COMPUTINGTENSOREIGENPAIRSUSINGHOMOTOPYMETHODS By LiangminZhou ADISSERTATION Submittedto MichiganStateUniversity inpartialentoftherequirements forthedegreeof AppliedMathematics|DoctorofPhilosophy 2015 ABSTRACT COMPUTINGTENSOREIGENPAIRSUSINGHOMOTOPYMETHODS By LiangminZhou Tensoreigenvalueproblemshavefoundimportantapplicationsinautomaticcontrol,sta- tisticaldataanalysis,tensorimaging,imageauthenticityvspectralhy- pergraphtheoryandquantumentanglement,etc.Theconceptofmode- k generalizedeigen- valuesandeigenvectorsofatensorisintroducedandsomepropertiesofsucheigenpairsare proved.Inparticular,anupperboundforthenumberofequivalenceclassesofgeneralized tensoreigenpairsusingmixedvolumeisderived.Basedonthisboundandthestructures oftensoreigenvalueproblems,twohomotopycontinuationtypealgorithmstosolvetensor eigenproblemsareproposed.Withproperimplementation,thesemethodscanallequiv- alenceclassesofisolatedgeneralizedeigenpairsandsomegeneralizedeigenpairscontained inthepositivedimensionalcomponents(ifthereareany).Analgorithmthatcombinesa straightforwardapproachandaNewtonhomotopymethodisintroducedtoextractrealgen- eralizedeigenpairsfromtheavailablecomplexgeneralizedeigenpairs.AMATLABsoftware package TenEig1.1 hasbeendevelopedtoimplementthesemethods.Numericalresultsare presentedtoillustratetheenessandof TenEig1.1 forcomputingcomplex orrealgeneralizedeigenpairs. Tomyparents iii ACKNOWLEDGMENTS IwouldliketoexpressmydeepestgratitudetomyadvisorProfessorTien-YienLiforhis well-timedguidance,encouragementandsupportallthetime.Withouthishelp,Icouldnot havebeenabletosurvivemyPh.Dlife.WhenIfailedmyattempttotakequalifying examsandreallythoughtIwassuchaloser,itwashimwhousedhisownstoryabout qualifyingexamstoremindmeinwhichwayIcandobetter.WhenIhadnoclueon myresearchandfeltIwassuchadullperson,itwashimwhocanalwayspointoutmy problemfromourdiscussionandgivemesomesmallenlighteningproblems.Workingon theseproblemsdiddeepenmyunderstandingonmyownproblemandwidenmyhorizon onmath.Overtheyears,Hisdiligent,persistentandoptimisticattitudetowardsresearch, teachingandlifehasbeeninspiringmeallthetime. IwouldliketothankDr.ChichiaChiu,Dr.KeithPromislow,Dr.JianliangQianand Dr.ZhengfangZhou,forservingasmembersofmythesiscommittee.Iamgratefulto ProfessorLixingHanforintroducingsuchaninterestingprojecttome,helpingmebuilda wholeviewofthedevelopmentofthetensoreigenproblemsandgivingmealotofadvice onthissubject.SpecialthanksgotoProfessorZhonggangZeng,withouthisencouragement onimplementingaMATLABversionofHOM4PS2.0,ProfessorLixingHanwouldhavenot beenabletousandwewouldnothavethischancetoworkonsuchawonderfulproject. IwanttotakethisopportunitytothankthosefriendswithwhomIhavesharedlaughters andtears,upsanddowns:YuChenandJunshanLin,forbeingsuchreliablefriends;Hongli Gao,forbeingaproblem-solver;QinfengGao,forbeingsohumorousandmakinglifeso joyful;XianfengHu,forbeingsuchafantasticcook;QiTangandZixuanWang,forsharing somuchtogether;XinYang,foralwaysstudyingdiligentlyandmotivatingme;QiongZheng, iv forhavingsuchawonderfulchancetobefriendstoyou. Lastbutnotleast,Ioweadebtofthankstomyfamily,whosealwaysbeingtherehas beenlightingupmywayawayfromhomeandbackintohome. v TABLEOFCONTENTS LISTOFTABLES .................................... vii Chapter1Introduction ............................... 1 Chapter2Tensoreigenvaluesandeigenvectors ................ 4 2.1ofgeneralizedmode- k tensoreigenvalueandeigenvectors.....4 2.2Propertiesofgeneralizedmode- k tensoreigenvalues..............8 Chapter3Theupperboundofthenumberofequivalenceclassesoftensor eigenpairs ................................. 12 3.1Introduction....................................12 3.2Preliminaries...................................13 3.3MainTheorembasedonBernstein'sTheorem.................15 Chapter4Computingcomplextensoreigenpairsbyhomotopymethods26 4.1Usinghomotopycontinuationmethodstosolvepolynomialsystems.....26 4.2Constructionofalinearhomotopytosolvetensoreigenpairs.........31 4.3Numericalalgorithmsofusinghomotopycontinuationmethodstosolvetensor eigenpairs.....................................37 Chapter5Computingrealtensoreigenpairsbyhomotopymethods ... 41 5.1UsingNewtonhomotopymethodtorealtensoreigenpairs........41 5.2Astraightforwardapproachtorealtensoreigenpairs...........45 5.3AnumericalalgorithmofrealtensoreigenpairsbasedontheNewton homotopyandthestraightforwardapproach..................53 Chapter6Implementationandnumericalresults ............... 54 6.1Examplesforcomputingcomplexeigenpairs..................55 6.2Computingsingulartensoreigenpairs......................59 6.3ExamplesforComputingRealEigenpairs....................64 APPENDIX ........................................ 68 BIBLIOGRAPHY .................................... 77 vi LISTOFTABLES Table5.1:IsolatedZ-eigenpairsofthetensorinExample5.2.1.........46 Table6.1:Comparisonof teig and eeig with PSOLVE and NSolve .......57 Table6.2:Performanceof teig oncomputingQi-eigenpairsofcomplexrandom tensors..................................57 Table6.3:Performanceof eeig oncomputingE-eigenpairsofcomplexrandom tensors..................................58 Table6.4:Performanceof teig and teneig oncomputinggeneralizedeigenpairs ofcomplexrandomtensors.......................59 Table6.5:EigenpairsofthetensorinExample6.2.1...............61 Table6.6:Z-eigenpairsofthetensorinExample6.3.1..............65 Table6.7:H-eigenpairsofthetensorinExample6.3.1..............65 Table6.8: zeig vsAlgorithm3.6([12]):CPUtime................66 Table6.9:Z-eigenvaluesofthetensorinExample6.3.3(*denotesthattheCPU timeusedbyAlgorithm3.6([12])whenitthethreelargest Z-eigenvalues)..............................67 TableA.1:Z-eigenpairsofthetensorinProblem1................69 TableA.2:Z-eigenvaluesofthetensorinProblem3................70 TableA.3:Z-eigenvaluesofthetensorinProblem4................71 TableA.4:Z-eigenpairsofthetensorinProblem5................72 TableA.5:Z-eigenpairsofthetensorinProblem7................73 TableA.6:Z-eigenpairsofthetensorinProblem8................73 TableA.7:Z-eigenpairsofthetensorinProblem9................74 vii TableA.8:Z-eigenpairsofthetensorinProblem10................75 TableA.9:Z-eigenpairsofthetensorinProblem11................75 TableA.10:Z-eigenpairsofthetensorinProblem12................76 viii Chapter1 Introduction EigenvaluesoftensorswereintroducedbyQi[35]andLim[29]in2005.Sincethen,tensor eigenvalueshavefoundapplicationsinautomaticcontrol,statisticaldataanalysis, tensorimaging,imageauthenticityvn,spectralhypergraphtheory,andquantum entanglement,etc.,seeforexample,[7,10,21,35,36,38,39,40]andthereferencestherein. Thetensoreigenvalueproblemhasbecomeanimportantsubjectofnumericalmultilinear algebra. Variousofeigenvaluesfortensorshavebeenproposedintheliterature,includ- ingE-eigenvaluesandeigenvaluesinthecomplexandZ-eigenvalues,H-eigenvalues,and D-eigenvaluesinthereal[29,35,38].In[6],Chang,Pearson,andZhangintroduceda notionofgeneraleigenvaluesfortensorsthatseveraltypesofeigenvalues.Recently thishasbeenfurthergeneralizedbyCui,Dai,andNie[12]. Unlikethematrixeigenvalueproblem,computingeigenvaluesofthethirdorhigherorder tensorsisstillinitsinfancy[18].Severalalgorithmswhichaimatcomputingoneorsome eigenvaluesofatensorhavebeendevelopedrecently.Thesealgorithmsaredesignedfor tensorsofcertaintype,suchasentry-wisenonnegativeorsymmetrictensors. Fornonnegativetensors,Ng,Qi,andZhou[33]proposedapower-typemethodforcom- putingthelargestH-eigenvalueofanonnegativetensor.MoversionsoftheNi-Qi-Zhou methodhavebeenproposedin[31,48,49]. Forrealsymmetrictensors,Hu,Huang,andQi[20]proposedasequential 1 programmingmethodforcomputingextremeZ-eigenvalues.KoldaandMayo[23]proposed ashiftedpowermethod(SSHOPM)forcomputingoneZ-eigenvalue.Theyhaveimproved SSHOPMin[24]byupdatingtheshiftparameteradaptively.Theresultingmethodcanbe usedtocomputearealgeneraleigenvalue.Han[17]proposedanunconstrainedoptimization methodforcomputingarealgeneraleigenvalueforevenorderrealsymmetrictensors.The methodsin[17,23,24]canmoreeigenvaluesofasymmetrictensoriftheyareexecuted multipletimesusingtstartingpoints.Recently,Cui,Dai,andNie[12]proposeda methodforcomputingallrealgeneralizedeigenvalues. Inthisarticle,wefocusoncomputing all eigenpairsofageneralrealorcomplextensor. Asindicatedinthenextchapter,eigenpairsofatensoractuallyamountstosolving asystemofpolynomials.Naturallyonewouldtempttousemethodsinalgebraicgeometry suchastheoebnerbasismethodandresultantmethod[9]forthispurpose.Thesemethods canobtainsymbolicsolutionsofapolynomialsystem,whichareaccurate.However,they areexpensiveintermsofcomputationalcostandstoragerequirement.Moreover,they aretobeparallelized.Aclassofnumericalmethods,thehomotopycontinuation methods,canreleasetheseshortcomings.Duringthepastfewdecades,remarkableprogresses havebeenmadeonhomotopycontinuationmethodsforsolvingpolynomialsystems,seefor example,[3,26,27,32,41]. Inthisthesiswewillcomputecomplexeigenpairsofgeneraltensorsbyusinghomotopy continuationmethodstosolvetheircorrespondingpolynomialsystems.Anattractingfeature ofthehomotopycontinuationmethodsistheircapabilityofallisolatedsolutionsof polynomialsystemsandsomesolutionsinthepositivedimensionalsolutioncomponents.We proposetwohomotopytypealgorithmsforcomputingcomplexeigenpairsofatensor.These algorithmsallowustoallequivalenceclassesofisolatedeigenpairsofageneraltensor 2 andsomeeigenpairsinpositivedimensionaleigenspaces(ifthereareany).Wealsopresenta homotopymethodandastraightforwardapproachtocomputerealeigenpairsbasedonthe availablecomplexeigenpairs.Numericalresultsexhibitedtheenessandof ourmethods. Thisdissertationisorganizedasfollows.InChapter2,wemode- k generalized eigenvaluesandeigenvectorswhichextendthematrixrighteigenpairsandlefteigenpairsto higherordertensors.Somepropertiesofsucheigenpairsareproved.InChapter3,anupper boundforthenumberofequivalenceclassesofgeneralizedtensoreigenpairsusingmixed volumeisderived.InChapter4,basedontheboundsderivedinChapter3twohomotopy methodsarepresentedtocomputemode- k generalizedcomplexeigenpairs.InChapter5, ahomotopymethodandastraightforwardapproachtocomputerealmode- k generalized eigenpairsareproposed.Finally,numericalresultsarepresentedinChapter6. 3 Chapter2 Tensoreigenvaluesandeigenvectors 2.1ofgeneralizedmode- k tensoreigenvalue andeigenvectors Let F = C or R bethecomplexortherealLet m 2, m 0 2,and n bepositive integers.Denotethesetofall m th-order, n -dimensionaltensorsonthe F by F [ m;n ] .A tensorin F [ m;n ] isindexedas A =( A i 1 i 2 i m ) ; where A i 1 i 2 i m 2 F ,for1 i 1 ;i 2 ; ;i m n . For x 2 C n ,thetensor A ascalarfunction A x m := n X i 1 ; ;i m =1 A i 1 i 2 i m x i 1 x i 2 x i m : (2.1) For1 k m , A ( k ) x m 1 isan n -vectorwhose j thentryisas ( A ( k ) x m 1 ) j = n X i 1 ; ;i k 1 ;i k +1 ; ;i m =1 A i 1 i k 1 ji k +1 i m x i 1 x i k 1 x i k +1 x i m : (2.2) When k =1,thenotationforthevector A (1) x m 1 isas A x m 1 . Arealtensor A2 R [ m;n ] is positive ifthescalarfunction A x m ispositiveforall 4 x 2 R n nf 0 g .Atensor A2 F [ m;n ] is symmetric ifitsentries A i 1 i 2 i m areinvariantunder anypermutationsoftheirindices i 1 ;i 2 ; ;i m .Atensor A2 F [ m;n ] iscalledthe m th-order unittensorif A i 1 ;:::;i m = 8 > < > : 1 ; when i 1 = = i m 0 ; otherwise. Wenowintroducethefollowingmode- k generalizedeigenvalueforageneral tensor A . DEFINITION2.1.1 Let A2 F [ m;n ] and B2 F [ m 0 ;n ] .Assumethat B x m 0 isnotidentically zero.For 1 k m ,ifthereexistascalar 2 C andavector x 2 C n nf 0 g suchthat when m 6 = m 0 , A ( k ) x m 1 = B x m 0 1 ; B x m 0 =1;(2.3) when m = m 0 , A ( k ) x m 1 = B x m 1 ; (2.4) thenwecall amode- k B -eigenvalueof A and x amode- k B -eigenvectorassociatedwith . Together ( x ) iscalledamode- k B -eigenpairof A . If 2 R ;x 2 R n ,then iscalledamode- k B R -eigenvalueof A and x amode- k B R - eigenvectorassociatedwith ,and ( x ) amode- k B R -eigenpairof A . Denotethesetofallmode- k B eigenvaluesof A by ˙ B ( A ( k ) ) . REMARK2.1.1 Let( x )beamode- k B -eigenpairof A .By(2.3)or(2.4),( x )isa solutionof A ( k ) x m 1 = B x m 0 1 ,sois( 0 ;x 0 )with 0 = t m m 0 and x 0 = tx for t 2 C nf 0 g . Hencethesolutionspaceof A ( k ) x m 1 = B x m 0 1 consistsoftequivalenceclasses. 5 Wedenotesuchanequivalenceclassby [( x )]:= f ( 0 ;x 0 ) j 0 = t m m 0 x 0 = tx;t 2 C nf 0 gg : When m 6 = m 0 ,takingarbitrary( 0 ;x 0 ) 2 [( x )]andsubstituting x 0 = tx into B x m 0 =1 in(2.3)yields t m 0 =1,whichgives m 0 tvaluesfor t .Hencethenormalization B x m 0 =1in(2.3)restrictsthechoicesof m 0 representativesolutionsfromeachequivalence class. Inlaterdiscussions,weoftenchooseonerepresentativefromeachequivalenceclass. REMARK2.1.2 Ifonlyonerepresentativeisneededfromeachequivalenceclassofeigen- pairs,wecansolve A ( k ) x m 1 = B x m 0 1 augmentedwithanadditionallinearequation a 1 x 1 + a 2 x 2 + + a n x n + b =0 ; (2.5) withgenericcomplexnumbers a 1 ;:::;a n ;b .Thennormalizetheresultingsolutionstosatisfy B x m 0 =1inthecase m 6 = m 0 . Theeigenvalues/eigenvectorsin[6,12,35,38]aremode-1eigenvalues/eigenvectors. Thetensorsconsideredinthosepapersareprimarilyrealsymmetrictensors.Forsymmet- rictensors,thesetsofmode- k B -eigenpairsandmode-1 B -eigenpairsarethesameforany k .Therefore,mode-1eigenvaluesservethepurposeofthosearticles.Nonetheless,non- symmetrictensorsalsoappearinapplicationsandtheoreticalstudies,see,forexample, [4,5,14,33,45,46].In[29],Limmode- k eigenvalues/eigenvectorsfornonsymmetric realtensors A when B isthe m 0 thorderunittensorforsome m 0 2.2.1.1 containsmoregeneral A and B . 6 Asin[6,12],2.1.1adaptsaapproachtotensoreigenvalues.It coversvarioustypesoftensoreigenvaluesintheliterature,including If A2 R [ m;n ] , m 0 =2,and B istheidentitymatrix I n 2 R n n ,themode-1 B -eigenpairs aretheE-eigenpairsandthemode-1 B R -eigenpairsaretheZ-eigenpairsdin[35], whichsatisfy A x m 1 = x T x =1 : (2.6) If A2 R [ m;n ] , m 0 =2and B = D ,where D 2 R n n isasymmetricpositive matrix,the B R -eigenpairsaretheD-eigenpairsin[38],whichsatisfy A x m 1 = x;x T Dx =1 : (2.7) If A2 R [ m;n ] , m = m 0 and B = I istheunittensor,thenmode-1 B -eigenpairs in[35]satisfying A x m 1 = [ m 1] ; (2.8) where x [ m 1] =[ x m 1 1 ;x m 1 2 ; ;x m 1 n ] T ,iscalledaQi-eigenpair. If A2 R [ m;n ] , m = m 0 and B = I istheunittensor,thenmode-1 B R -eigenvaluesare theH-eigenvaluesin[35]. 7 2.2Propertiesofgeneralizedmode- k tensoreigenval- ues When m = m 0 =2and B = I n (the n n identitymatrix),thenmode-1eigenvectorsare righteigenvectorsof A andmode-2eigenvectorsarelefteigenvectorsof A ,andthemode-1 andmode-2eigenvaluesaretheeigenvaluesofmatrix A ,i.e., ˙ B ( A (1) )= ˙ B ( A (2) ).However, when m 3, ˙ B ( A ( k ) )and ˙ B ( A ( l ) )arenotequalingeneralwhen k 6 = l ,unless A hasa certaintypeofsymmetry.Thefollowingexampleillustratesthis EXAMPLE2.2.1 Letthetensor A2 R [3 ; 2] be A 111 =1 ;A 121 =2 ;A 211 =3 ;A 221 =4 ; A 112 =5 ;A 122 =6 ;A 212 =7 ;A 222 =0 : Choose m 0 =2and B = I 2 (the2 2identitymatrix).Inthiscase,if( x )isan B -eigenpair of A ,sois( x ).Wefollow[4],taking( x )and( x )asthesameeigenpair.Then ˙ B ( A (1) )= f 0 : 4105 ; 4 : 3820 ; 9 : 8995 g ; ˙ B ( A (2) )= f 0 : 2851 ; 4 : 3536 ; 9 : 5652 g ; ˙ B ( A (3) )= f 0 : 2936 ; 4 : 3007 ; 9 : 4025 g : Clearly, ˙ B ( A ( k ) ) 6 = ˙ B ( A ( l ) )when k 6 = l . PROPOSITION2.2.1 Let A2 F [ m;n ] and B2 F [ m 0 ;n ] .If ( x ) isamode- k B -eigenpair and ( x ) isamode- l B -eigenpairof A suchthat B x m 0 6 =0 ,then = . 8 Proof :Since( x )isamode- k B -eigenpairof A , A ( k ) x m 1 = B x m 0 1 : (2.9) Recallthat A ( k ) x m 1 isby A ( k ) x m 1 = 0 B B B B B B B B B B B B B @ ( A ( k ) x m 1 ) 1 . . . ( A ( k ) x m 1 ) j . . . ( A ( k ) x m 1 ) n 1 C C C C C C C C C C C C C A with( A ( k ) x m 1 ) j givenby(2.2).Similarly, B x m 0 1 = 0 B B B B B B B B B B B B B @ ( B x m 0 1 ) 1 . . . ( B x m 0 1 ) j . . . ( B x m 0 1 ) n 1 C C C C C C C C C C C C C A ; where ( B x m 0 1 ) j = n X i 2 ;:::;i m' =1 B ji 2 :::i m' x i 2 :::x i m' : Thenmultiplyingbothsidesof(2.9)by x T fromtheleftyields x T A ( k ) x m 1 = T B x m 0 1 : 9 By x T A ( k ) x m 1 = A x m and x T B x m 0 1 = B x m 0 .So A x m = B x m 0 : Similarly,( x )isamode- l B -eigenpairof A willimplythat A x m = B x m 0 .Therefore, B x m 0 = B x m 0 .Henceif B x m 0 6 =0,then = . 2 Let A2 F [ m;n ] .For1 k > < > > : n ( m 1) n 1 ;m = m 0 ( m 1) n ( m 0 1) n m m 0 ;m 6 = m 0 : (3.4) Asaconsequence,when N isthenumberofequivalenceclassesofmode- k B -eigenpairsof A over C ,then N n ( m 1) n 1 for m = m 0 and N ( m 1) n ( m 0 1) n m m 0 for m 6 = m 0 .When A and B aregeneric,theaboveequalityholdsbyTheorem3.2.1and Theorem3.2.2. Toprove(3.4),let A2 C [ m;n ] and B2 C [ m 0 ;n ] begenerictensors.Similarto(3.1)the polynomialsystemcorrespondingtotheeigenproblem A ( k ) x m 1 = B x m 0 1 is T ( x )= 0 B B B B B B B B B @ ( A ( k ) x m 1 ) 1 ( B x m 0 1 ) 1 . . . ( A ( k ) x m 1 ) n ( B x m 0 1 ) n a 1 x 1 + a 2 x 2 + + a n x n + b 1 C C C C C C C C C A =0 : (3.5) Since A and B aregeneric,withoutlossofgeneralityonemayassumeallmonomials f x 1 1 x 2 2 :::x n n i 2 Z 0 ; 1 + 2 + + n = m 1 g 17 and f 1 1 x 2 2 :::x n n i 2 Z 0 ; 1 + 2 + + n = m 0 1 g willappearineachofthe n equationsin(3.5).Therefore,aftersubstituting(3.3)into the n equationsof(3.5),eachequationofthenewsystem T ( z ):= T ( x 1 ;:::;x n 1 ) containsallmonomials f x 1 1 x 2 2 :::x n 1 n 1 i 2 Z 0 ; 1 + 2 + + n 1 m 1 g and f 1 1 x 2 2 :::x n 1 n 1 i 2 Z 0 ; 1 + 2 + + n 1 m 0 1 g Sothesupports S 1 ;:::; S n of T areallequalto S := f (0 ; ) 2 ( Z n 1 0 ) T ; j j m 1 g[f (1 ; ) 2 ( Z n 1 0 ) T ; j j m 0 1 g : Let Q betheconvexhullof S .Thenverticesof Q aregivenbythefollowingpointsin( Z n ) T : z 0 =(0 ; 0 ;:::; 0) ; z 1 =(0 ; 0 ;:::; 0 ;m 1) ; z 2 =(0 ; 0 ;:::; 0 ;m 1 ; 0) ; . . . z n 1 =(0 ;m 1 ; 0 ;:::; 0) ; 18 z n =(1 ; 0 ;:::; 0) ; z n +1 =(1 ; 0 ;:::; 0 ;m 0 1) ; z n +2 =(1 ; 0 ;:::; 0 ;m 0 1 ; 0) ; . . . z 2 n 1 =(1 ;m 0 1 ; 0 ;:::; 0) : Denotethe i -thunitvectorin( R n ) T by e i for i =1 ;:::;n .Then z i = 8 > > > > > > > > > < > > > > > > > > > : 0 ;i =0 ( m 1) e n +1 i ; 1 i n 1 e 1 ;i = n e 1 +( m 0 1) e 2 n +1 i ;n +1 i 2 n 1 Tocomputethevolumeof Q ,wedivideitintofollowing n simplices Q 1 :=conv( z 0 ;z 1 ;:::;z n ) ; Q 2 :=conv( z 1 ;z 2 ;:::;z n +1 ) ; . . . Q i :=conv( z i 1 ;z i ;:::;z n + i 1 ) ; . . . Q n :=conv( z n 1 ;z n ;:::;z 2 n 1 ) : 19 Itfollowsthat Vol n ( Q 1 )= 1 n ! det 0 B B B B B B B B B @ z 1 z 0 z 2 z 0 . . . z n z 0 1 C C C C C C C C C A = 1 n ! det 0 B B B B B B B B B B B B B @ ( m 1) e n ( m 1) e n 1 . . . ( m 1) e 2 e 1 1 C C C C C C C C C C C C C A = ( m 1) n 1 n ! det 0 B B B B B B B B B B B B B @ e n e n 1 . . . e 2 e 1 1 C C C C C C C C C C C C C A = ( m 1) n 1 n ! det 0 B B B B B @ e 1 . . . e n 1 C C C C C A = ( m 1) n 1 n ! : (3.6) Obviouslythat z n iscontainedineach Q i for i =2 ;:::;n ,so Vol n ( Q i )= 1 n ! det 0 B B B B B B B B B B B B B B B B B B B B B @ z i 1 z n z i z n . . . z n 1 z n z n +1 z n . . . z n + i 1 z n 1 C C C C C C C C C C C C C C C C C C C C C A = 1 n ! det 0 B B B B B B B B B B B B B B B B B B B B B B B B @ ( m 1) e n +2 i e 1 ( m 1) e n +1 i e 1 . . . ( m 1) e 2 e 1 ( m 0 1) e n ( m 0 1) e n 1 . . . ( m 0 1) e n +2 i 1 C C C C C C C C C C C C C C C C C C C C C C C C A : 20 Furthercomputationgives Vol n ( Q i )= 1 n ! det 0 B B B B B B B B B B B B B B B B B B B B B B B B @ e 1 ( m 1) e n +1 i . . . ( m 1) e 2 ( m 0 1) e n ( m 0 1) e n 1 . . . ( m 0 1) e n +2 i 1 C C C C C C C C C C C C C C C C C C C C C C C C A = ( m 1) n i ( m 0 1) i 1 n ! det 0 B B B B B B B B B B B B B B B B B B B B B B B B @ e 1 e n +1 i . . . e 2 e n e n 1 . . . e n +2 i 1 C C C C C C C C C C C C C C C C C C C C C C C C A = ( m 1) n i ( m 0 1) i 1 n ! det 0 B B B B B @ e 1 . . . e n 1 C C C C C A = ( m 1) n i ( m 0 1) i 1 n ! : (3.7) Comparing(3.6)with(3.7),(3.7)actuallyalsoholdsfor i =1.Thus Vol n ( Q )= n X i =1 Vol n ( Q i )= n X i =1 ( m 1) n i ( m 0 1) i 1 n ! : Therefore,byLemma3.2.1 MV n ( S 1 ;:::; S n )= n !Vol n ( Q )= n X i =1 ( m 1) n i ( m 0 1) i 1 : Notethatfor i =1 ;:::;n , S i [f 0 gˆ S i .Hence MV n ( S 1 [f 0 g ;:::;S n [f 0 g ) MV n ( S 1 ;:::; S n )= n X i =1 ( m 1) n i ( m 0 1) i 1 : 21 Consequently, MV n ( S 1 [f 0 g ;:::;S n [f 0 g ) n X i =1 ( m 1) n 1 = n ( m 1) n 1 (3.8) for m = m 0 and MV n ( S 1 [f 0 g ;:::;S n [f 0 g ) ( m 1) n m 0 1 n X i =1 m 0 1 m 1 i = ( m 1) n m 0 1 m 0 1 m 1 1 m 0 1 m 1 n 1 m 0 1 m 1 =( m 1) n 1 m 0 1 m 1 n m m 0 = ( m 1) n ( m 0 1) n m m 0 (3.9) for m 6 = m 0 . Ontheotherhand,for m = m 0 ,letthediagonaltensors A2 C [ m;n ] and B2 C [ m;n ] be suchthat A ii:::i = i , B ii:::i =1andallotherentrieszero.Thenthenumberofequivalence classesofmode- k B -eigenpairsof A isequaltothenumberofsolutionstothefollowing systemofpolynomials 0 B B B B B B B B B B B B B @ x m 1 1 m 1 1 2 x m 1 2 m 1 2 . . . nx m 1 n m 1 n x 1 + x 2 + + x n 1 1 C C C C C C C C C C C C C A =0 : 22 Thezerosofthissystemare( x 1 ;:::;x n )=(1 ; 1 ; 0 ;:::; 0),(2 ; 0 ; 1 ; 0 ;:::; 0),...,( n; 0 ;:::; 0 ; 1) andeachzerohasmultiplicity( m 1) n 1 .ByTheorem3.2.2, MV n ( S 1 [f 0 g ;:::;S n [f 0 g ) n ( m 1) n 1 : Combiningwith(3.8),wehave MV n ( S 1 [f 0 g ;:::;S n [f 0 g )= n ( m 1) n 1 : For m 6 = m 0 ,considerthediagonaltensors A2 C [ m;n ] and B2 C [ m 0 ;n ] suchthat A ii:::i =1, B ii:::i =1andallotherentriesarezero.Assume m>m 0 .Thenthenumberof equivalenceclassesofmode- k B -eigenpairsof A isequaltothenumberofsolutionstothe followingsystemofpolynomials 0 B B B B B B B B B @ x m 1 1 m 0 1 1 . . . x m 1 n m 0 1 n a 1 x 1 + a 2 x 2 + + a n x n + b 1 C C C C C C C C C A = 0 B B B B B B B B B @ x m 0 1 1 ( x m m 0 1 ) . . . x m 0 1 n ( x m m 0 n ) a 1 x 1 + a 2 x 2 + + a n x n + b 1 C C C C C C C C C A =0 ; (3.10) where a 1 ;:::;a n ;b arerandomcomplexnumbers.Obviously x =0cannotbeasolution sinceitfailstosatisfytheaugmentedrandomhyperplane.AsdiscussedinRemark2.1.1, thehyperplaneisaddedtoensurethatonlyonerepresentativefromeachequivalenceclass canbeselected.Soweneedtothenumberofequivalenceclassesofeigenpairs( x ) fromthe n equationsof(3.10).Foreachd ,thereare m 1choicesforeach x i andamongthosechoices, m 0 1ofthemarezeros.Excludingthosecombinationswhich 23 make x =0,thereare( m 1) n ( m 0 1) n choicesfor x 1 ;:::;x n intotal.Furthermore, foreacheigenvalue ,let x bethecorrespondingsolutionofthe n equationsof(3.10). Then tx with t m m 0 =1isalsoasolutionassociatedwith .Thusforeacheigenvalue , solvingthe n equationsof(3.10)resultsin m m 0 correspondingsolutionsof x .By Remark2.1.1,these m m 0 solutionsareactuallyequivalent.Therefore,thereshouldbe (( m 1) n ( m 0 1) n ) = ( m m 0 )equivalenceclassesofeigenpairs,i.e.,(3.10)musthave (( m 1) n ( m 0 1) n ) = ( m m 0 )isolatedzerosin C n +1 intotal.ByTheorem3.2.2, MV n ( S 1 [f 0 g ;:::;S n [f 0 g ) (( m 1) n ( m 0 1) n ) = ( m m 0 ) : Combiningtheaboveinequalitywith(3.9)gives MV n ( S 1 [f 0 g ;:::;S n [f 0 g )=(( m 1) n ( m 0 1) n ) = ( m m 0 ) : 2 REMARK3.3.1 (a)ForQi-eigenpairs,wehave m 0 = m .Using(a)inTheorem3.3.1theupperboundof thenumberofequivalenceclassesofQi-eigenpairsis n ( m 1) n 1 .Thisresultincludes Theorem3.1.1provedin[35],inwhichthisboundofthenumberofQi-eigenvaluesis restrictedtoasymmetrictensorwhen m iseven. (b)ForE-eigenpairs, m 0 =2.Using(b)inTheorem3.3.1theupperboundofthenumber ofequivalenceclassesofE-eigenpairsis(( m 1) n 1) = ( m 2),whichagreeswith Theorem3.1.2provedin[4]. 24 (c)TheupperboundsprovidedinTheorem3.3.1isveryusefulindesigningeho- motopymethodsforcomputingmode- k generalizedeigenpairs.Infact,thehomotopy methoddescribedinAlgorithm4.3.1forthecase m = m 0 inthenextchapterrelieson thebound n ( m 1) n 1 . 25 Chapter4 Computingcomplextensoreigenpairs byhomotopymethods Let A2 C [ m;n ] and B2 C [ m 0 ;n ] .AsdiscussedinChapter2,theproblemofcomputingmode- k B -eigenpairsof A in(2.3)isequivalenttosolving(3.1),andwhen m 6 = m 0 ,wenormalize ( x )tosatisfy B x m 0 =1.Forpolynomialsystem(3.1),thehomotopycontinuationmethod iscommonlyusedtoitsnumericalsolutions. 4.1Usinghomotopycontinuationmethodstosolvepoly- nomialsystems Thebasicideaofusinghomotopycontinuationmethodtosolveageneralpolynomialsystem P ( x )=( p 1 ( x ) ;:::;p n ( x )) T =0asin(3.2)istodeform P ( x )toanotherpolynomial system Q ( x )withknownsolutionsintheplace.Thenundercertainconditions,asmooth curvethatemanatesfromasolutionof Q ( x )=0willleadtoasolutionof P ( x )=0. Fortheclassicallinearhomotopy[1]: H ( x;t )=(1 t ) Q ( x )+ tP ( x )=0 ;t 2 [0 ; 1] ; (4.1) where isagenericnonzerocomplexnumber,if Q ( x )ischosenproperly,thefollowing 26 propertieshold: Property0 (Triviality) Thesolutionsof Q ( x )=0areknownoreasytosolve. Property1 (Smoothness) Thesolutionsetof H ( x;t )=0for0 d i ((1 t ) c + tc ) x t h ^ ; ^ d i =((1 t ) c i 1 + tc i 1 ) x i 1 +((1 t ) c i 2 + tc i 2 ) x i 2 + X 2 S 0 i h ^ ; ^ i >d i ((1 t ) c + tc ) x t h ^ ; ^ d i ; wehave h i ( x; 0)= c i 1 x i 1 + c i 2 x i 2 29 foreach i =1 ;:::;n .Hence H ( x; 0)= 0 B B B B B B B B B B B B B @ c 1 11 x 11 + c 1 12 x 12 . . . c i 1 x i 1 + c i 2 x i 2 . . . c n 1 x n 1 + c n 2 x n 2 1 C C C C C C C C C C C C C A =0 : Thisisabinomialsystemhaving k :=det 0 B B B B B B B B B B B B B @ 11 12 . . . i 1 i 2 . . . n 1 n 2 1 C C C C C C C C C C C C C A nonsingularsolutionsin( C ) n .Averyetandstablenumericalmethodexistsfor allthosesolutions[27].Startingfromthesesolutionstotracksolutionpathsof(4.3), wewillreachsomesolutionsof P ( x )=0.tmixedcellswiththeircorresponding innernormals willdevotetpolyhedralhomotopies H ( x;t )asin(4.3).These thomotopieswillreachtisolatedzerosof P ( x )=0asprovedin[27].And 30 thetotalnumberofpathsneededtobetrackedhereis X k = X det 0 B B B B B B B B B B B B B @ 11 12 . . . i 1 i 2 . . . n 1 n 2 1 C C C C C C C C C C C C C A ; whichagreeswiththemixedvolumeMV n ( S 0 1 ;:::;S 0 n ). Althoughthepolyhedralhomotopycontinuationmethodusuallyfollowsfewerpathsthan theclassicallinearhomotopymethod,themixedcellcomputationinvolvedmaybecome veryexpensiveespeciallyforlargepolynomialsystems.Thusifalinearhomotopycanbe constructedsothatonlymixedvolumenumberofpathsneedtobetraced,thesystemshould besolvedbyusingalinearhomotopyinsteadofthepolyhedralhomotopy. 4.2Constructionofalinearhomotopytosolvetensor eigenpairs Tosolve(3.1),onecancertainlyusethepolyhedralhomotopycontinuationmethodimple- mentedinHOM4PS[26],PHCpack[42],PHoM[16],PSOLVE[47](whichisaMATLAB implementationofHOM4PS),orthetotaldegreehomotopycontinuationmethodimple- mentedinBertini[3].However,usingthesemethodstosolve(3.1)directlydoesnottake advantageofthespecialstructuresofatensoreigenproblem.Wewillintroducetwohomo- topytypealgorithmsherethatutilizesuchstructures. 31 Theorem3.3.1assertsthatthemixedvolumeof(3.1)is n ( m 1) n 1 for m = m 0 and (( m 1) n ( m 0 1) n ) = ( m m 0 )for m 6 = m 0 ,whichisfarlessthantheBezout'snumber, m n for m = m 0 ,andmax f ( m 1) n ; ( m 0 ) n g for m 6 = m 0 .Fromthisstandpoint,weconstruct alinearhomotopywithwhichonlymixedvolumenumberofpathsneedtobetraced. Forthislinearhomotopy,theknowledgeofmultihomogeneousBezout'snumber[41] isrequired.Forapolynomialsystem P ( x )=( p 1 ( x ) ;:::;p n ( x )) T in(3.2),where x = ( x 1 ;:::;x n ),wepartitionthevariables x 1 ;:::;x n into k groups y 1 =( x (1) 1 ;:::;x (1) l 1 ) ;y 2 = ( x (2) 1 ;:::;x (2) l 2 ) ;:::;y k =( x ( k ) 1 ;:::;x ( k ) l k )with l 1 + + l k = n .Let d ij bethedegreeof p i withrespectto y j for i =1 ;:::;n and j =1 ;:::;k .ThenthemultihomogeneousBezout's numberof P ( x )withrespectto( y 1 ;:::;y k )isthecotof l 1 1 l 2 2 ::: l k k intheproduct n Y i =1 ( d i 1 1 + + d ik k ) : Thefollowingtheoremwillplayaveryimportantroleinconstructingourlinearhomotopy. THEOREM4.2.1 [41] Let Q ( x ) beasystemofpolynomialshavingthesamemultihomoge- neousstructureas P ( x ) withrespecttocertainpartitionofthevariables ( x 1 ;:::;x n ) .Assume Q ( x )=0 hasexactlythemultihomogeneousBezout'snumberofnonsingularsolutionswith respecttothispartition,andlet H ( x;t )=(1 t ) Q ( x )+ tP ( x )=0 ; where t 2 [0 ; 1] and 2 C .For = re ,Properties1and2holdforallbutmany . 32 For(3.1),when m = m 0 thefollowingpolynomialsystem G ( x )= 0 B B B B B B B B B @ ( A ( k ) x m 1 ) 1 ( B x m 1 ) 1 . . . ( A ( k ) x m 1 ) n ( B x m 1 ) n a 1 x 1 + a 2 x 2 + + a n x n + b 1 C C C C C C C C C A =0(4.4) needstobesolved,where and x :=( x 1 ; ;x n ) T aretheunknowns, a 1 ;:::;a n ;b are randomlychosencomplexnumbers.Considerthesystem Q ( x )= 0 B B B B B B B B B B B B B @ ( 1 )( x m 1 1 1 ) ( 2 )( x m 1 2 2 ) . . . ( n )( x m 1 n n ) c 1 x 1 + :::c n x n + d 1 C C C C C C C C C C C C C A =0 ; (4.5) where d aswellas i ; i ;c i for i =1 ;:::;n aregenericnonzerocomplexnumbers. THEOREM4.2.2 For G ( x ) and Q ( x ) givenabove,allisolatedzeros ( x ) of G ( x ) in C n +1 canbefoundbythehomotopy H ( x;t )=(1 t ) Q ( x )+ tG ( x )=0 ;t 2 [0 ; 1](4.6) foralmostall 2 C . Proof :Evidently,withrespecttothepartition( )and( x 1 ;x 2 ;:::;x n )ofthevariables ( x 1 ;x 2 ;:::;x n ),bothsystems(4.4)and(4.5)havedegree1in( )anddegree m 1in 33 ( x 1 ;x 2 ;:::;x n )forthe n equations,anddegree0in( )anddegree1in( x 1 ;x 2 ;:::;x n ) forthelastequation.Hence(4.4)and(4.5)havethesamemultihomogeneousBezout's number,thatisthecotof 1 n 2 intheproduct [1 1 +( m 1) 2 ] n (0 1 +1 2 ) ; whichis 0 B @ n 1 1 C A ( m 1) n 1 = n ( m 1) n 1 : Wenowshow Q ( x )in(4.5)hasexactly n ( m 1) n 1 zeros.Intheplace,if is equaltononeof 1 ;:::; n ,then(4.5)becomesasystemof n +1equationsand n unknowns, whichisoverdeterminedwithgenericcots.Itthereforehasnosolutions.Thus must beequaltooneof 1 ;:::; n .Assume = 1 ,then x 1 ;:::;x n canbedeterminedby x m 1 2 2 =0 . . . x m 1 n n =0 c 1 x 1 + :::c n x n + d =0 Obviously,each x i for i =2 ;:::;n canbechosenasoneofthe( m 1)-throotof i and x 1 canbesolvedbysubstitutingthechosen x 2 ;:::;x n intothelastequation.Sothereare ( m 1) n 1 solutionscorrespondingto = 1 .Thisargumentholdsfor beinganyofthe i 's.Therefore,thereare n ( m 1) n 1 solutionsintotal. Itremainstoprovethateachsolutionof Q ( x )=0in(4.5)isnonsingular.Asdiscussed 34 above,anysolution( ;x )of(4.5) = i ( x 1 ) m 1 1 =0 . . . ( x i 1 ) m 1 i 1 =0(4.7) ( x i +1 ) m 1 i +1 =0 . . . ( x n ) m 1 n =0 c 1 x 1 + + c n x n + d =0 Let DQ ( x )betheJacobianof Q ( x )at( x ).Fornonsingularityof DQ ( ;x ),let A j ( x ):= x m 1 j j ;B j ( x ):=( j )( m 1) x m 2 j 35 for j =1 ;:::;n .Then DQ ( x )= 0 B B B B B B B B B B B B B B B B B B B B B B B B @ A 1 B 1 . . . . . . A i 1 B i 1 A i B i A i +1 B i +1 . . . . . . A n B n 0 c 1 :::c i 1 c i c i +1 :::c n 1 C C C C C C C C C C C C C C C C C C C C C C C C A : With A j ( ;x )=( x j ) m 1 j =0 ;j 6 = i and B i ( ;x )=( i )( m 1)( x i ) m 2 =( i i )( m 1)( x i ) m 2 =0 ; 36 wehave,by(4.7), DQ ( ;x )= 0 B B B B B B B B B B B B B B B B B B B B B B B B @ 0 B 1 . . . . . . 0 B i 1 A i 0 0 B i +1 . . . . . . 0 B n 0 c 1 :::c i 1 c i c i +1 :::c n 1 C C C C C C C C C C C C C C C C C C C C C C C C A ; where A j := A j ( ;x )and B j := B j ( ;x ).Itfollowsthat det( DQ ( ;x ))=( 1) i +1 A i ( 1) n + i c i Y j 6 = i B j 6 =0 by(4.7). 2 4.3Numericalalgorithmsofusinghomotopycontinu- ationmethodstosolvetensoreigenpairs Theorem4.2.2suggeststhat(4.6)canbeusedtosolve(3.1)when m = m 0 .Write u :=( x ), then(4.6)becomes H ( u;t )=(1 t ) Q ( u )+ tG ( u )=0 ;t 2 [0 ; 1](4.8) 37 where Q and G arein(4.5)and(4.4)respectively. Wenowpresentouralgorithmforcomputingmode- k generalizedeigenpairswhen m = m 0 ,i.e.,solving(4.4). ALGORITHM4.3.1 (Computecomplexmode- k B -eigenpairsof A ,where A ; B2 C [ m;n ] , i.e.,solving(4.4).) Step1. Computeallsolutionsof Q ( u ) asdin(4.5). Step2. Pathfollowing:Followthepathsfrom t =0 to t =1 usingtheprediction- correctionstrategy.Let ( u k ;t k ):=( u ( t k ) ;t k ) .Forthenextpoint ( u k +1 ;t k +1 ) onthe solutionpathof H ( u;t )=(1 t ) Q ( u )+ tG ( u )=0 ;t 2 [0 ; 1] asdin(4.8),thefollowingstepsareemployed: PredictionStep:Computethetangentvector du dt to H ( u;t )=0 at t k bysolvingthe linearsystem H u ( u k ;t k ) du dt = H t ( u k ;t k ) for du dt .Thencomputetheapproximation ~ u to u k +1 by ~ u = u k + t du dt ;t k +1 = t k + t; where t isthestepsize.Here u 0 ischosentobeonesolutionof Q ( u )=0 . CorrectionStep:UseNewton'siterations,i.e.,for i =0 ; 1 ; 2 ;::: ,compute v i +1 = v i [ H u ( v i ;t k +1 )] 1 H ( v i ;t k +1 ) with v 0 =~ u 38 until k H ( v J ;t J ) k isverysmall.Thenlet u k +1 = v J . Step3. Endgame.When t N isverycloseto1,thecorresponding u N shouldbevery closetoazero u of G ( u )= G ( x ) .SoNewton'siterations u ( k +1) = u ( k ) [ DG ( u ( k ) )] 1 G ( u ( k ) ) ;k =0 ; 1 ;::: willbeusedagaintorourapproximation ~ u to u .If DG ( u ) isnonsingular,then ~ u willbeaverygoodapproximationof u withmultiplicity1.If DG ( u ) issingular, ~ u is eitheranisolatedsingularzeroof G ( u ) withmultiplicity l> 1 orinapositivedimensional solutioncomponentof G ( u )=0 .WeuseastrategyprovidedinChapterVIIIof[27](see also[41])todeterminewhether ~ u isanisolatedzerowithmultiplicity l> 1 orinapositive dimensionalsolutioncomponentof G ( u )=0 . Step4. Foreachsolution u =( x ) obtainedinStep3,normalize x with i 0 := argmax 1 i n j x i j : y = x x i 0 (4.9) aneweigenpair ( y ) isobtained,sinceasmentionedin(2.4)andRemark2.1.1,if ( x ) is aneigenpair, ( tx ) for t 6 =0 isalsoaneigenpair. REMARK4.3.1 Noticethatif x isarealeigenvectorassociatedwitharealeigenvalue , tx forany t 2 C nf 0 g willbeacomplexeigenvectorassociatedwiththesameeigenvalue .If atanystageacomplexeigenvectorlike tx isobtainedinStep3ofAlgorithm4.3.1,applying (4.9)to tx willgiveanewrealeigenvector.Inthisregard,Step4isveryhelpfulfordetecting realeigenpairs. Tocomputemode- k generalizedtensoreigenpairswhen m 6 = m 0 ,weusetheequivalence 39 classstructureoftheeigenproblemasdescribedinRemark2.1.1.Wesolve(3.1)by PSOLVE[47]toarepresentative( x )fromeachequivalenceclassandthenall m 0 eigenpairsfromeachequivalenceclassbysimplyusing 0 = t m m 0 x 0 = tx ,where t isa rootof t m 0 =1. Onemaysolve(2.3)directlyfor m 0 eigenpairsfromeachequivalenceclass.However, thisalternativemustfollow m 0 timesmorepathsthantheaboveapproachcostingmore computations. Wenowpresentouralgorithmforcomputingmode- k generalizedeigenpairswhen m 6 = m 0 ,i.e.,solving(2.3). ALGORITHM4.3.2 (Computecomplexmode- k B -eigenpairsof A ,where A2 C [ m;n ] ; B2 C [ m 0 ;n ] with m 6 = m 0 ,i.e.,solving(2.3).) Step1. UsingPSOLVEtogetallsolutions ( x ) of(3.1). Step2. Foreach ( x ) obtainedinStep1,if B x m 0 6 =0 ,normalizeittogetaneigenpair ( ;x ) by = ( B x m 0 ) ( m m 0 ) =m 0 ;x = x ( B x m 0 ) 1 =m 0 tosatisfy(2.3). Step3. Compute m 0 equivalenteigenpairs ( 0 ;x 0 ) of ( ;x ) by 0 = t m m 0 and x 0 = tx with t beingarootof t m 0 =1 . 40 Chapter5 Computingrealtensoreigenpairsby homotopymethods Insomeapplications,tensor A isrealandonlyrealeigenpairs(orrealeigenvalues)of A areofinterest([12,35]).Inthissituation,onlyrealzerosofthepolynomialsystem(2.3)or (4.4)areneeded.Currentlythereisnoeemethodtoallrealzerosofapolynomial systemdirectly. 5.1UsingNewtonhomotopymethodtorealten- soreigenpairs Forarealtensor A ,arealeigenvaluemayhavecomplexeigenvectors.Toidentifyreal eigenvalues,wemaycomputecomplexzeros( x )of(4.4)byAlgorithm4.3.1or(2.3) byAlgorithm4.3.2,thenclassifytherealeigenvaluesbycheckingthesizeoftheirimaginary parts.Sp,let( ;x )beacomputedeigenpair.If j Im( ) j < 0 ; withthreshold 0 > 0,thenRe( )willbetakenasarealeigenvalue. Foraspoccasion,when m 6 = m 0 ,if m 0 m m 0 isanonzerointegermultipleof4(for 41 example, m =5 ;m 0 =4or m =10 ;m 0 =8,)and A hasaneigenpair( ;x )withapurely imaginaryeigenvalue = bi ,where b 2 R ,thenonecaneasilyshowthat( b; ( i ) 1 = ( m m 0 ) x ) and( b;i 1 = ( m m 0 ) x )areeigenpairswithrealeigenvalues.Therefore,when m 0 m m 0 isa nonzerointegermultipleof4,if( ;x )isaneigenpairfoundbyAlgorithm4.3.2suchthat j Re( ) j < 0 ; thenwetakeIm( )and Im( )asrealeigenvalues,withcorrespondingeigenvectors ( i ) 1 = ( m m 0 ) x and i 1 = ( m m 0 ) x . Whenlookforrealtensoreigenpairs(i.e.,botheigenvaluesandeigenvectorsarereal), thesituationbecomesmorecomplicated.Weproposeatwo-stepprocedure.First,compute complexzeros( x )of(4.4)byAlgorithm4.3.1or(2.3)byAlgorithm4.3.2.Thenextract allrealeigenpairs( x )fromthezerosjustcalculated. Forvector a =( a 1 ;:::;a n ) T 2 C n ,let Im( a )=(Im( a 1 ) ;:::; Im( a n )) T ; Re( a )=(Re( a 1 ) ;:::; Re( a n )) T : Suppose( ;x )isaneigenpairfoundinthestep.Therearepossiblytwocases:(i) ( ;x )isanisolatedeigenpair;(ii)( ;x )isaneigenpaircontainedinapositivedimen- sionalsolutioncomponentofsystem(4.4)or(2.3). When( ;x )isanisolatedeigenpair,take(Re( ) ; Re( x ))asarealeigenpairif k Im( ;x ) k 2 < 0 : If( ;x )isaneigenpairinapositivedimensionalsolutioncomponentofsystem(4.4) 42 or(2.3),ingeneralrealeigenvectorsinducedbywhicharenotwarrantedevenifthecorre- spondingeigenvalue isreal.Inthiscase,weusethefollowingNewtonhomotopy[1] H ( x;t ):= P ( x ) (1 t ) P ( ; Re( x )) ;t 2 [0 ; 1](5.1) tofollowhomotopycurvesof H ( x;t )=0in( x ) 2 R n +1 forarealeigenpair.Notice thatwhenfollowingcurvesinthecomplexspaceitisprovedin[27]thatthesolutioncurves of(5.1)canbeparameterizedby t ,butthesolutioncurvesof(5.1)maynotbeafunction of t whenrestrictedintherealspace.Soadtmethodtofollowcurvesisneeded.In thiscaseparameterizingthesolutioncurvesbythearclength s issuggestedin[27].For simplicity,write y ( s ):=( ( s ) ;x ( s ) ;t ( s )),then(5.1)becomes H ( y ( s ))=0. Wenowsummarizeouralgorithmforcomputingarealeigenpairfromacomplexeigenpair ( ;x )withreal ,whichisinapositivedimensionalsolutioncomponentof(4.4)or(2.3). ALGORITHM5.1.1 (Tracesolutionpathsof(5.1)intherealspacetogetarealeigen- pair.) Step1. Let y k := y ( s k ) andlet y 0 =( ; Re( x )) ,tothenextpointonthesolution pathof H ( y )=0 ,weusethefollowingstrategy: PredictionStep:Computethetangentvector dy ds to H ( y )=0 at y k bysolvingthe system DH ( y k ) dy ds =0 dy ds 2 =1 for dy ds .Here DH ( y k ) istheJacobianof H withrespectto y evaluatedat y k .Then 43 computetheapproximation ~ y to y k +1 by ~ y = y k + s dy ds ;s k +1 = s k + s; where s isastepsize. CorrectionStep:UseNewton'siterationstosolve ˘ from F ( ˘ ):= 0 B @ H ( ˘ ) ( ˘ ~ y ) dy ds 1 C A =0 withtheinitialpoint ˘ 0 =~ y ,i.e.,for i =0 ; 1 ; 2 ;::: ,compute ˘ i +1 = ˘ i [ DF ( ˘ i )] 1 F ( ˘ i ) until k F ( ˘ J ) k islessthanathresholdforsome J .Thenlet y k +1 = ˘ J . Step2. Endgame.DuringStep1ifaftermanyPrediction-CorrectionSteps t doesnotapproach1,thentheschemecannotprovidearealeigenpair,stop.Otherwise when t ( s N ) isverycloseto1forsome N ,thecorresponding ( ( s N ) ;x ( s N )) isverycloseto arealzero u R :=( R ;x R ) of P ( u ):= P ( x ) .SoNewton'siterations u ( k +1) = u ( k ) [ DP ( u ( k ) )] 1 P ( u ( k ) ) ;k =0 ; 1 ;::: with u (0) =( ( s N ) ;x ( s N )) willbeemployedtorourapproximation ~ u to u R .Take ~ u asarealeigenpair. 44 5.2Astraightforwardapproachtorealtensoreigen- pairs Aninterestingphenomenonisthatinsomespecialcases,therealeigenpairscanbeobtained inamorestraightforwardmannerfromthecomplexeigenpairsfoundbyAlgorithm4.3.1or Algorithm4.3.2,asillustratedinthefollowingexample. EXAMPLE5.2.1 Letthetensor A2 R [3 ; 5] (Example4.11in[12],seealso[34])begiven by A i;j;k = ( 1) i i + ( 1) j j + ( 1) k k ;i;j;k =1 ;:::; 5 : WewanttheZ-eigenpairsof A .Thecorrespondingpolynomialsystemtosolveis 3 x 2 1 3 x 1 x 2 25 6 x 1 x 3 7 2 x 1 x 4 22 5 x 1 x 5 7 6 x 2 x 3 1 2 x 2 x 4 7 5 x 2 x 5 7 6 x 2 3 13 6 x 3 x 4 181 60 x 3 x 5 1 2 x 2 4 19 10 x 4 x 5 7 5 x 2 5 1 =0 3 2 x 2 1 7 6 x 1 x 3 1 2 x 1 x 4 7 5 x 1 x 5 + 3 2 x 2 2 + 4 3 x 2 x 3 + 5 2 x 2 x 4 + 8 5 x 2 x 5 1 6 x 2 3 + 5 6 x 3 x 4 1 15 x 3 x 5 + x 2 4 + 11 10 x 4 x 5 + 1 10 x 2 5 2 =0 7 3 x 2 1 5 3 x 1 x 2 10 3 x 1 x 3 13 6 x 1 x 4 46 15 x 1 x 5 + 2 3 x 2 2 1 3 x 2 x 3 + 5 6 x 2 x 4 1 15 x 2 x 5 x 2 3 5 6 x 3 x 4 26 15 x 3 x 5 + 1 6 x 2 4 17 30 x 4 x 5 11 15 x 2 5 3 =0 7 4 x 2 1 1 2 x 1 x 2 13 6 x 1 x 3 x 1 x 4 19 10 x 1 x 5 + 5 4 x 2 2 + 5 6 x 2 x 3 +2 x 2 x 4 + x 2 x 5 5 12 x 2 3 + 1 3 x 3 x 4 17 30 x 3 x 5 + 3 4 x 2 4 + 3 5 x 4 x 5 3 20 x 2 5 4 =0 11 5 x 2 1 7 5 x 1 x 2 46 15 x 1 x 3 19 10 x 1 x 4 14 5 x 1 x 5 + 4 5 x 2 2 1 15 x 2 x 3 + 11 10 x 2 x 4 + 1 5 x 2 x 5 13 15 x 2 3 17 30 x 3 x 4 22 15 x 3 x 5 + 3 10 x 2 4 3 10 x 4 x 5 3 5 x 2 5 5 =0 x 2 1 + x 2 2 + x 2 3 + x 2 4 + x 2 5 1=0 45 Writetheabovepolynomialsystemas P ( x )=0.UsingAlgorithm4.3.2,62solutions ofthissystemarefound.Amongthem,4areisolatedsolutionswith k (Im( ) ; Im( x ) k assmall asthemachineepsilon.Therefore,theseisolatedzeroscanbeastheZ-eigenpairs. Notethatinthisexampleif( x )isaneigenpair,thensois( x ).Table5.1liststwo ofthesefourZ-eigenpairswithpositive . 4 : 2876 9 : 9779 x 1 0 : 1859 0 : 7313 x 2 0 : 7158 0 : 1375 x 3 0 : 2149 0 : 4674 x 4 0 : 5655 0 : 2365 x 5 0 : 2950 0 : 4146 Table5.1:IsolatedZ-eigenpairsofthetensorinExample5.2.1 Fortheremaining58solutionsof P ( x )=0,eachofthemhas =0andiscontained inapositivedimensionalsolutioncomponentof P ( x )=0.Foreachofthesezeros,it canbevthat( Re( x ) = k Re( x ) k 2 )and( Im( x ) = k Im( x ) k 2 )arebothsolutionsof P ( x )=0.Forexample,oneofthese58zerosis 0 B B B B B B B B B B B B B B B B B @ x 1 x 2 x 3 x 4 x 5 1 C C C C C C C C C C C C C C C C C A = 0 B B B B B B B B B B B B B B B B B @ 0 0 : 7136 0 : 4086 i 0 : 1425 0 : 0266 i 0 : 9941 0 : 2291 i 0 : 0180+0 : 1180 i 0 : 1200+0 : 5464 i 1 C C C C C C C C C C C C C C C C C A : 46 Takingtherealandimaginarypartsoftheabove x andlet ˘ =( 0 : 7136 ; 0 : 1425 ; 0 : 9941 ; 0 : 0180 ; 0 : 1200) T ; =( 0 : 4086 ; 0 : 0266 ; 0 : 2291 ; 0 : 1180 ; 0 : 5464) T : Normalizing ˘ and respectivelygives v =( 0 : 5764 ; 0 : 1151 ; 0 : 8030 ; 0 : 0146 ; 0 : 0969) T ; w =( 0 : 5599 ; 0 : 0365 ; 0 : 3139 ; 0 : 1617 ; 0 : 7487) T : Simplecomputationshows k P (0 ;v ) k 2 ˇ 5 : 8687 10 14 and k P (0 ;w ) k 2 ˇ 5 : 4921 10 14 . Therefore,(0 ;v )and(0 ;w )arebothZ-eigenpairs.Accordingly( Re( x ) = k Re( x ) k 2 )and ( Im( x ) = k Im( x ) k 2 )aretwoZ-eigenpairs. Theaboveexamplesuggeststhatif( ;x )isinapositivedimensionalsolutioncompo- nentof(2.3)and 2 R ,then( ; Re( x ) = k Re( x ) k 2 )and( ; Im( x ) = k Im( x ) k 2 )maybe mode- k B R eigenpairsof A .Thisprovidesastraightforwardapproachtorealeigenpairs fromeigenpairsbelongtopositivedimensionalcomponents.Actually,thisapproachhas beenquitesuccessfulforalltheexamples(e.g.,Example4.8,4.11,4.13,4.14)in[12]when arealZ-eigenvaluehasitelymanyrealZ-eigenvectors. Thisleadstothequestion:whatisthestructureoftheeigenspace f x 2 C n jA x m 1 = B x m 0 1 ; B x m 0 =1 g associatedwitharealeigenvalue ?Lookingintotheeigenspace of =0inExample5.2.1,i.e., f x 2 C 5 j P (0 ;x )=0 g ,allthe58eigenvectorsobtained fromAlgorithm4.3.2layonthehyperplane x 1 + x 2 + x 3 + x 4 + x 5 =0.Thisobservation stimulatesthefollowingproposition. 47 PROPOSITION5.2.1 Let A2 R [ m;n ] and B2 R [ m 0 ;n ] .Let k beanintegerwith 1 k m ,and 2 R bearealmode- k B eigenvalueof A .If V := f x 2 C n jA ( k ) x m 1 = B x m 0 1 g isacomplexlinearsubspaceof C n ,thenforany x = ˘ + 2 V suchthat ˘; 2 R n and ˘ 6 =0 ; 6 =0 , When m = m 0 , ˘ and arebothrealmode- k B eigenvectorsof A associatedwith . When m 6 = m 0 , B ˘ m 0 6 =0 and B m 0 6 =0 ,thenormalizedvectors v := ˘ ( B˘ m 0 ) 1 =m 0 ;w := ( B m 0 ) 1 =m 0 arerealmode- k B eigenvectorsof A associatedwith . Proof :Let x 2 V .Then A ( k ) x m 1 = B x m 0 1 : Takingtheconjugateoftheaboveequationyields A ( k ) x m 1 = B x m 0 1 : Since , A and B areallreal, A ( k ) x m 1 = B x m 0 1 : Itfollowsthat x 2 V .Since V isalinearsubspace, ˘ =( x + x ) = 2and =( x x ) = (2 i )are alsoin V .Thus,when m = m 0 , ˘ and arebothrealmode- k B eigenvectorsof A associated 48 with .If m 6 = m 0 ,wehave B v m 0 = n X i 1 ; ;i m 0 =1 B i 1 i 2 i m 0 v i 1 v i 2 v i m 0 = n X i 1 ; ;i m 0 =1 B i 1 i 2 i m 0 ˘ i 1 ( B˘ m 0 ) 1 =m 0 ˘ i 2 ( B˘ m 0 ) 1 =m 0 ˘ i m 0 ( B˘ m 0 ) 1 =m 0 = P n i 1 ; ;i m 0 =1 B i 1 i 2 i m 0 ˘ i 1 ˘ i 2 ˘ i m 0 B˘ m 0 = B˘ m 0 B˘ m 0 =1 : So v isarealmode- k B eigenvectorof A associatedwith .Similarly, Bw m 0 =1and w is alsoarealmode- k B eigenvectorof A associatedwith . 2 Consequently,whenZ-eigenpairsof A2 R [ m;n ] areinconcern,let 2 R bearealE- eigenvalueof A .If U := f x 2 C n jA x m 1 = g containsacomplexlinearsubspace V , thenforany x = ˘ + 2 V withnonzero ˘; 2 R n and x 2 V , ˘= k ˘ k 2 and = k k 2 areboth Z-eigenvectorsof A associatedwith . Anaturalquestioniswhentheeigenspace U inProposition5.2.1willcontaina linearsubspace V .Thefollowingpropositionssuggestsomepossibilities. PROPOSITION5.2.2 Let A beathird-orderand n -dimensionalrank-onetensor,i.e., A canbewrittenastheouterproductof 3 vectors a;b;c 2 C n : A := a b c with A ijk :=( a i b j c k ) ;i;j;k =1 ;:::;n: Let W :=span( a;b;c ) and V := W ? .Then U := f x 2 C n jA x 2 =0 g V ,i.e.,the 49 eigenspaceof A correspondingtotheeigenvalue0containsV. Proof :Ittoshowforany x 2 V , A x 2 =0.Actually,by A x 2 = 0 B B B B B B B B B B B B @ n P j;k =1 a 1 b j c k x j x k n P j;k =1 a 2 b j c k x j x k . . . n P j;k =1 a n b j c k x j x k 1 C C C C C C C C C C C C A : Simplecomputationgives A x 2 = 0 B B B B B B B B B B B B @ a 1 n P j;k =1 b j c k x j x k a 2 n P j;k =1 b j c k x j x k . . . a n n P j;k =1 b j c k x j x k 1 C C C C C C C C C C C C A = 0 B B B B B B B B B B B B B @ a 1 n P j =1 b j x j ! n P k =1 c k x k ! a 2 n P j =1 b j x j ! n P k =1 c k x k ! . . . a n n P j =1 b j x j ! n P k =1 c k x k ! 1 C C C C C C C C C C C C C A = 0 B B B B B B B B B @ a 1 ( b T x )( c T x ) a 2 ( b T x )( c T x ) . . . a n ( b T x )( c T x ) 1 C C C C C C C C C A = 0 B B B B B B B B B @ 0 0 . . . 0 1 C C C C C C C C C A ; wherethelastequalityholdssince x 2 span( a;b;c ) ? . 2 PROPOSITION5.2.3 Let A beathird-orderand n -dimensionaltensorwhichcanbe 50 decomposedas A = r X l =1 a ( l ) a ( l ) a ( l ) ; where r