On a conjecture of Ilmonen, Haukkanen and Merikoski concerning the smallest eigenvalues of certain GCD related matrices

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Contents lists available atScienceDirect

Linear

Algebra

and

its

Applications

www.elsevier.com/locate/laa

On

a

conjecture

of

Ilmonen,

Haukkanen

and

Merikoski

concerning

the

smallest

eigenvalues

of

certain

GCD

matrices

Ercan Altınışıka,∗_, _{Ali Keskin}a_, _{Mehmet Yıldız}a_,

Murat Demirbükenb

a_Department_of_Mathematics,_Faculty_of _Sciences,_Gazi_University,₀₆₅₀₀

Teknikokullar–Ankara,Turkey

b

DepartmentofComputerScience,FacultyofEngineering,İ.DoğramacıBilkent University,06800Bilkent–Ankara,Turkey

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received14July2015 Accepted23November2015 Availableonline15December2015 SubmittedbyR.Brualdi MSC: 15A18 15A23 15B36 15B48 11B39 11C20 Keywords: GCDmatrix (0,1)-matrix Positivematrix Eigenvalue Spectralradius Fibonaccinumber

LetKnbethesetofalln× n lowertriangular(0,1)-matrices

witheachdiagonalelementequalto1,Ln={Y YT: Y ∈ Kn}

andletcnbetheminimumofthesmallesteigenvalueofY YT

as Y goes throughKn.TheIlmonen–Haukkanen–Merikoski

conjecture(theIHMconjecture)statesthatcnisequaltothe

smallesteigenvalueofY0Y0T,whereY0 ∈ Kn with(Y0)ij = 1−(−1)i+j

2 fori> j.Inthispaper,wepresentaproofofthis

conjecture.Inourproofweuseaninequalityforspectralradii ofnonnegativematrices.

* Correspondingauthor.Tel.:+903122021070.

E-mailaddresses:ealtinisik@gazi.edu.tr(E. Altınışık),akeskin1729@gmail.com(A. Keskin),

yildizm78@mynet.com(M. Yıldız),murdem91@gmail.com(M. Demirbüken).

http://dx.doi.org/10.1016/j.laa.2015.11.023

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1. Introduction

Let S = {x1,x2,. . . ,xn} be a set of distinct positive integers, (xi,xj) denote the

greatest common divisor of xi and xj and let ε be apositive real number. The n× n

matrices (S) = ((xi,xj)) and (Sε) = ((xi,xj)ε) are called the GCD matrix and the

powerGCDmatrixonS,respectively.TheLCMmatrixandthepowerLCMmatrixare similarly deﬁned. In 1876, Smith [30] proved thatif S is factor closed, then det(S) = n

k=1ϕ(xk),where ϕ isEuler’stotient.Sincethen manyresultsonthese matriceshave

been publishedintheliterature,seee.g.[5,6,10,11,13,21].

AninterestingandactiveareainthestudyofGCDtypematricesistheir eigenstruc-ture. The ﬁrst results on this subject were published in the papers [32,7,23] but the paper ofHongand Loewy[16]canbe consideredasthe ﬁrstpaperon thestudyof the eigenvalues of GCD and related matrices due to the number theoretical aspect of the subject. Since their pioneeringpaper manyresults onthe subjecthave been published in the literature, see e.g. [1,3,4,14,15,17,19,24,26–28]. In that paper, Hong and Loewy investigatedtheasymptoticbehavioroftheeigenvaluesofpowerGCDmatricesbyusing sometoolsofnumbertheory.Besidetheirresultsonasymptoticbehaviorofthese matri-ces, inthesamepaperHongandLoewyintroducedaconstantcn andusedittopresent

alowerboundforthesmallesteigenvaluesofpowerGCDmatrices.LetKn bethesetof

all n× n lowertriangular (0,1)-matriceswith eachdiagonalelement equalto 1 and let

Ln={Y YT : Y ∈ Kn}.Theydeﬁnedthenumberscn dependingonlyonn asfollows:

cn= min Z∈Ln

μ(1)_n (Z) : μ(1)_n (Z) is the smallest eigenvalue of Z. (1.1) Then theyproved that

λ(1)_n ((Sε))≥ cn· min

1≤i≤n{Jε(xi)},

whereJεisJordan’sgeneralizationofEuler’stotientandλ(1)n ((Sε)) isthesmallest

eigen-valueofthepowerGCDmatrix(Sε_),_see_[16, _{Theorem 4.2]}_.

In2008,Ilmonen,HaukkanenandMerikoski[19]studiedeigenvaluesofmeetandjoin matriceswhichareabstractgeneralizationsofGCDandLCMmatrices,respectively,and they generalizedthe aboveresultconcerning the numberscn for positivedeﬁnite meet

matrices deﬁnedonlocally ﬁnite meetsemilattices.They alsoobtainedasimilar result for joinmatrices andhence,inparticular,forLCM matrices.Inthesamepaper,inthe light of their MATLAB calculations for n = 2,3,. . . ,7, they presented an interesting conjecture abouttheconstants cn.

Conjecture 1.1 (The IHM conjecture). (See [19,Conjecture 7.1].)Let Y0 = (y0ij)∈ Kn

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y_ij0 = 1− (−1)

i+j

2 (1.2)

ifi> j.Thencn isequaltothesmallesteigenvalue ofY0Y0T.

Wehave furthernumericalevidence of the IHM conjecture as follows. Recently,the ﬁrst, the second and the fourth author of the paper have investigated that the IHM conjectureholdsforn= 7 and n= 8 withthehelpofaMATLABcode.OurMATLAB code running on a computer1 _has _veriﬁed_the _truth _of _the _IHM _conjecture _for _n _{= 7}

in 23 minutes and for n = 8 in 3.5 days. Since |L8| = 228 = 268,435,456 and |L9| =

236_{= 68,719,476,736,}_it_would_take_about_{3 years}_to_verify_the_IHM_conjecture_for_n_{= 9}

withthehelpofourMATLABcode.Toovercomethisdiﬃcultyabouttime,wewritea diﬀerentcodeinCprogramminglanguage.WeuseNewton’sidentities(see[20])toobtain thecharacteristicpolynomialofamatrixZ inLnandwecalculatethesmallesteigenvalue

of Z by using Newton’s method (see [31]) to shorten therunning time. Indeed, our C coderunningonthesamecomputerhas veriﬁedthe truthof theIHM conjecturein30 minutesforn= 8 andin7daysforn= 9.Thus,wehaveconcludedthatConjecture 1.1 holdsfor n= 8 and n= 9. Thisinvestigationhasbeen presented bytheﬁrstauthor of thispaperin[2].

Afterobtainingenoughnumericalevidence thatthe IHMconjecture canbe true,we getthemotivation to ﬁndoutaproofof it.The strategyoftheproof isas follows. We provethatfor anymatrixY inKn, |Y−1|≤ |Y0−1|, wherethe matrixY0 given by(1.2)

and|Y−1| istheelement-wiseabsolutevalueofthematrixY−1.Secondly,weshowthat

|Z−1_| _{≤ |(Y}

0Y0T)−1| for all Z ∈ Ln. Then, by using an inequalityfor spectral radii of

nonnegativematrices, weobtainaproofoftheIHM conjecture.Weconcludethepaper with a conjecture on the uniqueness of the matrix Y0 and a discussion about further

studiesontheconstantcn.

2. PropertiesofmatricesinKn

Firstwepresentasimplefactaboutaparticularnilpotent(0,1)-matrixwhichweuse inthe courseof our proofs. Here we give theproof of this fact thoughone canﬁnd in theliterature.

Lemma 2.1. Let Y ∈ Kn and N := Y − I, where I is the n× n identity matrix. We

denoteby(Nk)ij theij-entryof thepositiveintegerk-th powerofN . Thenwehavethe

followingproperties.

i) (Nk₎

ij = 0 wheneveri− j < k,

ii) Y−1 = I− N + N2− · · · + (−1)n−1Nn−1.

1

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Proof. Since N is a strictly lower triangular (0,1)-matrix, the proof of the ﬁrst claim follows fromthematrixmultiplication.Fortheproofofthesecond claim,consider

In− (−N)n = (I + N )(I− N + N2− · · · + (−1)n−1Nn−1). Since Nn _{= 0 and}_{Y = I + N ,}_we_have

Y−1= I− N + N2− · · · + (−1)n−1Nn−1. 2

Now we investigate the inverse of any matrixY in Kn. In the following lemma we

obtainarecurrencerelationfortheentriesoftheinverseof Y .

Lemma2.2.LetY ∈ Kn andN := Y − I.Denote N = (nij) andY−1 = (aij).Thenwe

have thefollowingrecurrencerelationforakl:

akl= ⎧ ⎪ ⎨ ⎪ ⎩ 0 if k < l, 1 if k = l, − k−1i=l nkiail if k > l.

Proof. When we multiply both sides ofthe equality inLemma 2.1(ii) from the left by

−N,wehave

−NY−1 ₌_{−N + N}2_{− · · · + (−1)}n−1_Nn−1_{+ (}₋₁₎n_Nn_.

Since Nn _{= 0,}_one_can_easily _obtain

I− NY−1= Y−1. (2.1)

ByLemma 2.1,itisclearthatakl= 0 ifk < l andakl = 1 ifk = l.Now,from(2.1),we

have akl=− n i=1 nkiail

forallk > l and hence,byLemma 2.1,weobtain

akl=− k−1 i=l nkiail forallk > l. 2

Inthefollowing theorem,weﬁndanupper boundforeachaij interms ofFibonacci

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Theorem 2.1. LetY ∈ Kn and denoteY−1 = (aij). Then, for 1≤ j < i ≤ n, we have

|aij| ≤ fi−j,wherefi−j isthe(i− j)-th Fibonaccinumber.

Proof. LetN = (nij) beasinLemma 2.2.Letj = 1,2,. . . ,n−1.Weprovebyinduction

ont= 1,2,. . . ,n−j that|aj+t,j| ≤ ftforallY ∈ Kn.ByLemma 2.2,wehave|aj+1,j| =

nj+1,j,where nj+1,j canbe0 or1.Thus,|aj+1,j| ≤ 1= f1forallY ∈ Kn.Now,assume

that for each t = 1,2,. . . ,k− 1 we have |aj+t,j| ≤ ft for all Y ∈ Kn. We provethat

|aj+k,j| ≤ fk for allY ∈ Kn. ByLemma 2.2, we haveaj+k,j =− j+ki=j−1nj+k,iaij for

allY ∈ Kn.

Case 1. Assume nj+k,j+k−1 = 0. Then |aj+k,j| = j+k_i=j−2nj+k,iaij. Also, by

Lemma 2.2, aj+k−1,j = − j+k−2i=j nj+k−1,iaij. Since both of nj+k,i and nj+k−1,i for

each i = j,j + 1,. . . ,j + k− 2 can arbitrarily be 0 or 1, it is clear that aj+k,j and

aj+k−1,j gothrough thesamevalues,as Y goes throughthe setKn. Therefore, bythe

inductionhypothesis,weobtain|aj+k,j| ≤ fk−1≤ fk forallY ∈ Kn.

Case 2.Assumen_j+k,j+k−1 = 1.

Subcase i.Assumen_{j+k−1,j+k−2}= 0.ByLemma 2.2,wehave

|aj+k,j| ≤ j+k−2 i=j nj+k,iaij +|aj+k−1,j| .

Also, by Lemma 2.2, it is clear that a_j+k−1,j = − j+k−2_i=j n_j+k−1,iaij. Since both of

nj+k,iandnj+k−1,i foreachi= j,j + 1,. . . ,j + k− 2 canarbitrarilybe0 or1,itisclear

that− j+k−2_i=j nj+k,iaij and aj+k−1,j go through the samevalues, as Y goes through

the set Kn. Thus, by the inductionhypothesis, we obtain j+ki=j−2nj+k,iaij ≤ fk−1.

Beside this, by ourassumption in Subcase i, |aj+k−1,j| = j+ki=j−3nj+k−1,iaij. Since

both of n_j+k−1,i and n_j+k−2,i for each i = j,j + 1,. . . ,j + k− 3 can arbitrarily be 0 or 1,itisobviousthat j+k−3_i=j n_j+k−1,iaij andaj+k−2,j gothroughthesamevalues,as

Y goesthrough the set Kn. Bytheinduction hypothesis, we obtain|aj+k−1,j| ≤ fk−2

forallY ∈ Kn.Thus,|aj+k,j| ≤ fk−1+ fk−2= fk forallY ∈ Kn.

Subcase ii.Assumenj+k−1,j+k−2= 1. ByLemma 2.2,wehave

|aj+k,j| ≤ j+k−3 i=j nj+k,iaij + j+k−1 i=j+k−2 nj+k,iaij .

Since j+k_i=j−3nj+k,iaij andaj+k−2,j gothroughthesamevalues,asY goesthroughthe

setKn,bytheinductionhypothesis,wehave _i=jj+k−3nj+k,iaij ≤ fk−2forallY ∈ Kn.

Inadditiontothis, sincenj+k−1,j+k−2= 1 we obtain j+k−1 i=j+k−2 nj+k,iaij = (nj+k,j+k−2− 1)aj+k−2,j− j+k−3 i=j n_j+k−1,iaij.

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Here each(1− nj+k,j+k−2),nj+k−1,j,. . . ,nj+k−1,j+k−3 canarbitrarily be 0 or1.Thus, j+k−1

i=j+k−2nj+k,iaij andaj+k−1,j gothroughthesamevalues,asY goesthroughtheset

Knandhence,bytheinductionhypothesis, j+k−1_i=j+k−2nj+k,iaij ≤ fk−1forallY ∈ Kn.

Therefore, weobtain|aj+k,j| ≤ fk−2+ fk−1 = fk forallY ∈ Kn.

Theprincipleofinductioncompletestheproof. 2

LetA= (aij),B = (bij)∈ Mn(R),thatis,thesetofalln× n realmatrices.Wewrite

A ≥ 0 ifallaij ≥ 0.Also,we write A≥ B ifA− B ≥ 0.In additionto this, wedeﬁne

|A| = (|aij|),thatis,|A| istheelement-wiseabsolutevalueofA.Thelargest eigenvalue

of A inmodulusisdenotedbyρ(A) andcalledthespectral radiusofA.Nowweﬁx the notation fortherest ofthepaper.

Theorem 2.2. Let Y0 = (y0ij)∈ Kn be as in (1.2) andlet Z0 := Y0Y0T. ForallZ ∈ Ln,

we have|Z−1|≤ |Z₀−1|.

Proof. Firstly we obtaintheinverseofY0.LetN0= Y0− I andN0= (mij).Then

mij =

0 if i≤ j,

1−(−1)i+j

2 otherwise.

WeclaimthattheinverseofY0 isthen× n matrix(cij),where

cij= ⎧ ⎪ ⎨ ⎪ ⎩ 0 if i < j, 1 if i = j, (−1)i−jfi−j if i > j.

Since Y0∈ Kn byLemma 2.2, itisclearthatcij = 0 ifi< j andcij = 1 ifi= j.Also,

byLemma 2.2,wehave cij =− i−1 k=j mikckj

fori> j.Nowweprovethatcij = (−1)i−jfi−j wheneveri> j byinductionont= i−j.

Fort= 1,

cj+1,j =−mj+1,j=−1 = −f1.

Assumethatcj+t,j= (−1)tftforallt= 1,2,. . . ,k− 1.Recall that

cj+k,j=− j+k−1

s=j

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Bytheinductionhypothesis, ifk iseventhenwe have cj+k,j = k/2 s=1 f2s−1= fk

andifk isoddthen

cj+k,j =−1 −

(k−1)/2

s=1

f2s=−fk.

Here thelast equalities follow from the well-knownFibonacciidentities, see [22, Theo-rem 5.2andCorollary 5.1].Thus,cj+k,j= (−1)kfk.

Secondly,wecalculatetheinverseofZ0.SinceZ0= Y0Y0T andY0−1= (cij),wehave

(Z₀−1)ii = n k=1 c2_ki= 1 + n k=i+1 f_k−i2 (2.2)

foralli= 1,2,. . . ,n.Nowlet1≤ i< j≤ n.Then

(Z₀−1)ij= n t=1 ctictj = cji+ n t=j+1 ctictj = (−1)j−if_j−i+ n t=j+1 (−1)−i−jf_t−if_t−j = (−1)j−i(fj−i+ n t=j+1 ft−ift−j). (2.3)

SinceZ₀−1 issymmetric, forall1≤ j < i≤ n,

(Z₀−1)ij = (−1)i−j(fi−j+ n

t=i+1

ft−ift−j). (2.4)

Nowweprovetheclaimofthetheorem.ForeachZ ∈ Ln,there existsamatrixY in

Kn such thatZ = Y YT. Let Y−1 = (aij). Then, by Lemma 2.2 and Theorem 2.1, we

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_(Z−1₎ ii= n k=1 a2ki = n k=1 |aki|2 = n k=i |aki|2 ≤ 1 + n k=i+1 f_k−i2 =(Z₀−1)ii

foralli= 1,2,. . . ,n. Letj > i. ByLemma 2.2andTheorem 2.1,wehave _(Z−1₎ ij= n t=1 atiatj ≤ n t=1 |ati||atj| =|aji| + n t=j+1 |ati||atj| ≤ fj−i+ n t=j+1 ft−ift−j =(Z₀−1)ij.

Finally,sinceZ−1 andZ₀−1 aresymmetric,|Z−1|≤ |Z₀−1| forallZ∈ Ln. 2

3. ProofofConjecture 1.1

Thefollowing lemmaiscrucialintheproofofConjecture 1.1.

Lemma 3.1. (See [18, Theorem 8.1.18].) Let A,B ∈ Mn(R). If |A| ≤ B, then ρ(A) ≤

ρ(|A|)≤ ρ(B).

Proofoftheconjecture. LetZ0 beas inTheorem 2.2.Firstweprovethatthematrices Z₀−1 and|Z₀−1| have thesame characteristic polynomial.Bythedeﬁnition ofthe trace of asquarematrix,itisclearthat

tr((Z₀−1)k) = n i1,...,ik=1 (Z₀−1)i1i2. . . (Z −1 0 )ik−1ik(Z −1 0 )iki1

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foreachk = 1,2,. . . ,n.Also,fromformulae(2.2)–(2.4)intheproofofTheorem 2.2,one caneasily showthatsgn(Z₀−1)ij= (−1)i−j forall1≤ i,j≤ n.Thus,we have

sgn((Z₀−1)i1i2. . . (Z

−1

0 )ik−1ik(Z −1

0 )iki1) = 1

and hence tr(|Z₀−1|k) = tr((Z₀−1)k). By Newton’s identities [20], we obtain that Z₀−1

and |Z₀−1| have the same characteristic polynomial. Thus, ρ(|Z₀−1|) = ρ(Z₀−1). From

Theorem 2.2and Lemma 3.1,now weobtain

ρ(Z−1)≤ ρ(|Z−1|) ≤ ρ(|Z₀−1|) = ρ(Z₀−1),

forall Z inLn. Sinceall Z inLn are positivedeﬁnite,thesmallesteigenvalue ofZ0 is

lessthanor equalto thesmallesteigenvalueofZ forallZ inLn. 2

Finally, inour investigation [2], we cannot ﬁnd any matrixY other than Y0 in Kn

suchthat cn is equalto the smallesteigenvalue ofY YT foreach n= 2,3,. . . ,9.After

thisobservationwecanpresent thefollowing conjecture.

Conjecture3.1.Letn≥ 2.ThereisauniquematrixY inKn suchthatcn isequaltothe

smallesteigenvalue ofY YT.In otherwords,ifthesmallest eigenvalueof Y YT isequal tocn thenY = Y0,whereY0 isdeﬁned by(1.2).

4. Lowerboundsforcn

Intheliteraturetherearenotsomanyresultsonestimatingthevalueofcn.Recently,

Mattila[24] haspresented alower bound forcn.Indeed, heproved thatcn isbounded

belowby( 6

n4_+2n3_+2n2_+n)

n−1

2 .Thenheshowedthatthislowerboundcanbereplacedwith

( 48

n4_+56n2_+48n)

n−1

2 when n is even, and ( 48

n4_+50n2_+48n₋₅₁)

n−1

2 when n is odd. Recently,

beside Mattila’s results, Altınışık and Büyükköse [4] have obtaineda lower bound for thesmallest eigenvaluetn of then× n matrix EnTEn, where the ij-entry of En is 1if

j|i and0 otherwise,i.e.,tn≥

n n_k=1μ2_(k)−1_._Indeed,_this_bound_can_be_used_instead

oflower bounds includingcn for thesmallesteigenvaluesof GCDand related matrices

deﬁnedonS ={1,2,. . . ,n} intheliterature,see[14,16,19,24,27].Afterabovestudieson estimatingthevalueofcn,wenaturallyraise thefollowing problem.

Problem4.1.Canoneimprovethelowerboundsmentionedaboveforcn?

In the review process, Professor Jorma K. Merikoski proposed a solution to Prob-lem 4.1,whichispresented inAppendix A.

Acknowledgements

TheauthorswouldliketothankProfessorIlkerInam,ProfessorPenttiHaukkanenand two anonymous reviewers for their valuable suggestions inimproving the presentation

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of ourmanuscript.Wealsowouldlike tothanktoProfessorJormaK.Merikoskiforhis solutionto Problem 4.1.

Appendix A. AsolutiontoProblem 4.1(byJorma K.Merikoski) Recall thatthematrixZ₀−1= (ζij) satisﬁes

ζii= 1 + n k=i+1 f_k−i2 , |ζij| = f|j−i|+ n k=j+1 fk−ifk−j, i= j.

Alsorecallthatthespectralradius ofanonnegative squarematrixislessthan orequal to themaximalrowsum,andequalityholds ifand onlyifthematrixisirreducible and allrowsumsareequal(e.g., [9,Chapter 2,Theorem (2.35)]).Now

c−1_n = ρ(Z₀−1) = ρ(|Z₀−1|) < max i n j=1 |ζij| = n j=1 |ζ1j| =|ζ11| + n j=2 |ζ1j| = 1 + n k=2 f_k−12 + n j=2 f_j−1+ n k=j+1 f_k−1f_k−j = 1 + n−1 k=1 fk+ n j=1 n k=j+1 fk−1fk−j = 1 + S1+ S2.

It iswell-known[12,Eq. (22)]that

m k=1 fk = fm+2− 1; (A.1) so S1= fn+1− 1.

Moreworkmustbe donewith S2. Firstwehave

S2= (f12+ f22+· · · + fn−12 ) + (f2f1+ f3f2+· · · + fn−1fn−2) +· · ·

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= f1(f1+· · · + fn−1) + f2(f2+· · · + fn−1) +· · · + fj(fj+· · · + fn−1) +· · · + f_n−2(f_n−2+ f_n−1) + f_n2₋₁ = n−1 j=1 fj n−1 k=j fk. By(A.1), n−1 k=j fk = n−1 k=1 fk− j−1 k=1 fk= (fn+1− 1) − (fj+1− 1) = fn+1− fj+1, whichimplies S2= n−1 j=1 fj(fn+1− fj+1) = fn+1 n−1 j=1 fj− n−1 j=1 fjfj+1= T1− T2.

ItremainstosimplifyT1 and T2.By(A.1),

T1= fn+1(fn+1− 1).

Itiswell-known[29,SequenceA064831]that

m j=1fjfj+1= fm+12 − 1 ifn iseven, m j=1fjfj+1= fm+12 ifn isodd. Hence T2= fn2ifn iseven, T2= fn2− 1 ifn isodd. Inotherwords, T2= fn2− ηn, where ηn= 1− (−1)n 2 . Now c−1_n < 1 + S1+ T1− T2 = 1 + (fn+1− 1) + fn+1(fn+1− 1) − fn2+ ηn

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= fn+12 − fn2+ ηn

= (fn+1− fn)(fn+1+ fn) + ηn

= f_n−1fn+2+ ηn.

It iswell-known[8,Eq. (2)]that

fm−rfm+s− fmfm+s−r= (−1)m−r−1frfs. Hence fn−1fn+2= fnfn+1+ (−1)n, and so c−1_n < fnfn+1+ (−1)n+ ηn = fnfn+1+ θn, where θn= 1 + (−1)n 2 . Therefore cn> (fnfn+1+ θn)−1 = γn,

whichseemstobemuchbetterthanMattila’s[25]bounds.Forexample,c6= 0.0148 (in

three digits). Thebound γ6 = 0.00952 is rather good, while Mattila’s betterbound is

0.0000205. References

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