=∏ni=1∑sk=1bikcki=∑1≤j1,j2,?,jn≤s∏ni=1bijicjii.Sinces<nthereexistp,q(1≤p<q≤n)suchthatjp=jq.Thus∏ni=1ni=1bijicjii=0.Andweget∏ni=1ni=1aii=0,whichiscontradictorytothehypothesis.Thereforeρs(A)=n.Corollary1LetSbeazero-sum-freecommutativesemiringandA∈Mn(S).(1)IfAisanuppertriangularmatrixand∏ni=1ni=1aii≠0thenAisrank-full.(2)IfAisinvertiblethenAisrank-full.Adiagonalmatrixoverazero-sum-freecommutativesemiringisnotnecessarilyrank-full.Example1LetSbe4-elementBooleanalgebra{0,a,b,1}andA=[a00b].A=[a00b].Definition5LetSbeacommutativesemiringandA∈Mm,n(S)(m≤n).Thepermanentper(A)ofmatrixAisdefinedasfollows:per(A)=∑σ∈Sm,nm∏i=1aiσ(i),per(A)=∑σ∈Sm,n∏i=1maiσ(i),whereSm,ndenotesthesetofallinjectionsfromset{1,2,…,m}to{1,2,…,n}.Proposition2LetSbeacommutativesemiringandA∈Mm,n(S)(m≤n).Then(1)()per(AT)=per(A)ifm=n.(2)per(A)=∑nj=1nj=1aijper(Aij)(1≤i≤m),whereAijisthematrixformedbydeletingrowiandcolumnjfromA.Ρroof(2)per(A)=∑σ∈Sm,nm∏l=1alσ(l)=n∑j=1∑σ∈Sm,n,σ(i)=jm∏l=1alσ(l)=n∑j=1aij∑σ∈Sm,n,σ(i)=ja1σ(1)?ai-1,σ(i-1)ai+1,σ(i+1)?amσ(m)=n∑j=1aijper(Aij).Proof(2)per(A)=∑σ∈Sm,n∏l=1malσ(l)=∑j=1n∑σ∈Sm,n,σ(i)=j∏l=1malσ(l)=∑j=1naij∑σ∈Sm,n,σ(i)=ja1σ(1)?ai?1,σ(i?1)ai+1,σ(i+1)?amσ(m)=∑j=1naijper(Aij).Proposition2(2)istheformulathatper(A)expandsonarow.Butgenerallyspeaking,per(A)cannotexpandonacolumnasm<n.Theorem3IfSisapositivepartially-orderedcommutativesemiring,A∈Mm,m(S)andB∈Mm,n(S)(m≤n),thenper(AB)≥per(A)per(B).Ρroofper(AB)=∑σ∈Sm,nm∏i=1m∑k=1aikbkσ(i)=∑σ∈Sm,n∑1≤k1,?,km≤mm∏i=1aikibkiσ(i)=∑1≤k1,?,km≤ma1k1a2k2?amkm∑σ∈Sm,nbk1σ(1)bk2σ(2)?bkmσ(m)≥∑π∈Sm,ma1π(1)a2π(2)?amπ(m)∑σ∈Sm,nbπ(1)σ(1)bπ(2)σ(2)?bπ(m)σ(m)=∑π∈Sm,ma1π(1)a2π(2)?amπ(m)∑σ∈Sm,nm∏i=1biσ(π-1(i))=per(A)per(B).Proofper(AB)=∑σ∈Sm,n∏i=1m∑k=1maikbkσ(i)=∑σ∈Sm,n∑1≤k1,?,km≤m∏i=1maikibkiσ(i)=∑1≤k1,?,km≤ma1k1a2k2?amkm∑σ∈Sm,nbk1σ(1)bk2σ(2)?bkmσ(m)≥∑π∈Sm,ma1π(1)a2π(2)?amπ(m)∑σ∈Sm,nbπ(1)σ(1)bπ(2)σ(2)?bπ(m)σ(m)=∑π∈Sm,ma1π(1)a2π(2)?amπ(m)∑σ∈Sm,n∏i=1mbiσ(π?1(i))=per(A)per(B).ZhangprovedtheinequalityinTheorem3formatricesoveradistributivelatticeinthecaseofm=n.Definition6TheSchurproduct,A⊙B,ofm×nmatricesAandBoverasemiringisdefinedasmatrix(aijbij).Theorem4IfSisazero-sum-freecommutativesemiring,A,BT∈Mm,n(S)(m≤n)andAB=Imthenper(A⊙BT)=1.ProofWehave∏mi=1mi=1∑nk=1nk=1aikbki=1bythehypotheses.Thelefthandsidecanbeexpandedasthesumofnmterms,i.e.∑1≤j1,?,jm≤na1j1bj11a2j2bj22?amjmbjmm=1.∑1≤j1,?,jm≤na1j1bj11a2j2bj22?amjmbjmm=1.Inaddition,wehaveaikbkj=0forallk∈{1,…,n}andi≠jsince∑nk=1nk=1aikbkj=0fori≠j.Therefore,weget∑σ∈Sm,na1σ(1)bσ(1)1a2σ(2)bσ(2)2?amσ(m)bσ(m)m=1.∑σ∈Sm,na1σ(1)bσ(1)1a2σ(2)bσ(2)2?amσ(m)bσ(m)m=1.Thismeansper(A⊙BT)=1.Theorem5IfSisazero-sum-freecommutativesemiring,A,B∈Mn(S)andAB=Inthenaikajk=0andakiakj=0forallkandi≠j.ProofFirstly,bythehypotheseswehaveaikbkj=0forallkandi≠j.Usingtheorem4wegetaikajk=aikajkper(A⊙BΤ)=∑σ∈Sn,naikajkn∏l=1alσ(l)bσ(l)l.Sinceeachsummandintherighthandsidehasthefactoraikajkbkσ-1(k),wherei≠σ-1(k)orj≠σ-1(k),thusaikajk=0.Similarly,akiakj=∑σ∈Sn,nakiakjn∏l=1alσ(l)bσ(l)l.Eachsummandcontainsthefactorakiakjbiσ-1(i)bjσ-1(j)andthisisequalto0sincek≠σ-1(i)ork≠σ-1(j).Therefore,wehaveakiakj=0.Definition7ACommutativesemiringSiscalledaninclineifs+1=1foralls∈S.Example2I=(,max,×)isanincline.Herea×bisordinarymultiplication.ThedirectproductI2=(×,max,×)isalsoanincline.Hereoperationsmaxand×inI2aredefinedcomponentwise.Everydistributivelatticewiththeleastelement0andgreatestelement1isanincline.AninclineSisadistributivelattivewiththeleastelement0andgreatestelement1ifandonlyifs2=sforalls∈S().Proposition3IfSisaninclinethenforalla,b,c∈S(i)a+a=a.(ii)a+b=0impliesa=b=0.(iii)ab=1impliesa=b=1.(iv)a+b=1impliesa+b2=1.Proof(i)a+a=a(1+1)=a)=a.(ii)By(i)a=a+0=a+a+b=a+b=0.(iii)a=a)=a(1+b)=a+ab=a+1=1.InaninclineSdefiningthepartialorderrelation≤:a≤bifandonlyifa+b=b,wemakeSintoapositivepartially-orderedsemiring.Andwehave.Proposition4IfSisaninclinethenforalla,b∈S(i)0≤a≤1.(ii)ab≤b.(iii)a+b≥banda+bistheleastupperboundofaandb.ItfollowsfromTheorem3andProposition4(i)that:Theorem6IfSisaninclinethenthematrixset{A|per((A)=1,A∈Mn(S)}formsamonoidunderthematrixmultiplication.NotethatineachinclineinExample2,theequationax=bisalwayssolvableifa≥b.Inthiscasewegive:Theorem7LetSbeaninclineandA∈Mm,n(S).Iftheequationax=bissolvableinSforalla,b∈Ssatisfyinga≥b,thenρs(A)=1impliesρr(A)=ρc(A)=1.ProofLetA=uTv,u=(u1,u2,…,um)∈Vmandv∈Vn.Since∑mk=1uk≥ui(i=1,2,…,m),wehaveR(A)=〈(∑mk=1uk)v〉bythehypothesis.Thereforeρr(A)=1.Similarlywehaveρc(A)=1.Example3TakethesubsetS={0}∪{12n+k|n,k∈{0,1,2,?}}oftherealinterval.Thus(S,max,×)isanincline.TakethematrixA=overS.Wehaveρs(A)=1,butρc(A)=2.Proposition5IfAisanivertibelmatrixoveraninclineSthenper(A)=1.ProofLetBbetheinverseofA.Thenper(A⊙BT)=1byTheorem4.SinceA⊙BT≤Awehaveper(A⊙BT)≤per(A).Thereforeper(A)=1.Proposition6LetSbeaninclineandA∈Mn(S).Ifper(A)=1then∑nk=1aik=1and∑nk=1aki=1foralli∈{1,2,…,n}.ProofSinceper(A)=∑nk=1aikper(Aik)≤∑nk=1aikwehave∑nk=1aik=1(i=1,2,…,n).Sinceper(AT)=per(A)=1weobtain∑nk=1aki=1.Definition8LetSbeaninclineandA∈Mn(S).Aissaidtobeorthogonalrelativetotherowsif∑nk=1aik=1foreachi∈{1,2,…,n}andaikajk=0foranyi≠jandk∈{1,2,…,n}.Theorthogonalityrelativetothecolumnsisdefinedinasimilarmanner.Theorem8LetSbeaninclineandA∈Mn(S).AisinvertibleifandonlyifAisorthogonalrelativetotherowsifandonlyifAisorthogonalrelativetothecolumns.Andinthiscase,ATistheuniqueinverseofA.ProofIfAisinvertiblethenAisorthogonalrelativetotherowsbyProposition5and6andTheorem5.Inversely,letAbeorthogonalrelativetotherows.ApplyingProposition3(iv)totheequality∑nk=1aik=1weget∑nk=1a2ik=1.Therefore,AAT=In.Thisleadstoakiakj=0forallkandi≠jbyTheorem5andper(A⊙A)=1byTheor