The Electronic Journal of Linear Algebra.
A publication of the International Linear Algebra Society.
Volume 7, pp. 30-40, March 2000.
ISSN 1081-3810. http://math.technion.ac.il/iic/ela
ELA
DIGRAPHS WITHLARGE EXPONENT S.KIRKLAND
y
, D. D. OLESKY z
, AND P. VAN DEN DRIESSCHE x
Abstrat. Primitivedigraphs on nverties with exponents at least b!n=2+2, where !n = (n 1)
2
+1, are onsidered. For n 3, allsuh digraphs ontaining a Hamilton yle are haraterized;andforn6,allsuhdigraphsontainingayleoflengthn 1areharaterized. Eaheigenvalueofanystohastimatrixhavinga digraphinoneofthesetwolassesisprovedto begeometriallysimple.
Keywords.primitivediretedgraph,exponent. AMSsubjet lassiations.15A48,05C20
1. Introdution. Adiretedgraph(digraph)Disprimitive ifforsomepositive integerm there is a(direted) walkof length m between any twovertiesu and v (inludingu=v). Theminimumsuhm istheexponent of D, denoted byexp(D). It is well known that D is primitive i it is strongly onneted and the gd of its yle lengthsis 1. A nonnegativematrix A is primitive if A
m
is entrywisepositive forsomepositiveintegerm. IfD=D(A),thedigraphofaprimitivematrixA, then exp(D)=exp(A),whihistheminimummsuh thatA
m
isentrywisepositive. Denoting(n 1)
2
+1by! n
,thebestupperbound forexp(D)when aprimitive digraph D has n 2 verties is given by exp(D) !
n
, with equality holding i D = D(W
n
) where W n
is a Wielandt matrix; see, e.g., [2, Theorem 3.5.6℄. When n= 2,then D(W
2
), onsisting of a1yle and a2yle, has exponent equalto 2. Heneforth weassumethatn3. ThedigraphD(W
n
)onsistsofaHamiltonyle (i.e.,ayleoflengthn)andonemorear,betweenapairofvertiesthataredistane twoapartontheHamiltonyle,givingayleoflengthn 1.
The followingresult ofLewin and Vitek[6, Theorem 3.1℄,see also [2, Theorem 3.5.8℄,isthebasisforourdisussionofdigraphswithlargeexponent.
Theorem 1.1. If D has n 3 verties and is primitive with suÆiently large exponent,namely
exp(D)b! n
=2+2;with ! n
=(n 1) 2
+1; (1)
thenD has ylesof exatlytwodierent lengthsj;kwith nk>j.
WesaythataprimitivedigraphDonnvertiessatisfying(1)hasalargeexponent. Note that in Theorem 1.1, gd(j;k)=1sineD is primitive. Ifgd(j;k)=1, then
Reeivedbythe editors on 26January 2000. Aeptedfor publiation on21 February2000. Handlingeditor:RihardA.Brualdi.
y
Department of Mathematis and Statistis, University of Regina, Regina, Saskathewan, S4S0A2,Canada(kirklandmath.uregina.a). ResearhsupportedinpartbyanNSERCResearh Grant.
z
DepartmentofComputerSiene,UniversityofVitoria,Vitoria,BritishColumbia,V8W3P6, Canada(doleskysr.uvi.a).ResearhsupportedinpartbyanNSERCResearhGrant.
x
ELA
DigraphswithLargeExponent 31 everyintegergreaterthanorequalto(j 1)(k 1)anbewrittenas
1 j+
2
k,where
i
arenonnegativeintegers. Thevalue(j 1)(k 1)isthesmallestsuhinteger,and isalledtheFrobenius-Shurindexforthetworelativelyprimeintegersj andk;see, e.g.,[2,Lemma 3.5.5℄.
The Frobenius-Shur index is used to prove the following result that gives a neessaryand suÆientonditionfortheexisteneofaprimitivedigraphwithlarge exponentandylesoftwospeiedlengths.
Theorem 1.2. Let k and j be suhthat gd(j;k)= 1 and n k > j. There exists a primitive digraph D on n verties having only yle lengths k and j and exp(D)b!
n
=2+2ij(k 2)b! n
=2+2 n.
Proof. Suppose thatDisadigraphwithlargeexponentandylelengthskand j <kn. Welaim thatforanypairof vertiesuandv,thereis awalkfrom uto v oflengthatmostk+n j 1nthat goesthroughavertexonakyleanda vertexona j yle. Toprovethis laim,note that from the proof ofTheorem 1in [4℄,there arenopairsofvertexdisjointylesin D;that is, anypairof ylesshare atleastoneommonvertex. Ifthereisawalkfromutovoflengthlessthanorequal to nthat passesthroughat leastonevertexonak yleand atleast onevertexon aj yle, thenthelaimisproved.
Sosupposethatthisisnotthease. Inpartiular,assumethatuandv areonly onk (resp. j) yles,andanypathfrom uto v passesonly throughvertiesnoton any j (resp. k) yle. Consider therst ase. Let lbe thenumber of verties not on aj yle, and note that 2l n j. Sine ashortest pathfrom uto v goes onlythroughvertiesnotonaj yle,thelengthpofsuhapathsatisespl 1. Considerthewalkfrom utov formed byrsttraversingak yle atu(neessarily going throughavertex onaj yle),then taking thepath oflength pfrom uto v. This generatesawalkfrom uto v that goesthroughavertex onak yle andone on aj yle, and itslength isk+pk+l 1k+n j 1. Theseond ase follows by interhangingk and j and noting that j+n k 1 < k+n j 1. Thus thelaim is proved. By theFrobenius-Shur index, there is awalkfrom uto v of length k+n j 1+(k 1)(j 1) = n+j(k 2) for any pairu;v. Thus n+j(k 2)exp(D)b!
n
=2+2,givingtheonditiononkandj.
For the onverse, assume the ondition on k and j, and onsider the digraph D onsistingof thek yle 1!k !k 1! ! k+j n+1! k+j n ! k+j n 1!!2!1,andars1!k+1!k+2!!n 1!n!k+j n. ThusD hasexatlyonekyleandonej yle. Considerthelengthofawalkfrom ktok+j n+1. Suhawalkhaslengthn j 1ork+n j 1+
1 k+
2 j for somenonnegativeintegers
i
,and (fromtheFrobenius-Shurindex)thereisnowalk oflengthk+n j 1+(k 1)(j 1) 1. Thus
exp(D)k+n j 1+(k 1)(j 1)=n+j(k 2)b! n
=2+2:
ELA
32 S.Kirkland,D.D.OleskyandP.vandenDriesshe
WeassumethatDhasalargeexponentandfousonthegraphtheoretiaspets of this ondition. In Setion 2, we haraterize the ase when D has a Hamilton yle (k = n 3); and in Setion 3, we haraterize the ase k = n 1. Our haraterizations give some information on the ase for general k n when n 4, sine a result of Beasleyand Kirkland [1, Theorem 1℄ implies that any indued subdigraphonkvertiesthatisprimitivealsohaslargeexponent(relativetob!
k =2+ 2), so thestrutureof somesuhinduedsubdigraphsis knownfrom ourresults. It is known from results in [6℄ exatlywhih numbers b!
n
=2+2are attainableas exponents of primitive digraphs. (Note that there are somegaps in this exponent set.) Ourwork inSetions2and3fousesondesribingtheorrespondingdigraphs whenkn 1.
Somealgebraionsequenesofthelargeexponentondition(1)forastohasti matrixA with D(A) = D have been investigated in [4℄ and [5℄. The harateristi polynomial of A has a simple form (see [4, Theorem 1℄), and, if n is suÆiently large,thenabouthalf oftheeigenvaluesofAhavemodulusloseto1. Kirklandand Neumann[5℄onsideredthemagnitudesoftheentriesinthegroupgeneralizedinverse ofI A(whihmeasuresstabilityoftheleftPerronvetorofAunderperturbations). In Setion 4 we use results of Setions 2 and 3 to investigate the multipliities of eigenvaluesofstohastimatrieswithlargeexponents.
2. TheHamiltonianCase. AssumingthatDhaslargeexponentanda Hamil-tonyle,webeginbyndingpossiblelengthsforotherylesinD.
Lemma 2.1. Suppose that D is a primitive digraph on n 3 verties with exp(D)b!
n
=2+2 and that D has a Hamilton yle. Then D has preisely one Hamilton yle,andallother yleshavelengthj,wheren>jd(n 1)=2e.
Proof. By Theorem 1.1, D ontainsyles of exatlytwo lengths, n = k > j. W.l.o.g.takethegivenHamiltonyleas1!n!n 1!!2!1,andassume that the ar 1 ! j lies on a seond Hamilton yle. Note that the only possible ars from any vertex i are i ! i 1 (mod n) and i ! i+j 1 (mod n). Sine thear j+1!j is noton theseond Hamilton yle, this yle must inlude the ar j+1! (j+1)+j 1=2j (mod n). Similarly, there is anar onthe seond Hamiltonylefrom(m 1)j+1tomj(modn),form=1;:::;n. Asgd(j;n)=1,D ontainsthedigraphofaprimitiveirulant. By[3,Theorem2.1℄,exp(D)(n 1) orexp(D)bn=2,thusexp(D)<b!
n
=2+2. Hene,there isnoseond Hamilton yleinD. Forthelowerboundonj,takek=ninTheorem1.2;seealso[4,Theorem 1℄.
IfD haslargeexponentandk=n=3,then Lemma2.1impliesthat j2f1;2g: Forj=1,Donsistingofa3yleanda1ylehasexponentequalto4=b!
3 =2+2. Forj=2=n 1,eitherD=D(W
3
)withexponentequalto5=! 3
,orDonsistsof a3ylewithtwo2ylesandhasexponentequalto4. Thislastaseisanexample oftheresultthatadigraphDonnvertieshasexp(D)=(n 1)
2
iD isisomorphi to an n ylewith two additionalars from onseutive verties forming twon 1 yles;see,e.g.,[2, pp.82{83℄.
Hamilto-ELA
DigraphswithLargeExponent 33 isodd,thentheasej=(n 1)=2isslightlydierentandisgiveninTheorem2.3.
Theorem 2.2. Supposethat j n=2. Then D isa primitive digraph on n3 verties withexp(D)b!
n
=2+2and ylelengths nand j iD isisomorphi to a(primitive)subdigraph ofthe digraphformedby taking the yle1!n!n 1! !2!1, andadding inthe ars i!i+j 1for 1in j+1.
Proof. AssumethatDisprimitivewithlargeexponentandhasaHamiltonyle. Then by Lemma 2.1, D hasonly oneHamilton yle and other yles of length j, whih by assumption is at least n=2. W.l.o.g. assume that the Hamilton yle is 1!n ! n 1! !2! 1;and that D ontainsthe ar 1!j. SineD has ylesofjust twodierentlengths,eahvertexiof D hasoutdegree2,andifthe outdegreeis 2,then theoutarsfrom vertexi arei!i 1andi!i+j 1. Here and throughouttheproof, allindies aremod n. As 1!j,theoutdegree ofvertex i is 1for eah i 2 fn j+2;:::;jg,sine otherwise1! j ! j 1! !i ! i+j 1 n!i+j 2 n!!2!1isayleoflengthlessthanj. Consequently iftheoutdegreeofvertexi2f2;:::;jgis2,theninfati2f2;:::;n j+1g. Ifthere isnosuhi,thenDhasthedesiredstruture,sineDhasatmostn j+1onseutive vertiesontheHamiltonyle(namely1andj+1;:::;n)ofoutdegree2. Heneforth suppose that there exists i2 f2;:::;n j+1gwith outdegree 2,and leti
1 be the maximumsuhi;thusi
1 n j+1onseutiveverties,asdesired. Sosupposeheneforththatn j+i
1
>j,that isi
1
>2j n0. Supposealsothat thereexistsi 2
havingoutdegree2. Theni 2 oneHamiltonyle(Lemma2.1),thisimpliesthat3j n=j,givingaontradition, sinegd(n;j)=1. Thusagaineahofvertiesi
1
+1;:::;j+i 1
1hasoutdegree1, andsoat mostn j+1onseutivevertieshaveoutdegree2,asdesired.
Fortheonverse,onsiderthemaximalsuhdigraphDwiththeaboveHamilton yleandthen j+1additionalars. Notethateahofthevertiesn j+2;:::;n D ismaximal,anyprimitivesubdigraphhasexponentatleastaslargeasexp(D).
Theorem 2.3. Suppose that n 3 is odd and j = (n 1)=2. Then D is a primitive digraph on n verties with exp(D) b!
n
=2+2 and yle lengths n and j i D isisomorphi toa(primitive)subdigraphof the digraph formedby takingthe yle 1 ! n ! n 1 ! ! 2 ! 1, and adding in the ars i ! i+j 1 for 1i(n 1)=2=j.
Proof. First assume that exp(D) b! n
=2+2 = (n 1) 2
ELA
34 S.Kirkland,D.D.OleskyandP.vandenDriesshe (n 1)+(n 1)(j 1) =j(n 1) =(n 1)
2
=2 from i to eah vertex of D. It followsthat there mustbe avertex with distane2 tothe nearestj yle. W.l.o.g. thatvertexisn,withvertexn 2onaj yle. Infatthatj yleisn 2!n 3! !(n 1)=2=j !n 2,otherwisen 1ornisonaj yle. Noneoftheverties j+1;j+2;;nan haveoutdegree2(otherwiseoneofn 1or nisonaj yle). However,thej 1additionalarsi!i+j 1fori=1;2;:::;j 1maybeinluded inD. ThusitfollowsthatD isasubdigraphofthedigraphthat hasthen 1yle andtheadditionalj arsasin thetheoremstatement.
For the onverse, note that if D is isomorphi to a subdigraph of the speied digraph, then a walk from n to n 1 of length greater than 1 must traverse the entireHamiltonyle, sowalksfrom nto n 1havelength1or n+1+
1 n+
2 j where
1 and
2
arenonnegativeintegers. Thus(byFrobenius-Shur)thereisnowalk from n to n 1 of length n+1+(n 1)(n 3)=2 1 = (n 1)
2
=2+1, so that exp(D)(n 1)
2
=2+2,asdesired.
Using the strutures of Hamiltonian digraphs D with largeexponents given in Theorems2.2and2.3,wedeterminetheexatvalueofexp(D)intermsofaparameter athatdependsonwhih j ylesourinD.
Corollary 2.4. Supposethat D isa primitive digraph on n3verties with exp(D)b!
n
=2+2, a Hamilton yleand all other yles of length j,where n> jd(n 1)=2e. Supposethat the Hamilton yleis1!n!n 1!!2!1. Let1an j+1if jn=2,and1aj ifj=(n 1)=2. SupposethatD also ontainsthe ar(s) 1!j anda!a+j 1,andthat ifi isavertexof outdegree 2, then1ia. Then exp(D)=n a+1+(n 2)j.
Proof. Theshortestwalkfromntoa+jthatpassesthroughavertexonajyle haslengthn a j+n,soitfollows(byFrobeniusShur)thatthereisnowalkfrom ntoa+joflengthn a j+n+(n 1)(j 1) 1.Thusexp(D)n a+1+(n 2)j. Further,sinethereisawalkbetweenanytwovertiesoflengthatmostn a j+n thatgoesthroughavertexonaj yle,itfollowsthatexp(D)n a+1+(n 2)j, andthusexp(D)=n a+1+(n 2)j.
Ifjn=2,notethatexp(D)=n a+1+(n 2)jj(n 1)for1an j+1, givingtheresultof[6,Corollary3.1℄whenk=nwithouttheadditionalassumption. Alsonote that if j =n 1and a=1,then exp(D)ahievesits maximumvalue of !
n
, and D=D(W n
),as desribed inSetion 1. Itis interestingto notethat in the aboveorollary,itisonlythevalueofathat inuenesthevalueoftheexponent;if 2i a 1,the presene or abseneof the ari !i+j 1does notaet the exponent. Forxed n andj, thisresult givesarange of valuesof exp(D) in whih therearenogaps;see[6℄.
ELA
DigraphswithLargeExponent 35 indies are taken mod (n 1). Vertex n repliates vertex v 2 f1;:::;n 1g in a digraph D on n verties iffor all a;b 2 f1;:::;n 1g;a! n i a! v and n !b iv !b. Thusin theadjaenymatrixA withD=D(A), therows(andolumns) orrespondingtovertiesnand varethesame.
Lemma 3.1. LetD beastronglyonneteddigraphonn5verties, withyle lengthsn 1and j,where n 1>j3. Supposethat 1!n 1!!2!1is ann 1yle,andthat!n. Thennhasoutdegreeatmost2,witheithern! 2 orn!+j 2orboth. Furthermore,if theoutdegreeofnis2,thentheindegree of nis1.
Proof. Firstsupposethat thereisanarn!a. Thenthere isaylen!a! a 1!!!noflengtha +2ifa>,orlengthn+1+a if>a. In theformer ase,a +2=j orn 1, from whih itfollowsthat a=+j 2or 2;inthelatterasesimilarlya=+j 2or 2. Thisestablishesthepossible outarsfrom n. Finally, assumethat n! 2and n !+j 2. Suppose that d!nforsomed6=. Asabovethetwooutarsfromnanbewritten asd 2and d+j 2. As d6=,it followsthat d 2=+j 2and 2=d+j 2. Hene d =j and d=j,givingaontradition. Thus theindegreeofnis1.
Corollary 3.2. LetD beasinLemma3.1. If n!,theneither +2!nor +2 j!norboth. Furthermore, ifthe indegreeof nis2,thenthe outdegreeof n is1.
Proof. FormD 0
byreversingtheorientationofeahar inD. ThenLemma 3.1 appliestoD
0
,andtheresultfollows.
Theorem 3.3. Supposethat n6andn 1>j>n=2. Then D isaprimitive digraph onnverties withexp(D)b!
n
=2+2andylelengthsn 1andj i(up to relabeling of verties andreversal of eah ar) D is a (primitive)subdigraph of a digraph formedby taking an n 1 yle1!n 1!n 2!!2!1, adding inthe ars a!a+j 1for 1an j,andone ofthe following:
(a)ars sothatn repliatesi 0
for axedi 0
2f1;:::;n 1g, (b)ars1!n;n!n 2andn!j 1.
Proof. First suppose that D is primitive with exp(D) b! n
=2+2and yle lengths n 1 and j. By relabeling the verties and/or reversing eah ar in D if neessary, we mayassume that the n 1 yle is asabove, and that vertex n has indegree1(Lemma3.1andCorollary3.2). Ifthesubdigraphinduedbyf1;:::;n 1g is not primitive, then this subdigraph is just the n 1 yle, and without loss of generality1!n,sobyLemma3.1theoutarsofnareasubsetofthosegivenin(b). Sosupposethatthesubdigraphinduedbyf1;:::;n 1gisprimitive. Itfollowsfrom aresult ofBeasley andKirkland [1, Theorem 1℄,that theexponentof this indued subdigraphisatleastb!
n
=2,whihinturnisatleastb! n 1
=2+2. Henewithout lossof generality, take the subdigraph to ontain thear 1 !j, and (by Theorem 2.2 ) to have the property that if a ! a+j 1, then 1 a n j. Let a
0 be themaximum suh a. Supposethat i!n and notefrom Lemma 3.1 that theonly possibleoutarsfromnaren!i 2andn!i+j 2. Considerthetwoases: (i) n6!i+j 2,(ii)n!i+j 2.
ELA
36 S.Kirkland,D.D.OleskyandP.vandenDriesshe above),D isasubdigraphofoneonstrutedasin(a)(withi
0
=i 1). Case(ii): n!i+j 2: If1i 1n jorn 1i 1a
0
+j 1,thenDisa subdigraphofoneoftheonesonstrutedin(a)(ifi6=1,withi
0
2:NotethatDontains thelosedwalka for some nonnegativeintegers
1
=1). Thelasttwooftheseimplythat j=n 1(aontradition). Therstofthese threeanonlyourif3j=2(n 1), and sinej andn 1arerelativelyprime, thisis also impossible. Consequently, it mustbetheasethat1i 1n j orn 1i 1a
0
+j 1,sothat Disa subgraphofoneoftheonesonstrutedin (a)or(b).
For theonverse,onsider amaximal digraph H onstrutedasin (a). Sine n repliates i
is formed from H by deleting n and its inidentars. NowH
0
isHamiltonianonn 1vertiesandhasthedigraphstruture ofTheorem 2.2,thusexp(H
0 )b!
n 1
=2+2:ApplyingCorollary2.4toH 0 (bytheusualFrobenius-Shur argument),sothat theexponentis atleast(n 2)j, givingtherequiredresultasin(a).
Note that the result of Theorem 3.3 does not hold for small values of n. For example,ifn=5adigraphasin(a)ofTheorem3.3withexponentequalto9<10= b!
5
=2+2an be onstruted bytaking a Hamiltonian digraphon 4 verties with twoadditionalarsfrom onseutivevertiesformingtwo3 yles(see,e.g.,[2,pp. 82-83℄)andvertex5repliatingvertex1.
Theorem 3.4. Supposethat n6iseven andj =n=2. Then D isaprimitive digraph onnverties withexp(D)b!
n
=2+2andylelengthsn 1andj i(up to relabeling of verties andreversal of eah ar) D is a (primitive)subdigraph of a digraph formedby taking an n 1 yle1!n 1!n 2!!2!1, adding inthe arsi!i+j 1for 1in=2 3, andoneofthe onstrutions(a)or (b) inTheorem3.3.
Proof. First suppose that D is primitive with exp(D) b! n
=2+2 and yle lengthsn 1and j. Asin theproofof Theorem3.3, assumethatthen 1yle is asabove,thatthesubdigraphinduedbyf1;:::;n 1gisprimitive,with1!j,and with theproperty that ifa! a+j 1,then 1a n j. Finally, alsosuppose that i!n. ByLemma 3.1andCorollary3.2there aretwoasestoonsider: (i)D ontainsexatlyoneofthearsn!i+j 2andi j!n,(ii)Dontainsneither thearn!i+j 2northeari j!n.
ELA
DigraphswithLargeExponent 37 n!n i+j. Withn i+2replaedbyi,thisdigraphontainsthearn!i+j 2. So withoutlossof generality, weassumethat thear n !i+j 2is in D. Sine vertexnisonaj-yleandsineD hasdiameter atmostn 1,itfollowsthatthere isawalkfromntoanyvertexoflengthn 1+(n 2)(n=2 1)=(n
2
2n+2)=2,and similarlythatfromanyvertexinDthereisawalktonoflength(n
2
2n+2)=2. Sine exp(D)b!
n
=2+2=(n 2
2n+6)=2,itmustbetheasethattherearevertiesu andv2f1;:::;n 1gsuhthatthereisnowalkfromutovoflength(n
2
2n+4)=2. Observethat for anyvertexw 2f1;:::;n 1gthat is onaj-yle, there is awalk fromwtoeveryvertexinf1;:::;n 1goflengthn 2+(n 2)(n=2 1)=(n
2
2n)=2. Asaresult,theshortestwalkfromutoavertexinf1;:::;n 1gthat isonaj-yle must havelength at least 3. It followsfrom this that in fat vertex n 1 mustbe at least3 stepsfrom thenearestj-yle, sothat in partiular, noneof n 1,n 2 and n 3an be onaj-yle. Thus in D, n j 6!n 1,n j 16!n 2and n j 26!n 3,andsoifa!a+j 1,thenan j 3=n=2 3. Further,it mustbetheasethat1in j 2,otherwiseoneofvertiesn 1,n 2andn 3 isonaj-yle(involvingvertiesiandn). Consequently,Danberelabeledtoyield asubdigraphofoneofthoseonstrutedin (a)withi
0
=i 1(if 2in j 2), or(b)(if i=1).
Case (ii): If D ontains neither the ar n ! i+j 2nor the ar i j ! n, then nhas both indegreeand outdegree 1,with i!n!i 2. Nowif D ontains either of the ars i 1 ! i+j 2 or i j ! i 1, then the labels of verties i 1and n anbeexhanged and ase (i)above applies. On the other hand if D ontains neither of those two ars, then i 1 has indegree and outdegree 1, with i!i 1!i 2,sothat vertexnrepliatesvertexi 1. Thus exp(D)=exp(D
0 ) whereD
0
isformed fromD bydeletingvertexnandthearsinidentwith it. From Corollary2.4 with n replaed byn 1, exp(D
0
)=n 1 a+1+(n 3)j where a = maxfb isavertexin D
0
: thear b ! b+j 1 isin D 0
g. Thus exp(D 0
) = exp(D)= n a+(n 3)n=2 (n
2
2n+6)=2,whih implies that a n=2 3. ConsequentlyDisasubdigraphofoneofthoseonstrutedin(a)withi
0
=i 1. Fortheonverse,onsider adigraph H onstrutedasin (a). Sinenrepliates i
0
, exp(H) = exp(H 0
), where H 0
is formed from H by deleting n and its inident ars. AppealingtoCorollary2.4withnreplaedbyn 1,a=n=2 3;andj=n=2, exp(H
0 )=(n
2
2n+6)=2=b! n
=2+2ifnis even. Finally, onsider thedigraph H onstrutedin(b). Evidentlythewalksfrom vertexn 1to n 3anonlyhave lengthsequalto2,orto2+n 1+
1
(n 1)+ 2
j fornonnegativeintegers 1
and
2
. Itfollowsthat thereisnowalkfrom n 1to n 3oflength(n 2
2n+4)=2,so thatexp(H)(n
2
2n+6)=2.
ELA
38 S.Kirkland,D.D.OleskyandP.vandenDriesshe
Theorem4.1. LetAbeaprimitive,rowstohastin-by-nmatrixwithn3and exp(A)b!
n
=2+2. Letk andj bethe twoylelengths inD(A)with nk>j. ThenAhasamultiplenonzeroeigenvaluei= r,whereristheuniquepositive rootof kx
j +jx
k
=k j. Whenthis isthe ase, kisodd andj iseven. Proof. ByTheorem1in[4℄,theharateristiequationofAisz
n
=0,forsome2(0;1). Thusanonzeroeigenvaluesatises z
Notethat1isalwaysaneigenvalue,and(byDesartes'ruleofsigns)thereisnoother positiveeigenvalue. Let=e
i
be aneigenvaluewith >0and 0< <2:By dierentiating, if is a multiple eigenvalue, then it also satises
j
=(k j)=k, giving
j
= (k j)=k and = 2l=j for some positive integer l < j. Further dierentiation showsthat thealgebraimultipliity of is 2. Bytaking imaginary partsoftheharateristiequation,
k
=(k j). Substitutinginto(2)givesk j
+j k
=k j. Theonverseisstraightforward.
From theharateristiequation, amatrixsatisfyingtheonditionsof Theorem 4.1haszeroasaneigenvalueik<n,anditsalgebraimultipliityisn k.
Thedigraphharaterizationsin Setions 2and 3leadto resultsaboutthe geo-metrimultipliitiesof eigenvaluesofprimitive,stohastimatrieswithlarge expo-nent.
Theorem 4.2. Let A be aprimitive, row stohasti n-by-n matrix with n 3 and exp(A) b!
n
=2+2. If D(A) is Hamiltonian, then eah eigenvalue of A is geometriallysimple.
Proof. Letthelengthoftheshorteryle(s)inD(A)bej d(n 1)=2ebyLemma 2.1. Forjn=2takep=n j+1,andforj=(n 1)=2takep=(n 1)=2. Thenby Theorems2.2and 2.3,withoutlossofgeneralitybypermutation similarityA=[a
ij ℄ hasthe following form: a
1;n unredued Hessenberg matrix. Bydeleting row1and olumnn, itanbeseenthat rankA(n 1)[7, Exerise22,p. 274℄. Similarly, rank(A I)=n 1foreah eigenvalueofA. This impliesthateaheigenvaluehasgeometri multipliityone. Asanexampleoftheaboveeigenvalueresults,onsiderthe3-by-3rowstohasti matrixAhavingk=3andj=2asintheproofofTheorem4.2with
1 =
2 =1=2. Note that exp(A)= 4. The harateristiequation of A is z
3
z (1 ) =0, with =3=4; thus A has eigenvalues1; 1=2; 1=2. Here 1=2isan eigenvalueof algebraimultipliity2(as preditedby Theorem4.1), but geometri multipliity1 (aspreditedbyTheorem4.2).
ELA
DigraphswithLargeExponent 39 eigenvalueof Aisgeometrially simple.
Proof. Sine k=n 1,=0isasimpleeigenvalueof A. Letthelengthofthe shorteryle(s)inD(A)bejdn=2ebyTheorem1.2. Forsimpliity,onlytheproof fortheasej >n=2is given,theasej =n=2is essentiallythe same. Forj>n=2, byTheorem3.3,withoutlossofgeneralitybypermutationsimilarityA=[a
ij ℄,orits transpose,musthaveoneoftwoformsorrespondingto(a)or(b).
Inase(a), without lossofgeneralitynanbetaken to repliateavertex with outdegree 1. (Thisis beause,byLemma 3.1, n haseither indegree oroutdegree 1, so,if neessary,takeA
T
.) Let vertex nrepliatevertexi where n 1i>n j. Consider the matrix A I, where 6= 0 and the digraph of A is as in Theorem 3.3(a). FormB from A I bydeleting therstrowand thelastolumn. Then B isblokuppertriangularwitha(1;1)blokoforderi 2anda(2;2)blokoforder n i+1. Sine the (1;1) blok is uppertriangular with positive diagonal entries, it is nonsingular. The (2;2) blok has the rst n i diagonal entries positive, in eah superdiagonalentry, anda1in thelast rowrstolumn. Everyother entry in the(2;2)blokiszero. Byexpandingabouttherstrow,thedeterminantofthe (2;2)blokhasmagnitude
n i
. Asaresult,B isnonsingular,sothat A I hasa submatrixofrankn 1.
In ase(b), a that A is primitive. Deleting rown and olumn n 1, theremaining submatrix of A I is upperHessenberg,andhasrankn 1forallvaluesof, beauseithasa uniquenonzerotransversaloflengthn 1(fromthesubdiagonaland(1;n)entriesof A I).
Thus rank (A I) = n 1 for every eigenvalue of A, and the geometri multipliityofeaheigenvalueisone.
We lose the paper with a lass of examples to show that for k n 2, a rowstohastimatrixwithlargeexponentanhaveaneigenvalueoflargegeometri multipliity. suh that0<<1,and formtheprimitiverowstohasti n-by-nmatrixA with nonzeroentriesasfollows:a
1;k 1 digraphofA anbeformedbystartingfrom D(W
k
)andtaking eahof theverties k+1;:::;nrepliating vertex3. Sinevertex3is repliatedn ktimes, there isa walk involvingany of the vertiesk+1;:::;n in D(A) i there is a orresponding walkinvolvingvertex3in D(W
ELA
40 S.Kirkland,D.D.OleskyandP.vandenDriesshe REFERENCES
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