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Research ArticleCirculant Type Matrices with the Sum and Product ofFibonacci and Lucas Numbers

Zhaolin Jiang1 Yanpeng Gong2 and Yun Gao2

1 Department of Mathematics Linyi University Linyi Shandong 276000 China2 Institute of Applied Mathematics Shandong University of Technology Zibo Shandong 255049 China

Correspondence should be addressed to Zhaolin Jiang jzh1208sinacom

Received 28 April 2014 Accepted 4 June 2014 Published 19 June 2014

Copyright copy 2014 Zhaolin Jiang et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Circulant type matrices have become an important tool in solving differential equations In this paper we consider circulanttype matrices including the circulant and left circulant and 119892-circulant matrices with the sum and product of Fibonacci andLucas numbers Firstly we discuss the invertibility of the circulant matrix and present the determinant and the inverse matrixby constructing the transformation matrices Furthermore the invertibility of the left circulant and 119892-circulant matrices is alsodiscussedWe obtain the determinants and the inversematrices of the left circulant and 119892-circulantmatrices by utilizing the relationbetween left circulant and 119892-circulant matrices and circulant matrix respectively

1 Introduction

Circulant matrices may play a crucial role for solving variousdifferential equations In [1] Ruiz-Claeyssen and dos SantosLeal introduced factor circulant matrices matrices with thestructure of circulants butwith the entries below the diagonalbeing multiplied by the same factor The diagonalizationof a circulant matrix and the spectral decomposition areconveniently generalized to block matrices with the structureof factor circulants Matrix and partial differential equationsinvolving factor circulants are considered Wu and Zou in[2] discussed the existence and approximation of solutionsof asymptotic or periodic boundary value problems of mixedfunctional differential equations They focused on (513) in[2] with a circulant matrix whose principal diagonal entriesare zeroes In [3] some Routh-Hurwitz stability conditionsare generalized to the fractional order case The authorsconsidered the 1-system CML (10) They selected a circu-lant matrix which reads a tridiagonal matrix Ahmed andElgazzar used coupled map lattices (CML) as an alternativeapproach to include spatial effects in fractional order systems(FOS) Consider the 1-system CML (10) in [4] They claimedthat the system is stable if all the eigenvalues of the circulantmatrix satisfy (2) in [4] Trench considered nonautonomous

systems of linear differential equations (1) in [5] with someconstraint on the coefficientmatrix119860(119905) One case is that119860(119905)is a variable block circulant matrix Kloeden et al adoptedthe simplest approximation schemes for (1) in [6] with theEuler method which reads (5) in [6] They exploited thatthe covariance matrix of the increments can be embeddedin a circulant matrix The total loops can be done by fastFourier transformation which leads to a total computationalcost of 119874(119898 log119898) = 119874(119899 log 119899) Guo et al concerned ongeneric Dn-Hopf bifurcation to a delayed Hopfield-Cohen-Grossberg model of neural networks (517) in [7] where 119879denoted an interconnection matrix They especially assumed119879 is a symmetric circulant matrix Lin and Yang discretizedthe partial integrodifferential equation (PIDE) in pricingoptions with the preconditioned conjugate gradient (PCG)method which constructed the circulant preconditionersBy using FFT the cost for each linear system is 119874(119899 log 119899)where 119899 is the size of the system in [8] Lee et al investigateda high-order compact (HOC) scheme for the general two-dimensional (2D) linear partial differential equation (11) in[9] with a mixed derivative Meanwhile in order to establishthe 2D combined compact difference (CCD2) scheme theyrewrote (11) in [9] into (21) in [9] Towrite theCCD2 systemin a concise style they employed circulant matrix to obtain

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 375251 12 pageshttpdxdoiorg1011552014375251

2 Abstract and Applied Analysis

the corresponding whole CCD2 linear system (210) in [9]whose entries are circulant block

Circulant type matrices have important applications invarious disciplines including image processing communica-tions signal processing encoding solving Toeplitz matrixproblems and least squares problemsThey have been put onfirm basis with the work of Davis [10] Jiang and Zhou [11]and Gray [12]

In [13] the authors pointed out the processes based on theeigenvalue of circulant type matrices with iid entries Thereare discussions about the convergence in probability and indistribution of the spectral norm of circulant typematrices in[14] The 119892-circulant matrices play an important role in vari-ous applications as well For details please refer to [15 16] andthe references therein Ngondiep et al showed the singularvalues of 119892-circulants in [17] In [18 19] the authors gave thelimiting spectral distributions of left circulant matrices

The Fibonacci and Lucas sequences are defined by thefollowing recurrence relations [20 21] respectively

119865119899+2

= 119865119899+1

+ 119865119899

where 1198650= 0 119865

1= 1

119871119899+2

= 119871119899+1

+ 119871119899

where 1198710= 2 119871

1= 1

(1)

For 119899 ge 0 the first few values of the sequences are givenby the following equation

119899 0 1 2 3 4 5 6 7 8 sdot sdot sdot

1198651198990 1 1 2 3 5 8 13 21 sdot sdot sdot

1198711198992 1 3 4 7 11 18 29 47 sdot sdot sdot

(2)

Let 120572 120573 be the roots of characteristic equation 1199092minus119909minus1 =0 then the Binet formulas of the sequences 119865

119899 and 119871

119899 have

the form

119865119899=120572119899

minus 120573119899

120572 minus 120573 119871

119899= 120572119899

+ 120573119899

2 120573 =

2

(3)

LetF119899= 119865119899sdot119871119899andL

119899= 119865119899+119871119899 so we can get two new

sequencesF119899andL

119899[22]The two sequences are defined by

the following recurrence relations respectively

F119899+2

= 3F119899+1

+F119899 where F

0= 0 F

1= 1

L119899+2

=L119899+1

+L119899 where L

0= 2 L

1= 2

(4)

For 119899 ge 0 the first few values of the sequences are givenby the following equation

119899 0 1 2 3 4 5 6 7 8 sdot sdot sdot

F1198990 1 3 8 21 55 144 377 987 sdot sdot sdot

L1198992 2 4 6 10 16 26 42 68 sdot sdot sdot

(5)

TheF119899is given by the formulaF

119899= (120572119899

1minus120573119899

1)(1205721minus1205731)

where 1205721 1205731are the roots of 1199092 minus 3119909 + 1 = 0L

119899is given by

the formula L119899= 119865119899+ 119871119899= (120572119899

minus 120573119899

)(120572 minus 120573) + (120572119899

+ 120573119899

)where 120572 120573 are the roots of 1199092 minus 119909 minus 1 = 0

Besides some scholars have given various algorithmsfor the determinants and inverses of nonsingular circulant

matrices [10 11] Unfortunately the computational com-plexities of these algorithms are very amazing with theorder of matrix increasing However some authors gave theexplicit determinants and inverse of circulant and skew-circulant involving Fibonacci and Lucas numbers For exam-ple Dazheng gave the determinant of the Fibonacci-Lucasquasicyclic matrices in [20] Shen et al considered circulantmatrices with Fibonacci and Lucas numbers and presentedtheir explicit determinants and inverses by constructingthe transformation matrices [21] Jaiswal evaluated somedeterminants of circulant whose elements are the generalizedFibonacci numbers [23] Lind presented the determinants ofcirculant and skew-circulant involving Fibonacci numbers[24] Bozkurt and Tam gave determinants and inversesof circulant matrices with Jacobsthal and Jacobsthal-Lucasnumbers [25]

In [22] the authors gave some determinantal and perma-nental representations of F

119899and L

119899and complex factor-

ization formulas The purpose of this paper is to obtain theexplicit determinants and inverse of circulant type matricesby some perfect properties ofF

119899andL

119899

In this paper we adopt the following two conventions00

= 1 and for any sequence 119886119899 sum119899119896=119894119886119896= 0 in the case

119894 gt 119899

Definition 1 (see [10 11]) In a circulant matrix (or rightcirculant matrix [26])

Circ (1198861 1198862 119886

119899) =

[[[[

[

11988611198862sdot sdot sdot 119886

119899

1198861198991198861sdot sdot sdot 119886119899minus1

11988621198863sdot sdot sdot 119886

1

]]]]

]

(6)

each row is a cyclic shift of the row above to the right

Circulant matrix is a special case of a Toeplitz matrix It isevidently determined by its first row (or column)

Definition 2 (see [11 26]) In a left circulantmatrix (or reversecirculant matrix [13 14 18 19])

LCirc (1198861 1198862 119886

119899) =

[[[[

[

11988611198862sdot sdot sdot 119886

119899

11988621198863sdot sdot sdot 119886

1

1198861198991198861sdot sdot sdot 119886119899minus1

]]]]

]

(7)

each row is a cyclic shift of the row above to the left

Left circulant matrix is a special Hankel matrix

Definition 3 (see [14 27]) A 119892-circulant matrix is an 119899 times 119899complex matrix with the following form

119860119892119899=(

1198861

1198862

sdot sdot sdot 119886119899

119886119899minus119892+1

119886119899minus119892+2

sdot sdot sdot 119886119899minus119892

119886119899minus2119892+1

119886119899minus2119892+2

sdot sdot sdot 119886119899minus2119892

d

119886119892+1

119886119892+2

sdot sdot sdot 119886119892

) (8)

Abstract and Applied Analysis 3

where 119892 is a nonnegative integer and each of the subscripts isunderstood to be reduced modulo 119899

The first row of 119860119892119899

is (1198861 1198862 119886

119899) its (119895 + 1)th row

is obtained by giving its 119895th row a right circular shift by 119892positions (equivalently 119892 mod 119899 positions) Note that 119892 = 1or 119892 = 119899 + 1 yields the standard circulant matrix If 119892 = 119899 minus 1then we obtain the left circulant matrix

Lemma4 (see [21]) Let119860 = Circ(1198861 1198862 119886

119899) be a circulant

matrix then one has

(i) 119860 is invertible if and only if 119891(120596119896) = 0 (119896 = 0 1

2 119899 minus 1) where 119891(119909) = sum119899

119895=1119886119895119909119895minus1 and 120596 =

exp(2120587119894119899)

(ii) If119860 is invertible then the inverse119860minus1 of119860 is a circulantmatrix

Lemma 5 Define

Δ =((

(

1 0 0 sdot sdot sdot 0 0

0 0 0 sdot sdot sdot 0 1

0 0 0 sdot sdot sdot 1 0

d

0 0 1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

))

)

(9)

the matrix Δ is an orthogonal cyclic shift matrix (and aleft circulant matrix) It holds that LCirc(119886

1 1198862 119886

119899) =

ΔCirc(1198861 1198862 119886

119899)

Lemma6 (see [27]) The 119899times119899matrixQ119892is unitary if and only

if (119899 119892) = 1 whereQ119892is a 119892-circulant matrix with the first row

119890lowast

= [1 0 0]

Lemma 7 (see [27]) 119860119892119899

is a 119892-circulant matrix with the firstrow [119886

1 1198862 119886

119899] if and only if 119860

119892119899= Q119892119862 where 119862 =

Circ(1198861 1198862 119886

119899)

2 Determinant and Inverse of a CirculantMatrix with the Product of the Fibonacciand Lucas Numbers

In this section letA119899= Circ(F

1F2 F

119899) be a circulant

matrix Firstly we give the determinant equation of matrixA119899 Afterwards we prove that A

119899is an invertible matrix for

119899 gt 2 and then we find the inverse of the matrixA119899

Theorem 8 Let A119899= Circ(F

1F2 F

119899) be a circulant

matrix then one has

detA119899= (1 minusF

119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(10)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Proof Obviously detA1= 1 satisfies (10) In the case 119899 gt 1

let

Γ =

((((((

(

1

minus3 1

1 1 minus3

0 0 1 minus3 1

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 1

))))))

)119899times119899

Π1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (minusF119899

F1minusF119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (minusF119899

F1minusF119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0minusF119899

F1minusF119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(11)

We can obtain

ΓA119899Π1

=((

(

F11198911015840

119899F119899minus1

F119899minus2

sdot sdot sdot F2

0 119891119899

minusF119899minus2

minusF119899minus3

sdot sdot sdot minusF1

0 0 F1minusF119899+1

0 0 F119899

F1minusF119899+1

F

119899

0 0 0 F1minusF119899+1

))

)

(12)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(13)

We obtain

det Γ detA119899detΠ1

= F1[F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

]

times (F1minusF119899+1)119899minus2

= F1[F1minusF119899+1

+

119899minus1

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

]

4 Abstract and Applied Analysis

times (F1minusF119899+1)119899minus2

= (1 minusF119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(14)

while

det Γ = detΠ1= (minus1)

(119899minus1)(119899minus2)2

(15)

we have

detA119899= (1 minusF

119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(16)

Thus the proof is completed

Theorem 9 Let A119899= Circ(F

1F2 F

119899) be a circulant

matrix if 119899 gt 2 thenA119899is an invertible matrix

Proof When 119899 = 3 in Theorem 8 we have detA3= 468 = 0

henceA3is invertible In the case 119899 gt 3 sinceF

119899= (120572119899

1minus120573119899

1)

(1205721minus 1205731) where 120572

1+ 1205731= 3 120572

1sdot 1205731= 1We have

119891 (120596119896

) =

119899

sum

119895=1

F119895(120596119896

)119895minus1

=1

1205721minus 1205731

119899

sum

119895=1

(120572119895

1minus 120573119895

1) (120596119896

)119895minus1

=1

1205721minus 1205731

[1205721(1 minus 120572

119899

1)

1 minus 1205721120596119896

minus1205731(1 minus 120573

119899

1)

1 minus 1205731120596119896

]

=1

1205721minus 1205731

[(1205721minus 1205731) minus (120572

119899+1

1minus 120573119899+1

1)

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

+1

1205721minus 1205731

[12057211205731(120572119899

1minus 120573119899

1) 120596119896

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

=1 minusF

119899+1+F119899120596119896

1 minus 3120596119896 + 1205962119896(119896 = 1 2 119899 minus 1)

(17)

If there exists 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 1minusF

119899+1+F119899120596119897

= 0 for 1minus3120596119897+1205962119897 = 0 thus 120596119897 =(F119899+1

minus 1)F119899is a real number While 120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899) hence sin(2119897120587119899) = 0 so we have120596119897

= minus1 for 0 lt 2119897120587119899 lt 2120587 But 119909 = minus1 is not the root ofequation 1 minusF

119899+1+F119899119909 = 0 (119899 gt 3) We obtain 119891(120596119896) = 0

for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) = sum119899

119895=1F119895=

F119899+1

minusF119899minus 1 = 0 By Lemma 4 the proof is completed

Lemma 10 Let the matrixG = [119892119894119895]119899minus2

119894119895=1

be of the form

119892119894119895=

F1minusF119899+1 119894 = 119895

F119899 119894 = 119895 + 1

0 otherwise(18)

and then the inverse Gminus1 = [1198921015840119894119895]119899minus2

119894119895=1

of the matrix G is equalto

1198921015840

119894119895=

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(19)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Obviously 119888

119894119895= 0 for 119894 lt 119895 In

the case 119894 = 119895 we obtain 119888119894119894= 1198921198941198941198921015840

119894119894= (F1minusF119899+1) sdot (1(F

1minus

F119899+1)) = 1 For 119894 ge 119895 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 119892119894119894minus1

1198921015840

119894minus1119895+ 1198921198941198941198921015840

119894119895

= F119899sdot

(minusF119899)119894minus119895minus1

(F1minusF119899+1)119894minus119895

+ (F1minusF119899+1) sdot

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

= 0

(20)

We verify GGminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verify Gminus1G = 119868

119899minus2 Thus

the proof is completed

Theorem 11 Let A119899= Circ(F

1F2 F

119899) (119899 gt 2) be a

circulant matrix then one has

Aminus1

119899=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(21)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(22)

Abstract and Applied Analysis 5

Proof Let

Π2=

(((((

(

1 minus1198911015840

119899minus1198911015840

119899

119891119899

F119899minus2

minusF119899minus1

sdot sdot sdot minus1198911015840

119899

119891119899

F1

0 1F119899minus2

119891119899

sdot sdot sdotF1

119891119899

0 0 1 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 1

)))))

)

(23)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(24)

We have

ΓA119899Π1Π2= D1oplusG (25)

whereD1= diag(F

1 119891119899) is a diagonal matrix andD

1oplusG is

the direct sum ofD1andG If we denoteΠ = Π

1Π2 then we

obtain

Aminus1

119899= Π (D

minus1

1oplusGminus1

) Γ (26)

Since the last row elements of the matrix Π are

0 1F119899minus2

119891119899

F119899minus3

119891119899

F2

119891119899

F1

119891119899

(27)

By Lemma 10 if we letAminus1119899= Circ(119909

1 1199092 119909

119899) its last

row elements are given by the following equations

1199092= minus

3

119891119899

+1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199093=

F1

119891119899(F1minusF119899+1)

1199094=1

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus

3F1

119891119899(F1minusF119899+1)

1199095=1

119891119899

3

sum

119894=1

F4minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

+F1

119891119899(F1minusF119899+1)

119909119899=1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus4

sum

119894=1

F119899minus3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199091=1

119891119899

minus3

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

(28)

Let 119862(119895)119899

= sum119895

119894=1(F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

) (119895 = 1

2 119899 minus 2) we have

119862(2)

119899minus 3119862(1)

119899

= minus3F1

F1minusF119899+1

+

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=minusF119899

(F1minusF119899+1)2

minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

= minus3

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894+

119899minus3

sum

119894=1

F119899minus119894minus2

(minusF119899)119894minus1

(F1minusF119899+1)119894

=(minus3F

1) (minusF

119899)119899minus3

(F1minusF119899+1)119899minus2

+

119899minus3

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=

119899minus2

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

119862(119895+2)

119899minus 3119862(119895+1)

119899+ 119862(119895)

119899

=

119895+2

sum

119894=1

F119895+3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus 3

119895+1

sum

119894=1

F119895+2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+

119895

sum

119894=1

F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

=F2(minusF119899)119895

(F1minusF119899+1)119895+1

+F1(minusF119899)119895+1

(F1minusF119899+1)119895+2

minus3F1(minusF119899)119895

(F1minusF119899+1)119895+1

6 Abstract and Applied Analysis

+

119895

sum

119894=1

(F119895+3minus119894

minus 3F119895+2minus119894

+F119895+1minus119894

) (minusF119899)119894minus1

(F1minusF119899+1)119895+1

=(minusF119899)119895+1

(F1minusF119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(29)

We obtain

Aminus1

119899= Circ(

1 minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

119891119899

119862(119899minus2)

119899minus 3

119891119899

119862(1)

119899

119891119899

119862(2)

119899minus 3119862(1)

119899

119891119899

119862(3)

119899minus 3119862(2)

119899+ 119862(1)

119899

119891119899

119862(119899minus2)

119899minus 3119862(119899minus3)

119899+ 119862(119899minus4)

119899

119891119899

)

=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(30)

3 Determinant and Inverse of a CirculantMatrix with the Sum of the Fibonacci andLucas Numbers

In this section let B119899= Circ(L

1L2 L

119899) be a circulant

matrix Firstly we give an explicit determinant formula ofmatrix B

119899 Afterwards we prove that B

119899is an invertible

matrix for any positive integer 119899 and then we find its inverse

Theorem 12 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

detB119899= 2[(2 minusL

119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(31)

whereL119899is the 119899th 119865

119899+ 119871119899number

Proof ObviouslyB1= 2 satisfies (31) when 119899 gt 1 Let

Σ =

((((((

(

1

minus2 1

minus1 1 minus1

0 0 1 minus1 minus1

c c c0 1 c c0 1 minus1 c 0

0 1 minus1 minus1

))))))

)119899times119899

Ω1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (L119899minus 2

L1minusL119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (L119899minus 2

L1minusL119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0L119899minus 2

L1minusL119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(32)

Then

ΣB119899Ω1

=((

(

L11198971015840

119899L119899minus1

sdot sdot sdot L3

L2

0 119897119899minus2L119899minus1

+L119899sdot sdot sdot sdot sdot sdot minus2L

2+L3

0 0 L1minusL119899+1

0 0 2 minusL119899

d

0 0 2 minusL119899

L1minusL119899+1

))

)

(33)

where

119897119899=L1minus 2L119899+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(34)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(35)

We can obtain

detΣ detB119899detΩ1

=L1[L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

Abstract and Applied Analysis 7

=L1[L1minusL119899+1

+

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(36)

while

detΣ = detΩ1= (minus1)

(119899minus1)(119899minus2)2

(37)

We have

detB119899

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(38)

Theorem 13 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix thenB119899is invertible for any positive integer 119899

Proof SinceL119899= (120572119899

minus120573119899

)(120572minus120573)+ 120572119899

+ 120573119899 where120572+120573 = 1

120572 sdot 120573 = minus1 We have

119891 (120596119896

) =

119899

sum

119895=1

L119895(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573+ 120572119895

+ 120573119895

)(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573) (120596119896

)119895minus1

+

119899

sum

119895=1

(120572119895

+ 120573119895

) (120596119896

)119895minus1

=1

120572 minus 120573[120572 (1 minus 120572

119899

)

1 minus 120572120596119896minus120573 (1 minus 120573

119899

)

1 minus 120573120596119896]

+120572 (1 minus 120572

119899

)

1 minus 120572120596119896+120573 (1 minus 120573

119899

)

1 minus 120573120596119896

=1 minus 119865119899+1

minus 119865119899120596119896

1 minus 120596119896 minus 1205962119896+1 minus 119871119899+1

+ (2 minus 119871119899) 120596119896

1 minus 120596119896 minus 1205962119896

=2 minus (119865

119899+1+ 119871119899+1) minus (119865

119899+ 119871119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896

=2 minusL

119899+1minus (L119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896(119896 = 1 2 119899 minus 1)

(39)

If there exist 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 2 minusL

119899+1minus (L119899minus 2)120596119897

= 0 for 1 minus120596119897 minus1205962119897 = 0 120596119897 =(2minusL

119899+1)(L119899minus2) is a real number while120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899)Hence sin(2119897120587119899) = 0 so we have 120596119897 = minus1 for 0 lt

2119897120587119899 lt 2120587 But 119909 = minus1 is not the root of the equation2 minus L

119899+1minus (L119899minus 2)119909 = 0 for any positive integer 119899 We

obtain 119891(120596119896) = 0 for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) =sum119899

119895=1L119895= L119899+1

+ L119899minus 4 = 0 By Lemma 4 the proof is

completed

Lemma 14 Let matrixH = [ℎ119894119895]119899minus2

119894119895=1

be of the form

ℎ119894119895=

L1minusL119899+1 119894 = 119895

2 minusL119899 119894 = 119895 + 1

0 otherwise(40)

and then inverseHminus1 = [ℎ1015840119894119895]119899minus2

119894119895=1

of the matrixH is equal to

ℎ1015840

119894119895=

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(41)

Proof Let 119903119894119895= sum119899minus2

119896=1ℎ119894119896ℎ1015840

119896119895 Obviously 119903

119894119895= 0 for 119894 lt 119895 In the

case 119894 = 119895 we obtain

119903119894119894= ℎ119894119894ℎ1015840

119894119894= (L1minusL119899+1) sdot

1

L1minusL119899+1

= 1 (42)

For 119894 ge 119895 + 1 we obtain

119903119894119895=

119899minus2

sum

119896=1

ℎ119894119896ℎ1015840

119896119895= ℎ119894119894minus1

ℎ1015840

119894minus1119895+ ℎ119894119894ℎ1015840

119894119895

= (2 minusL119899) sdot

(L119899minus 2)119894minus119895minus1

(L1minusL119899+1)119894minus119895

+ (L1minusL119899+1) sdot

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

= 0

(43)

We verify HHminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyHminus1H = 119868

119899minus2 Thus

the proof is completed

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

the corresponding whole CCD2 linear system (210) in [9]whose entries are circulant block

Circulant type matrices have important applications invarious disciplines including image processing communica-tions signal processing encoding solving Toeplitz matrixproblems and least squares problemsThey have been put onfirm basis with the work of Davis [10] Jiang and Zhou [11]and Gray [12]

In [13] the authors pointed out the processes based on theeigenvalue of circulant type matrices with iid entries Thereare discussions about the convergence in probability and indistribution of the spectral norm of circulant typematrices in[14] The 119892-circulant matrices play an important role in vari-ous applications as well For details please refer to [15 16] andthe references therein Ngondiep et al showed the singularvalues of 119892-circulants in [17] In [18 19] the authors gave thelimiting spectral distributions of left circulant matrices

The Fibonacci and Lucas sequences are defined by thefollowing recurrence relations [20 21] respectively

119865119899+2

= 119865119899+1

+ 119865119899

where 1198650= 0 119865

1= 1

119871119899+2

= 119871119899+1

+ 119871119899

where 1198710= 2 119871

1= 1

(1)

For 119899 ge 0 the first few values of the sequences are givenby the following equation

119899 0 1 2 3 4 5 6 7 8 sdot sdot sdot

1198651198990 1 1 2 3 5 8 13 21 sdot sdot sdot

1198711198992 1 3 4 7 11 18 29 47 sdot sdot sdot

(2)

Let 120572 120573 be the roots of characteristic equation 1199092minus119909minus1 =0 then the Binet formulas of the sequences 119865

119899 and 119871

119899 have

the form

119865119899=120572119899

minus 120573119899

120572 minus 120573 119871

119899= 120572119899

+ 120573119899

2 120573 =

2

(3)

LetF119899= 119865119899sdot119871119899andL

119899= 119865119899+119871119899 so we can get two new

sequencesF119899andL

119899[22]The two sequences are defined by

the following recurrence relations respectively

F119899+2

= 3F119899+1

+F119899 where F

0= 0 F

1= 1

L119899+2

=L119899+1

+L119899 where L

0= 2 L

1= 2

(4)

For 119899 ge 0 the first few values of the sequences are givenby the following equation

119899 0 1 2 3 4 5 6 7 8 sdot sdot sdot

F1198990 1 3 8 21 55 144 377 987 sdot sdot sdot

L1198992 2 4 6 10 16 26 42 68 sdot sdot sdot

(5)

TheF119899is given by the formulaF

119899= (120572119899

1minus120573119899

1)(1205721minus1205731)

where 1205721 1205731are the roots of 1199092 minus 3119909 + 1 = 0L

119899is given by

the formula L119899= 119865119899+ 119871119899= (120572119899

minus 120573119899

)(120572 minus 120573) + (120572119899

+ 120573119899

)where 120572 120573 are the roots of 1199092 minus 119909 minus 1 = 0

Besides some scholars have given various algorithmsfor the determinants and inverses of nonsingular circulant

matrices [10 11] Unfortunately the computational com-plexities of these algorithms are very amazing with theorder of matrix increasing However some authors gave theexplicit determinants and inverse of circulant and skew-circulant involving Fibonacci and Lucas numbers For exam-ple Dazheng gave the determinant of the Fibonacci-Lucasquasicyclic matrices in [20] Shen et al considered circulantmatrices with Fibonacci and Lucas numbers and presentedtheir explicit determinants and inverses by constructingthe transformation matrices [21] Jaiswal evaluated somedeterminants of circulant whose elements are the generalizedFibonacci numbers [23] Lind presented the determinants ofcirculant and skew-circulant involving Fibonacci numbers[24] Bozkurt and Tam gave determinants and inversesof circulant matrices with Jacobsthal and Jacobsthal-Lucasnumbers [25]

In [22] the authors gave some determinantal and perma-nental representations of F

119899and L

119899and complex factor-

ization formulas The purpose of this paper is to obtain theexplicit determinants and inverse of circulant type matricesby some perfect properties ofF

119899andL

119899

In this paper we adopt the following two conventions00

= 1 and for any sequence 119886119899 sum119899119896=119894119886119896= 0 in the case

119894 gt 119899

Definition 1 (see [10 11]) In a circulant matrix (or rightcirculant matrix [26])

Circ (1198861 1198862 119886

119899) =

[[[[

[

11988611198862sdot sdot sdot 119886

119899

1198861198991198861sdot sdot sdot 119886119899minus1

11988621198863sdot sdot sdot 119886

1

]]]]

]

(6)

each row is a cyclic shift of the row above to the right

Circulant matrix is a special case of a Toeplitz matrix It isevidently determined by its first row (or column)

Definition 2 (see [11 26]) In a left circulantmatrix (or reversecirculant matrix [13 14 18 19])

LCirc (1198861 1198862 119886

119899) =

[[[[

[

11988611198862sdot sdot sdot 119886

119899

11988621198863sdot sdot sdot 119886

1

1198861198991198861sdot sdot sdot 119886119899minus1

]]]]

]

(7)

each row is a cyclic shift of the row above to the left

Left circulant matrix is a special Hankel matrix

Definition 3 (see [14 27]) A 119892-circulant matrix is an 119899 times 119899complex matrix with the following form

119860119892119899=(

1198861

1198862

sdot sdot sdot 119886119899

119886119899minus119892+1

119886119899minus119892+2

sdot sdot sdot 119886119899minus119892

119886119899minus2119892+1

119886119899minus2119892+2

sdot sdot sdot 119886119899minus2119892

d

119886119892+1

119886119892+2

sdot sdot sdot 119886119892

) (8)

Abstract and Applied Analysis 3

where 119892 is a nonnegative integer and each of the subscripts isunderstood to be reduced modulo 119899

The first row of 119860119892119899

is (1198861 1198862 119886

119899) its (119895 + 1)th row

is obtained by giving its 119895th row a right circular shift by 119892positions (equivalently 119892 mod 119899 positions) Note that 119892 = 1or 119892 = 119899 + 1 yields the standard circulant matrix If 119892 = 119899 minus 1then we obtain the left circulant matrix

Lemma4 (see [21]) Let119860 = Circ(1198861 1198862 119886

119899) be a circulant

matrix then one has

(i) 119860 is invertible if and only if 119891(120596119896) = 0 (119896 = 0 1

2 119899 minus 1) where 119891(119909) = sum119899

119895=1119886119895119909119895minus1 and 120596 =

exp(2120587119894119899)

(ii) If119860 is invertible then the inverse119860minus1 of119860 is a circulantmatrix

Lemma 5 Define

Δ =((

(

1 0 0 sdot sdot sdot 0 0

0 0 0 sdot sdot sdot 0 1

0 0 0 sdot sdot sdot 1 0

d

0 0 1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

))

)

(9)

the matrix Δ is an orthogonal cyclic shift matrix (and aleft circulant matrix) It holds that LCirc(119886

1 1198862 119886

119899) =

ΔCirc(1198861 1198862 119886

119899)

Lemma6 (see [27]) The 119899times119899matrixQ119892is unitary if and only

if (119899 119892) = 1 whereQ119892is a 119892-circulant matrix with the first row

119890lowast

= [1 0 0]

Lemma 7 (see [27]) 119860119892119899

is a 119892-circulant matrix with the firstrow [119886

1 1198862 119886

119899] if and only if 119860

119892119899= Q119892119862 where 119862 =

Circ(1198861 1198862 119886

119899)

2 Determinant and Inverse of a CirculantMatrix with the Product of the Fibonacciand Lucas Numbers

In this section letA119899= Circ(F

1F2 F

119899) be a circulant

matrix Firstly we give the determinant equation of matrixA119899 Afterwards we prove that A

119899is an invertible matrix for

119899 gt 2 and then we find the inverse of the matrixA119899

Theorem 8 Let A119899= Circ(F

1F2 F

119899) be a circulant

matrix then one has

detA119899= (1 minusF

119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(10)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Proof Obviously detA1= 1 satisfies (10) In the case 119899 gt 1

let

Γ =

((((((

(

1

minus3 1

1 1 minus3

0 0 1 minus3 1

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 1

))))))

)119899times119899

Π1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (minusF119899

F1minusF119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (minusF119899

F1minusF119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0minusF119899

F1minusF119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(11)

We can obtain

ΓA119899Π1

=((

(

F11198911015840

119899F119899minus1

F119899minus2

sdot sdot sdot F2

0 119891119899

minusF119899minus2

minusF119899minus3

sdot sdot sdot minusF1

0 0 F1minusF119899+1

0 0 F119899

F1minusF119899+1

F

119899

0 0 0 F1minusF119899+1

))

)

(12)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(13)

We obtain

det Γ detA119899detΠ1

= F1[F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

]

times (F1minusF119899+1)119899minus2

= F1[F1minusF119899+1

+

119899minus1

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

]

4 Abstract and Applied Analysis

times (F1minusF119899+1)119899minus2

= (1 minusF119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(14)

while

det Γ = detΠ1= (minus1)

(119899minus1)(119899minus2)2

(15)

we have

detA119899= (1 minusF

119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(16)

Thus the proof is completed

Theorem 9 Let A119899= Circ(F

1F2 F

119899) be a circulant

matrix if 119899 gt 2 thenA119899is an invertible matrix

Proof When 119899 = 3 in Theorem 8 we have detA3= 468 = 0

henceA3is invertible In the case 119899 gt 3 sinceF

119899= (120572119899

1minus120573119899

1)

(1205721minus 1205731) where 120572

1+ 1205731= 3 120572

1sdot 1205731= 1We have

119891 (120596119896

) =

119899

sum

119895=1

F119895(120596119896

)119895minus1

=1

1205721minus 1205731

119899

sum

119895=1

(120572119895

1minus 120573119895

1) (120596119896

)119895minus1

=1

1205721minus 1205731

[1205721(1 minus 120572

119899

1)

1 minus 1205721120596119896

minus1205731(1 minus 120573

119899

1)

1 minus 1205731120596119896

]

=1

1205721minus 1205731

[(1205721minus 1205731) minus (120572

119899+1

1minus 120573119899+1

1)

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

+1

1205721minus 1205731

[12057211205731(120572119899

1minus 120573119899

1) 120596119896

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

=1 minusF

119899+1+F119899120596119896

1 minus 3120596119896 + 1205962119896(119896 = 1 2 119899 minus 1)

(17)

If there exists 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 1minusF

119899+1+F119899120596119897

= 0 for 1minus3120596119897+1205962119897 = 0 thus 120596119897 =(F119899+1

minus 1)F119899is a real number While 120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899) hence sin(2119897120587119899) = 0 so we have120596119897

= minus1 for 0 lt 2119897120587119899 lt 2120587 But 119909 = minus1 is not the root ofequation 1 minusF

119899+1+F119899119909 = 0 (119899 gt 3) We obtain 119891(120596119896) = 0

for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) = sum119899

119895=1F119895=

F119899+1

minusF119899minus 1 = 0 By Lemma 4 the proof is completed

Lemma 10 Let the matrixG = [119892119894119895]119899minus2

119894119895=1

be of the form

119892119894119895=

F1minusF119899+1 119894 = 119895

F119899 119894 = 119895 + 1

0 otherwise(18)

and then the inverse Gminus1 = [1198921015840119894119895]119899minus2

119894119895=1

of the matrix G is equalto

1198921015840

119894119895=

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(19)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Obviously 119888

119894119895= 0 for 119894 lt 119895 In

the case 119894 = 119895 we obtain 119888119894119894= 1198921198941198941198921015840

119894119894= (F1minusF119899+1) sdot (1(F

1minus

F119899+1)) = 1 For 119894 ge 119895 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 119892119894119894minus1

1198921015840

119894minus1119895+ 1198921198941198941198921015840

119894119895

= F119899sdot

(minusF119899)119894minus119895minus1

(F1minusF119899+1)119894minus119895

+ (F1minusF119899+1) sdot

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

= 0

(20)

We verify GGminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verify Gminus1G = 119868

119899minus2 Thus

the proof is completed

Theorem 11 Let A119899= Circ(F

1F2 F

119899) (119899 gt 2) be a

circulant matrix then one has

Aminus1

119899=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(21)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(22)

Abstract and Applied Analysis 5

Proof Let

Π2=

(((((

(

1 minus1198911015840

119899minus1198911015840

119899

119891119899

F119899minus2

minusF119899minus1

sdot sdot sdot minus1198911015840

119899

119891119899

F1

0 1F119899minus2

119891119899

sdot sdot sdotF1

119891119899

0 0 1 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 1

)))))

)

(23)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(24)

We have

ΓA119899Π1Π2= D1oplusG (25)

whereD1= diag(F

1 119891119899) is a diagonal matrix andD

1oplusG is

the direct sum ofD1andG If we denoteΠ = Π

1Π2 then we

obtain

Aminus1

119899= Π (D

minus1

1oplusGminus1

) Γ (26)

Since the last row elements of the matrix Π are

0 1F119899minus2

119891119899

F119899minus3

119891119899

F2

119891119899

F1

119891119899

(27)

By Lemma 10 if we letAminus1119899= Circ(119909

1 1199092 119909

119899) its last

row elements are given by the following equations

1199092= minus

3

119891119899

+1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199093=

F1

119891119899(F1minusF119899+1)

1199094=1

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus

3F1

119891119899(F1minusF119899+1)

1199095=1

119891119899

3

sum

119894=1

F4minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

+F1

119891119899(F1minusF119899+1)

119909119899=1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus4

sum

119894=1

F119899minus3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199091=1

119891119899

minus3

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

(28)

Let 119862(119895)119899

= sum119895

119894=1(F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

) (119895 = 1

2 119899 minus 2) we have

119862(2)

119899minus 3119862(1)

119899

= minus3F1

F1minusF119899+1

+

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=minusF119899

(F1minusF119899+1)2

minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

= minus3

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894+

119899minus3

sum

119894=1

F119899minus119894minus2

(minusF119899)119894minus1

(F1minusF119899+1)119894

=(minus3F

1) (minusF

119899)119899minus3

(F1minusF119899+1)119899minus2

+

119899minus3

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=

119899minus2

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

119862(119895+2)

119899minus 3119862(119895+1)

119899+ 119862(119895)

119899

=

119895+2

sum

119894=1

F119895+3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus 3

119895+1

sum

119894=1

F119895+2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+

119895

sum

119894=1

F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

=F2(minusF119899)119895

(F1minusF119899+1)119895+1

+F1(minusF119899)119895+1

(F1minusF119899+1)119895+2

minus3F1(minusF119899)119895

(F1minusF119899+1)119895+1

6 Abstract and Applied Analysis

+

119895

sum

119894=1

(F119895+3minus119894

minus 3F119895+2minus119894

+F119895+1minus119894

) (minusF119899)119894minus1

(F1minusF119899+1)119895+1

=(minusF119899)119895+1

(F1minusF119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(29)

We obtain

Aminus1

119899= Circ(

1 minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

119891119899

119862(119899minus2)

119899minus 3

119891119899

119862(1)

119899

119891119899

119862(2)

119899minus 3119862(1)

119899

119891119899

119862(3)

119899minus 3119862(2)

119899+ 119862(1)

119899

119891119899

119862(119899minus2)

119899minus 3119862(119899minus3)

119899+ 119862(119899minus4)

119899

119891119899

)

=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(30)

3 Determinant and Inverse of a CirculantMatrix with the Sum of the Fibonacci andLucas Numbers

In this section let B119899= Circ(L

1L2 L

119899) be a circulant

matrix Firstly we give an explicit determinant formula ofmatrix B

119899 Afterwards we prove that B

119899is an invertible

matrix for any positive integer 119899 and then we find its inverse

Theorem 12 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

detB119899= 2[(2 minusL

119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(31)

whereL119899is the 119899th 119865

119899+ 119871119899number

Proof ObviouslyB1= 2 satisfies (31) when 119899 gt 1 Let

Σ =

((((((

(

1

minus2 1

minus1 1 minus1

0 0 1 minus1 minus1

c c c0 1 c c0 1 minus1 c 0

0 1 minus1 minus1

))))))

)119899times119899

Ω1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (L119899minus 2

L1minusL119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (L119899minus 2

L1minusL119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0L119899minus 2

L1minusL119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(32)

Then

ΣB119899Ω1

=((

(

L11198971015840

119899L119899minus1

sdot sdot sdot L3

L2

0 119897119899minus2L119899minus1

+L119899sdot sdot sdot sdot sdot sdot minus2L

2+L3

0 0 L1minusL119899+1

0 0 2 minusL119899

d

0 0 2 minusL119899

L1minusL119899+1

))

)

(33)

where

119897119899=L1minus 2L119899+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(34)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(35)

We can obtain

detΣ detB119899detΩ1

=L1[L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

Abstract and Applied Analysis 7

=L1[L1minusL119899+1

+

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(36)

while

detΣ = detΩ1= (minus1)

(119899minus1)(119899minus2)2

(37)

We have

detB119899

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(38)

Theorem 13 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix thenB119899is invertible for any positive integer 119899

Proof SinceL119899= (120572119899

minus120573119899

)(120572minus120573)+ 120572119899

+ 120573119899 where120572+120573 = 1

120572 sdot 120573 = minus1 We have

119891 (120596119896

) =

119899

sum

119895=1

L119895(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573+ 120572119895

+ 120573119895

)(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573) (120596119896

)119895minus1

+

119899

sum

119895=1

(120572119895

+ 120573119895

) (120596119896

)119895minus1

=1

120572 minus 120573[120572 (1 minus 120572

119899

)

1 minus 120572120596119896minus120573 (1 minus 120573

119899

)

1 minus 120573120596119896]

+120572 (1 minus 120572

119899

)

1 minus 120572120596119896+120573 (1 minus 120573

119899

)

1 minus 120573120596119896

=1 minus 119865119899+1

minus 119865119899120596119896

1 minus 120596119896 minus 1205962119896+1 minus 119871119899+1

+ (2 minus 119871119899) 120596119896

1 minus 120596119896 minus 1205962119896

=2 minus (119865

119899+1+ 119871119899+1) minus (119865

119899+ 119871119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896

=2 minusL

119899+1minus (L119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896(119896 = 1 2 119899 minus 1)

(39)

If there exist 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 2 minusL

119899+1minus (L119899minus 2)120596119897

= 0 for 1 minus120596119897 minus1205962119897 = 0 120596119897 =(2minusL

119899+1)(L119899minus2) is a real number while120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899)Hence sin(2119897120587119899) = 0 so we have 120596119897 = minus1 for 0 lt

2119897120587119899 lt 2120587 But 119909 = minus1 is not the root of the equation2 minus L

119899+1minus (L119899minus 2)119909 = 0 for any positive integer 119899 We

obtain 119891(120596119896) = 0 for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) =sum119899

119895=1L119895= L119899+1

+ L119899minus 4 = 0 By Lemma 4 the proof is

completed

Lemma 14 Let matrixH = [ℎ119894119895]119899minus2

119894119895=1

be of the form

ℎ119894119895=

L1minusL119899+1 119894 = 119895

2 minusL119899 119894 = 119895 + 1

0 otherwise(40)

and then inverseHminus1 = [ℎ1015840119894119895]119899minus2

119894119895=1

of the matrixH is equal to

ℎ1015840

119894119895=

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(41)

Proof Let 119903119894119895= sum119899minus2

119896=1ℎ119894119896ℎ1015840

119896119895 Obviously 119903

119894119895= 0 for 119894 lt 119895 In the

case 119894 = 119895 we obtain

119903119894119894= ℎ119894119894ℎ1015840

119894119894= (L1minusL119899+1) sdot

1

L1minusL119899+1

= 1 (42)

For 119894 ge 119895 + 1 we obtain

119903119894119895=

119899minus2

sum

119896=1

ℎ119894119896ℎ1015840

119896119895= ℎ119894119894minus1

ℎ1015840

119894minus1119895+ ℎ119894119894ℎ1015840

119894119895

= (2 minusL119899) sdot

(L119899minus 2)119894minus119895minus1

(L1minusL119899+1)119894minus119895

+ (L1minusL119899+1) sdot

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

= 0

(43)

We verify HHminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyHminus1H = 119868

119899minus2 Thus

the proof is completed

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

where 119892 is a nonnegative integer and each of the subscripts isunderstood to be reduced modulo 119899

The first row of 119860119892119899

is (1198861 1198862 119886

119899) its (119895 + 1)th row

is obtained by giving its 119895th row a right circular shift by 119892positions (equivalently 119892 mod 119899 positions) Note that 119892 = 1or 119892 = 119899 + 1 yields the standard circulant matrix If 119892 = 119899 minus 1then we obtain the left circulant matrix

Lemma4 (see [21]) Let119860 = Circ(1198861 1198862 119886

119899) be a circulant

matrix then one has

(i) 119860 is invertible if and only if 119891(120596119896) = 0 (119896 = 0 1

2 119899 minus 1) where 119891(119909) = sum119899

119895=1119886119895119909119895minus1 and 120596 =

exp(2120587119894119899)

(ii) If119860 is invertible then the inverse119860minus1 of119860 is a circulantmatrix

Lemma 5 Define

Δ =((

(

1 0 0 sdot sdot sdot 0 0

0 0 0 sdot sdot sdot 0 1

0 0 0 sdot sdot sdot 1 0

d

0 0 1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

))

)

(9)

the matrix Δ is an orthogonal cyclic shift matrix (and aleft circulant matrix) It holds that LCirc(119886

1 1198862 119886

119899) =

ΔCirc(1198861 1198862 119886

119899)

Lemma6 (see [27]) The 119899times119899matrixQ119892is unitary if and only

if (119899 119892) = 1 whereQ119892is a 119892-circulant matrix with the first row

119890lowast

= [1 0 0]

Lemma 7 (see [27]) 119860119892119899

is a 119892-circulant matrix with the firstrow [119886

1 1198862 119886

119899] if and only if 119860

119892119899= Q119892119862 where 119862 =

Circ(1198861 1198862 119886

119899)

2 Determinant and Inverse of a CirculantMatrix with the Product of the Fibonacciand Lucas Numbers

In this section letA119899= Circ(F

1F2 F

119899) be a circulant

matrix Firstly we give the determinant equation of matrixA119899 Afterwards we prove that A

119899is an invertible matrix for

119899 gt 2 and then we find the inverse of the matrixA119899

Theorem 8 Let A119899= Circ(F

1F2 F

119899) be a circulant

matrix then one has

detA119899= (1 minusF

119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(10)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Proof Obviously detA1= 1 satisfies (10) In the case 119899 gt 1

let

Γ =

((((((

(

1

minus3 1

1 1 minus3

0 0 1 minus3 1

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 1

))))))

)119899times119899

Π1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (minusF119899

F1minusF119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (minusF119899

F1minusF119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0minusF119899

F1minusF119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(11)

We can obtain

ΓA119899Π1

=((

(

F11198911015840

119899F119899minus1

F119899minus2

sdot sdot sdot F2

0 119891119899

minusF119899minus2

minusF119899minus3

sdot sdot sdot minusF1

0 0 F1minusF119899+1

0 0 F119899

F1minusF119899+1

F

119899

0 0 0 F1minusF119899+1

))

)

(12)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(13)

We obtain

det Γ detA119899detΠ1

= F1[F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

]

times (F1minusF119899+1)119899minus2

= F1[F1minusF119899+1

+

119899minus1

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

]

4 Abstract and Applied Analysis

times (F1minusF119899+1)119899minus2

= (1 minusF119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(14)

while

det Γ = detΠ1= (minus1)

(119899minus1)(119899minus2)2

(15)

we have

detA119899= (1 minusF

119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(16)

Thus the proof is completed

Theorem 9 Let A119899= Circ(F

1F2 F

119899) be a circulant

matrix if 119899 gt 2 thenA119899is an invertible matrix

Proof When 119899 = 3 in Theorem 8 we have detA3= 468 = 0

henceA3is invertible In the case 119899 gt 3 sinceF

119899= (120572119899

1minus120573119899

1)

(1205721minus 1205731) where 120572

1+ 1205731= 3 120572

1sdot 1205731= 1We have

119891 (120596119896

) =

119899

sum

119895=1

F119895(120596119896

)119895minus1

=1

1205721minus 1205731

119899

sum

119895=1

(120572119895

1minus 120573119895

1) (120596119896

)119895minus1

=1

1205721minus 1205731

[1205721(1 minus 120572

119899

1)

1 minus 1205721120596119896

minus1205731(1 minus 120573

119899

1)

1 minus 1205731120596119896

]

=1

1205721minus 1205731

[(1205721minus 1205731) minus (120572

119899+1

1minus 120573119899+1

1)

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

+1

1205721minus 1205731

[12057211205731(120572119899

1minus 120573119899

1) 120596119896

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

=1 minusF

119899+1+F119899120596119896

1 minus 3120596119896 + 1205962119896(119896 = 1 2 119899 minus 1)

(17)

If there exists 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 1minusF

119899+1+F119899120596119897

= 0 for 1minus3120596119897+1205962119897 = 0 thus 120596119897 =(F119899+1

minus 1)F119899is a real number While 120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899) hence sin(2119897120587119899) = 0 so we have120596119897

= minus1 for 0 lt 2119897120587119899 lt 2120587 But 119909 = minus1 is not the root ofequation 1 minusF

119899+1+F119899119909 = 0 (119899 gt 3) We obtain 119891(120596119896) = 0

for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) = sum119899

119895=1F119895=

F119899+1

minusF119899minus 1 = 0 By Lemma 4 the proof is completed

Lemma 10 Let the matrixG = [119892119894119895]119899minus2

119894119895=1

be of the form

119892119894119895=

F1minusF119899+1 119894 = 119895

F119899 119894 = 119895 + 1

0 otherwise(18)

and then the inverse Gminus1 = [1198921015840119894119895]119899minus2

119894119895=1

of the matrix G is equalto

1198921015840

119894119895=

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(19)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Obviously 119888

119894119895= 0 for 119894 lt 119895 In

the case 119894 = 119895 we obtain 119888119894119894= 1198921198941198941198921015840

119894119894= (F1minusF119899+1) sdot (1(F

1minus

F119899+1)) = 1 For 119894 ge 119895 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 119892119894119894minus1

1198921015840

119894minus1119895+ 1198921198941198941198921015840

119894119895

= F119899sdot

(minusF119899)119894minus119895minus1

(F1minusF119899+1)119894minus119895

+ (F1minusF119899+1) sdot

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

= 0

(20)

We verify GGminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verify Gminus1G = 119868

119899minus2 Thus

the proof is completed

Theorem 11 Let A119899= Circ(F

1F2 F

119899) (119899 gt 2) be a

circulant matrix then one has

Aminus1

119899=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(21)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(22)

Abstract and Applied Analysis 5

Proof Let

Π2=

(((((

(

1 minus1198911015840

119899minus1198911015840

119899

119891119899

F119899minus2

minusF119899minus1

sdot sdot sdot minus1198911015840

119899

119891119899

F1

0 1F119899minus2

119891119899

sdot sdot sdotF1

119891119899

0 0 1 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 1

)))))

)

(23)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(24)

We have

ΓA119899Π1Π2= D1oplusG (25)

whereD1= diag(F

1 119891119899) is a diagonal matrix andD

1oplusG is

the direct sum ofD1andG If we denoteΠ = Π

1Π2 then we

obtain

Aminus1

119899= Π (D

minus1

1oplusGminus1

) Γ (26)

Since the last row elements of the matrix Π are

0 1F119899minus2

119891119899

F119899minus3

119891119899

F2

119891119899

F1

119891119899

(27)

By Lemma 10 if we letAminus1119899= Circ(119909

1 1199092 119909

119899) its last

row elements are given by the following equations

1199092= minus

3

119891119899

+1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199093=

F1

119891119899(F1minusF119899+1)

1199094=1

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus

3F1

119891119899(F1minusF119899+1)

1199095=1

119891119899

3

sum

119894=1

F4minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

+F1

119891119899(F1minusF119899+1)

119909119899=1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus4

sum

119894=1

F119899minus3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199091=1

119891119899

minus3

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

(28)

Let 119862(119895)119899

= sum119895

119894=1(F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

) (119895 = 1

2 119899 minus 2) we have

119862(2)

119899minus 3119862(1)

119899

= minus3F1

F1minusF119899+1

+

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=minusF119899

(F1minusF119899+1)2

minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

= minus3

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894+

119899minus3

sum

119894=1

F119899minus119894minus2

(minusF119899)119894minus1

(F1minusF119899+1)119894

=(minus3F

1) (minusF

119899)119899minus3

(F1minusF119899+1)119899minus2

+

119899minus3

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=

119899minus2

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

119862(119895+2)

119899minus 3119862(119895+1)

119899+ 119862(119895)

119899

=

119895+2

sum

119894=1

F119895+3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus 3

119895+1

sum

119894=1

F119895+2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+

119895

sum

119894=1

F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

=F2(minusF119899)119895

(F1minusF119899+1)119895+1

+F1(minusF119899)119895+1

(F1minusF119899+1)119895+2

minus3F1(minusF119899)119895

(F1minusF119899+1)119895+1

6 Abstract and Applied Analysis

+

119895

sum

119894=1

(F119895+3minus119894

minus 3F119895+2minus119894

+F119895+1minus119894

) (minusF119899)119894minus1

(F1minusF119899+1)119895+1

=(minusF119899)119895+1

(F1minusF119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(29)

We obtain

Aminus1

119899= Circ(

1 minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

119891119899

119862(119899minus2)

119899minus 3

119891119899

119862(1)

119899

119891119899

119862(2)

119899minus 3119862(1)

119899

119891119899

119862(3)

119899minus 3119862(2)

119899+ 119862(1)

119899

119891119899

119862(119899minus2)

119899minus 3119862(119899minus3)

119899+ 119862(119899minus4)

119899

119891119899

)

=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(30)

3 Determinant and Inverse of a CirculantMatrix with the Sum of the Fibonacci andLucas Numbers

In this section let B119899= Circ(L

1L2 L

119899) be a circulant

matrix Firstly we give an explicit determinant formula ofmatrix B

119899 Afterwards we prove that B

119899is an invertible

matrix for any positive integer 119899 and then we find its inverse

Theorem 12 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

detB119899= 2[(2 minusL

119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(31)

whereL119899is the 119899th 119865

119899+ 119871119899number

Proof ObviouslyB1= 2 satisfies (31) when 119899 gt 1 Let

Σ =

((((((

(

1

minus2 1

minus1 1 minus1

0 0 1 minus1 minus1

c c c0 1 c c0 1 minus1 c 0

0 1 minus1 minus1

))))))

)119899times119899

Ω1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (L119899minus 2

L1minusL119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (L119899minus 2

L1minusL119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0L119899minus 2

L1minusL119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(32)

Then

ΣB119899Ω1

=((

(

L11198971015840

119899L119899minus1

sdot sdot sdot L3

L2

0 119897119899minus2L119899minus1

+L119899sdot sdot sdot sdot sdot sdot minus2L

2+L3

0 0 L1minusL119899+1

0 0 2 minusL119899

d

0 0 2 minusL119899

L1minusL119899+1

))

)

(33)

where

119897119899=L1minus 2L119899+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(34)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(35)

We can obtain

detΣ detB119899detΩ1

=L1[L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

Abstract and Applied Analysis 7

=L1[L1minusL119899+1

+

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(36)

while

detΣ = detΩ1= (minus1)

(119899minus1)(119899minus2)2

(37)

We have

detB119899

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(38)

Theorem 13 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix thenB119899is invertible for any positive integer 119899

Proof SinceL119899= (120572119899

minus120573119899

)(120572minus120573)+ 120572119899

+ 120573119899 where120572+120573 = 1

120572 sdot 120573 = minus1 We have

119891 (120596119896

) =

119899

sum

119895=1

L119895(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573+ 120572119895

+ 120573119895

)(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573) (120596119896

)119895minus1

+

119899

sum

119895=1

(120572119895

+ 120573119895

) (120596119896

)119895minus1

=1

120572 minus 120573[120572 (1 minus 120572

119899

)

1 minus 120572120596119896minus120573 (1 minus 120573

119899

)

1 minus 120573120596119896]

+120572 (1 minus 120572

119899

)

1 minus 120572120596119896+120573 (1 minus 120573

119899

)

1 minus 120573120596119896

=1 minus 119865119899+1

minus 119865119899120596119896

1 minus 120596119896 minus 1205962119896+1 minus 119871119899+1

+ (2 minus 119871119899) 120596119896

1 minus 120596119896 minus 1205962119896

=2 minus (119865

119899+1+ 119871119899+1) minus (119865

119899+ 119871119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896

=2 minusL

119899+1minus (L119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896(119896 = 1 2 119899 minus 1)

(39)

If there exist 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 2 minusL

119899+1minus (L119899minus 2)120596119897

= 0 for 1 minus120596119897 minus1205962119897 = 0 120596119897 =(2minusL

119899+1)(L119899minus2) is a real number while120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899)Hence sin(2119897120587119899) = 0 so we have 120596119897 = minus1 for 0 lt

2119897120587119899 lt 2120587 But 119909 = minus1 is not the root of the equation2 minus L

119899+1minus (L119899minus 2)119909 = 0 for any positive integer 119899 We

obtain 119891(120596119896) = 0 for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) =sum119899

119895=1L119895= L119899+1

+ L119899minus 4 = 0 By Lemma 4 the proof is

completed

Lemma 14 Let matrixH = [ℎ119894119895]119899minus2

119894119895=1

be of the form

ℎ119894119895=

L1minusL119899+1 119894 = 119895

2 minusL119899 119894 = 119895 + 1

0 otherwise(40)

and then inverseHminus1 = [ℎ1015840119894119895]119899minus2

119894119895=1

of the matrixH is equal to

ℎ1015840

119894119895=

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(41)

Proof Let 119903119894119895= sum119899minus2

119896=1ℎ119894119896ℎ1015840

119896119895 Obviously 119903

119894119895= 0 for 119894 lt 119895 In the

case 119894 = 119895 we obtain

119903119894119894= ℎ119894119894ℎ1015840

119894119894= (L1minusL119899+1) sdot

1

L1minusL119899+1

= 1 (42)

For 119894 ge 119895 + 1 we obtain

119903119894119895=

119899minus2

sum

119896=1

ℎ119894119896ℎ1015840

119896119895= ℎ119894119894minus1

ℎ1015840

119894minus1119895+ ℎ119894119894ℎ1015840

119894119895

= (2 minusL119899) sdot

(L119899minus 2)119894minus119895minus1

(L1minusL119899+1)119894minus119895

+ (L1minusL119899+1) sdot

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

= 0

(43)

We verify HHminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyHminus1H = 119868

119899minus2 Thus

the proof is completed

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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OptimizationJournal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

times (F1minusF119899+1)119899minus2

= (1 minusF119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(14)

while

det Γ = detΠ1= (minus1)

(119899minus1)(119899minus2)2

(15)

we have

detA119899= (1 minusF

119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 minusF119899+1

minusF119899

)

119896minus1

(16)

Thus the proof is completed

Theorem 9 Let A119899= Circ(F

1F2 F

119899) be a circulant

matrix if 119899 gt 2 thenA119899is an invertible matrix

Proof When 119899 = 3 in Theorem 8 we have detA3= 468 = 0

henceA3is invertible In the case 119899 gt 3 sinceF

119899= (120572119899

1minus120573119899

1)

(1205721minus 1205731) where 120572

1+ 1205731= 3 120572

1sdot 1205731= 1We have

119891 (120596119896

) =

119899

sum

119895=1

F119895(120596119896

)119895minus1

=1

1205721minus 1205731

119899

sum

119895=1

(120572119895

1minus 120573119895

1) (120596119896

)119895minus1

=1

1205721minus 1205731

[1205721(1 minus 120572

119899

1)

1 minus 1205721120596119896

minus1205731(1 minus 120573

119899

1)

1 minus 1205731120596119896

]

=1

1205721minus 1205731

[(1205721minus 1205731) minus (120572

119899+1

1minus 120573119899+1

1)

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

+1

1205721minus 1205731

[12057211205731(120572119899

1minus 120573119899

1) 120596119896

1 minus (1205721+ 1205731) 120596119896 + 120572

112057311205962119896

]

=1 minusF

119899+1+F119899120596119896

1 minus 3120596119896 + 1205962119896(119896 = 1 2 119899 minus 1)

(17)

If there exists 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 1minusF

119899+1+F119899120596119897

= 0 for 1minus3120596119897+1205962119897 = 0 thus 120596119897 =(F119899+1

minus 1)F119899is a real number While 120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899) hence sin(2119897120587119899) = 0 so we have120596119897

= minus1 for 0 lt 2119897120587119899 lt 2120587 But 119909 = minus1 is not the root ofequation 1 minusF

119899+1+F119899119909 = 0 (119899 gt 3) We obtain 119891(120596119896) = 0

for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) = sum119899

119895=1F119895=

F119899+1

minusF119899minus 1 = 0 By Lemma 4 the proof is completed

Lemma 10 Let the matrixG = [119892119894119895]119899minus2

119894119895=1

be of the form

119892119894119895=

F1minusF119899+1 119894 = 119895

F119899 119894 = 119895 + 1

0 otherwise(18)

and then the inverse Gminus1 = [1198921015840119894119895]119899minus2

119894119895=1

of the matrix G is equalto

1198921015840

119894119895=

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(19)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Obviously 119888

119894119895= 0 for 119894 lt 119895 In

the case 119894 = 119895 we obtain 119888119894119894= 1198921198941198941198921015840

119894119894= (F1minusF119899+1) sdot (1(F

1minus

F119899+1)) = 1 For 119894 ge 119895 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 119892119894119894minus1

1198921015840

119894minus1119895+ 1198921198941198941198921015840

119894119895

= F119899sdot

(minusF119899)119894minus119895minus1

(F1minusF119899+1)119894minus119895

+ (F1minusF119899+1) sdot

(minusF119899)119894minus119895

(F1minusF119899+1)119894minus119895+1

= 0

(20)

We verify GGminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verify Gminus1G = 119868

119899minus2 Thus

the proof is completed

Theorem 11 Let A119899= Circ(F

1F2 F

119899) (119899 gt 2) be a

circulant matrix then one has

Aminus1

119899=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(21)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(22)

Abstract and Applied Analysis 5

Proof Let

Π2=

(((((

(

1 minus1198911015840

119899minus1198911015840

119899

119891119899

F119899minus2

minusF119899minus1

sdot sdot sdot minus1198911015840

119899

119891119899

F1

0 1F119899minus2

119891119899

sdot sdot sdotF1

119891119899

0 0 1 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 1

)))))

)

(23)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(24)

We have

ΓA119899Π1Π2= D1oplusG (25)

whereD1= diag(F

1 119891119899) is a diagonal matrix andD

1oplusG is

the direct sum ofD1andG If we denoteΠ = Π

1Π2 then we

obtain

Aminus1

119899= Π (D

minus1

1oplusGminus1

) Γ (26)

Since the last row elements of the matrix Π are

0 1F119899minus2

119891119899

F119899minus3

119891119899

F2

119891119899

F1

119891119899

(27)

By Lemma 10 if we letAminus1119899= Circ(119909

1 1199092 119909

119899) its last

row elements are given by the following equations

1199092= minus

3

119891119899

+1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199093=

F1

119891119899(F1minusF119899+1)

1199094=1

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus

3F1

119891119899(F1minusF119899+1)

1199095=1

119891119899

3

sum

119894=1

F4minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

+F1

119891119899(F1minusF119899+1)

119909119899=1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus4

sum

119894=1

F119899minus3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199091=1

119891119899

minus3

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

(28)

Let 119862(119895)119899

= sum119895

119894=1(F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

) (119895 = 1

2 119899 minus 2) we have

119862(2)

119899minus 3119862(1)

119899

= minus3F1

F1minusF119899+1

+

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=minusF119899

(F1minusF119899+1)2

minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

= minus3

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894+

119899minus3

sum

119894=1

F119899minus119894minus2

(minusF119899)119894minus1

(F1minusF119899+1)119894

=(minus3F

1) (minusF

119899)119899minus3

(F1minusF119899+1)119899minus2

+

119899minus3

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=

119899minus2

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

119862(119895+2)

119899minus 3119862(119895+1)

119899+ 119862(119895)

119899

=

119895+2

sum

119894=1

F119895+3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus 3

119895+1

sum

119894=1

F119895+2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+

119895

sum

119894=1

F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

=F2(minusF119899)119895

(F1minusF119899+1)119895+1

+F1(minusF119899)119895+1

(F1minusF119899+1)119895+2

minus3F1(minusF119899)119895

(F1minusF119899+1)119895+1

6 Abstract and Applied Analysis

+

119895

sum

119894=1

(F119895+3minus119894

minus 3F119895+2minus119894

+F119895+1minus119894

) (minusF119899)119894minus1

(F1minusF119899+1)119895+1

=(minusF119899)119895+1

(F1minusF119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(29)

We obtain

Aminus1

119899= Circ(

1 minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

119891119899

119862(119899minus2)

119899minus 3

119891119899

119862(1)

119899

119891119899

119862(2)

119899minus 3119862(1)

119899

119891119899

119862(3)

119899minus 3119862(2)

119899+ 119862(1)

119899

119891119899

119862(119899minus2)

119899minus 3119862(119899minus3)

119899+ 119862(119899minus4)

119899

119891119899

)

=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(30)

3 Determinant and Inverse of a CirculantMatrix with the Sum of the Fibonacci andLucas Numbers

In this section let B119899= Circ(L

1L2 L

119899) be a circulant

matrix Firstly we give an explicit determinant formula ofmatrix B

119899 Afterwards we prove that B

119899is an invertible

matrix for any positive integer 119899 and then we find its inverse

Theorem 12 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

detB119899= 2[(2 minusL

119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(31)

whereL119899is the 119899th 119865

119899+ 119871119899number

Proof ObviouslyB1= 2 satisfies (31) when 119899 gt 1 Let

Σ =

((((((

(

1

minus2 1

minus1 1 minus1

0 0 1 minus1 minus1

c c c0 1 c c0 1 minus1 c 0

0 1 minus1 minus1

))))))

)119899times119899

Ω1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (L119899minus 2

L1minusL119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (L119899minus 2

L1minusL119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0L119899minus 2

L1minusL119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(32)

Then

ΣB119899Ω1

=((

(

L11198971015840

119899L119899minus1

sdot sdot sdot L3

L2

0 119897119899minus2L119899minus1

+L119899sdot sdot sdot sdot sdot sdot minus2L

2+L3

0 0 L1minusL119899+1

0 0 2 minusL119899

d

0 0 2 minusL119899

L1minusL119899+1

))

)

(33)

where

119897119899=L1minus 2L119899+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(34)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(35)

We can obtain

detΣ detB119899detΩ1

=L1[L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

Abstract and Applied Analysis 7

=L1[L1minusL119899+1

+

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(36)

while

detΣ = detΩ1= (minus1)

(119899minus1)(119899minus2)2

(37)

We have

detB119899

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(38)

Theorem 13 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix thenB119899is invertible for any positive integer 119899

Proof SinceL119899= (120572119899

minus120573119899

)(120572minus120573)+ 120572119899

+ 120573119899 where120572+120573 = 1

120572 sdot 120573 = minus1 We have

119891 (120596119896

) =

119899

sum

119895=1

L119895(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573+ 120572119895

+ 120573119895

)(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573) (120596119896

)119895minus1

+

119899

sum

119895=1

(120572119895

+ 120573119895

) (120596119896

)119895minus1

=1

120572 minus 120573[120572 (1 minus 120572

119899

)

1 minus 120572120596119896minus120573 (1 minus 120573

119899

)

1 minus 120573120596119896]

+120572 (1 minus 120572

119899

)

1 minus 120572120596119896+120573 (1 minus 120573

119899

)

1 minus 120573120596119896

=1 minus 119865119899+1

minus 119865119899120596119896

1 minus 120596119896 minus 1205962119896+1 minus 119871119899+1

+ (2 minus 119871119899) 120596119896

1 minus 120596119896 minus 1205962119896

=2 minus (119865

119899+1+ 119871119899+1) minus (119865

119899+ 119871119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896

=2 minusL

119899+1minus (L119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896(119896 = 1 2 119899 minus 1)

(39)

If there exist 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 2 minusL

119899+1minus (L119899minus 2)120596119897

= 0 for 1 minus120596119897 minus1205962119897 = 0 120596119897 =(2minusL

119899+1)(L119899minus2) is a real number while120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899)Hence sin(2119897120587119899) = 0 so we have 120596119897 = minus1 for 0 lt

2119897120587119899 lt 2120587 But 119909 = minus1 is not the root of the equation2 minus L

119899+1minus (L119899minus 2)119909 = 0 for any positive integer 119899 We

obtain 119891(120596119896) = 0 for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) =sum119899

119895=1L119895= L119899+1

+ L119899minus 4 = 0 By Lemma 4 the proof is

completed

Lemma 14 Let matrixH = [ℎ119894119895]119899minus2

119894119895=1

be of the form

ℎ119894119895=

L1minusL119899+1 119894 = 119895

2 minusL119899 119894 = 119895 + 1

0 otherwise(40)

and then inverseHminus1 = [ℎ1015840119894119895]119899minus2

119894119895=1

of the matrixH is equal to

ℎ1015840

119894119895=

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(41)

Proof Let 119903119894119895= sum119899minus2

119896=1ℎ119894119896ℎ1015840

119896119895 Obviously 119903

119894119895= 0 for 119894 lt 119895 In the

case 119894 = 119895 we obtain

119903119894119894= ℎ119894119894ℎ1015840

119894119894= (L1minusL119899+1) sdot

1

L1minusL119899+1

= 1 (42)

For 119894 ge 119895 + 1 we obtain

119903119894119895=

119899minus2

sum

119896=1

ℎ119894119896ℎ1015840

119896119895= ℎ119894119894minus1

ℎ1015840

119894minus1119895+ ℎ119894119894ℎ1015840

119894119895

= (2 minusL119899) sdot

(L119899minus 2)119894minus119895minus1

(L1minusL119899+1)119894minus119895

+ (L1minusL119899+1) sdot

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

= 0

(43)

We verify HHminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyHminus1H = 119868

119899minus2 Thus

the proof is completed

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Proof Let

Π2=

(((((

(

1 minus1198911015840

119899minus1198911015840

119899

119891119899

F119899minus2

minusF119899minus1

sdot sdot sdot minus1198911015840

119899

119891119899

F1

0 1F119899minus2

119891119899

sdot sdot sdotF1

119891119899

0 0 1 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 1

)))))

)

(23)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

1198911015840

119899=

119899minus1

sum

119896=1

F119896+1(

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(24)

We have

ΓA119899Π1Π2= D1oplusG (25)

whereD1= diag(F

1 119891119899) is a diagonal matrix andD

1oplusG is

the direct sum ofD1andG If we denoteΠ = Π

1Π2 then we

obtain

Aminus1

119899= Π (D

minus1

1oplusGminus1

) Γ (26)

Since the last row elements of the matrix Π are

0 1F119899minus2

119891119899

F119899minus3

119891119899

F2

119891119899

F1

119891119899

(27)

By Lemma 10 if we letAminus1119899= Circ(119909

1 1199092 119909

119899) its last

row elements are given by the following equations

1199092= minus

3

119891119899

+1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199093=

F1

119891119899(F1minusF119899+1)

1199094=1

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus

3F1

119891119899(F1minusF119899+1)

1199095=1

119891119899

3

sum

119894=1

F4minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

+F1

119891119899(F1minusF119899+1)

119909119899=1

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus3

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus4

sum

119894=1

F119899minus3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1199091=1

119891119899

minus3

119891119899

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+1

119891119899

119899minus3

sum

119894=1

F119899minus2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

(28)

Let 119862(119895)119899

= sum119895

119894=1(F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

) (119895 = 1

2 119899 minus 2) we have

119862(2)

119899minus 3119862(1)

119899

= minus3F1

F1minusF119899+1

+

2

sum

119894=1

F3minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=minusF119899

(F1minusF119899+1)2

minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

= minus3

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894+

119899minus3

sum

119894=1

F119899minus119894minus2

(minusF119899)119894minus1

(F1minusF119899+1)119894

=(minus3F

1) (minusF

119899)119899minus3

(F1minusF119899+1)119899minus2

+

119899minus3

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

=

119899minus2

sum

119894=1

minusF119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

119862(119895+2)

119899minus 3119862(119895+1)

119899+ 119862(119895)

119899

=

119895+2

sum

119894=1

F119895+3minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894minus 3

119895+1

sum

119894=1

F119895+2minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

+

119895

sum

119894=1

F119895+1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

=F2(minusF119899)119895

(F1minusF119899+1)119895+1

+F1(minusF119899)119895+1

(F1minusF119899+1)119895+2

minus3F1(minusF119899)119895

(F1minusF119899+1)119895+1

6 Abstract and Applied Analysis

+

119895

sum

119894=1

(F119895+3minus119894

minus 3F119895+2minus119894

+F119895+1minus119894

) (minusF119899)119894minus1

(F1minusF119899+1)119895+1

=(minusF119899)119895+1

(F1minusF119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(29)

We obtain

Aminus1

119899= Circ(

1 minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

119891119899

119862(119899minus2)

119899minus 3

119891119899

119862(1)

119899

119891119899

119862(2)

119899minus 3119862(1)

119899

119891119899

119862(3)

119899minus 3119862(2)

119899+ 119862(1)

119899

119891119899

119862(119899minus2)

119899minus 3119862(119899minus3)

119899+ 119862(119899minus4)

119899

119891119899

)

=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(30)

3 Determinant and Inverse of a CirculantMatrix with the Sum of the Fibonacci andLucas Numbers

In this section let B119899= Circ(L

1L2 L

119899) be a circulant

matrix Firstly we give an explicit determinant formula ofmatrix B

119899 Afterwards we prove that B

119899is an invertible

matrix for any positive integer 119899 and then we find its inverse

Theorem 12 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

detB119899= 2[(2 minusL

119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(31)

whereL119899is the 119899th 119865

119899+ 119871119899number

Proof ObviouslyB1= 2 satisfies (31) when 119899 gt 1 Let

Σ =

((((((

(

1

minus2 1

minus1 1 minus1

0 0 1 minus1 minus1

c c c0 1 c c0 1 minus1 c 0

0 1 minus1 minus1

))))))

)119899times119899

Ω1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (L119899minus 2

L1minusL119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (L119899minus 2

L1minusL119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0L119899minus 2

L1minusL119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(32)

Then

ΣB119899Ω1

=((

(

L11198971015840

119899L119899minus1

sdot sdot sdot L3

L2

0 119897119899minus2L119899minus1

+L119899sdot sdot sdot sdot sdot sdot minus2L

2+L3

0 0 L1minusL119899+1

0 0 2 minusL119899

d

0 0 2 minusL119899

L1minusL119899+1

))

)

(33)

where

119897119899=L1minus 2L119899+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(34)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(35)

We can obtain

detΣ detB119899detΩ1

=L1[L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

Abstract and Applied Analysis 7

=L1[L1minusL119899+1

+

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(36)

while

detΣ = detΩ1= (minus1)

(119899minus1)(119899minus2)2

(37)

We have

detB119899

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(38)

Theorem 13 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix thenB119899is invertible for any positive integer 119899

Proof SinceL119899= (120572119899

minus120573119899

)(120572minus120573)+ 120572119899

+ 120573119899 where120572+120573 = 1

120572 sdot 120573 = minus1 We have

119891 (120596119896

) =

119899

sum

119895=1

L119895(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573+ 120572119895

+ 120573119895

)(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573) (120596119896

)119895minus1

+

119899

sum

119895=1

(120572119895

+ 120573119895

) (120596119896

)119895minus1

=1

120572 minus 120573[120572 (1 minus 120572

119899

)

1 minus 120572120596119896minus120573 (1 minus 120573

119899

)

1 minus 120573120596119896]

+120572 (1 minus 120572

119899

)

1 minus 120572120596119896+120573 (1 minus 120573

119899

)

1 minus 120573120596119896

=1 minus 119865119899+1

minus 119865119899120596119896

1 minus 120596119896 minus 1205962119896+1 minus 119871119899+1

+ (2 minus 119871119899) 120596119896

1 minus 120596119896 minus 1205962119896

=2 minus (119865

119899+1+ 119871119899+1) minus (119865

119899+ 119871119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896

=2 minusL

119899+1minus (L119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896(119896 = 1 2 119899 minus 1)

(39)

If there exist 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 2 minusL

119899+1minus (L119899minus 2)120596119897

= 0 for 1 minus120596119897 minus1205962119897 = 0 120596119897 =(2minusL

119899+1)(L119899minus2) is a real number while120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899)Hence sin(2119897120587119899) = 0 so we have 120596119897 = minus1 for 0 lt

2119897120587119899 lt 2120587 But 119909 = minus1 is not the root of the equation2 minus L

119899+1minus (L119899minus 2)119909 = 0 for any positive integer 119899 We

obtain 119891(120596119896) = 0 for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) =sum119899

119895=1L119895= L119899+1

+ L119899minus 4 = 0 By Lemma 4 the proof is

completed

Lemma 14 Let matrixH = [ℎ119894119895]119899minus2

119894119895=1

be of the form

ℎ119894119895=

L1minusL119899+1 119894 = 119895

2 minusL119899 119894 = 119895 + 1

0 otherwise(40)

and then inverseHminus1 = [ℎ1015840119894119895]119899minus2

119894119895=1

of the matrixH is equal to

ℎ1015840

119894119895=

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(41)

Proof Let 119903119894119895= sum119899minus2

119896=1ℎ119894119896ℎ1015840

119896119895 Obviously 119903

119894119895= 0 for 119894 lt 119895 In the

case 119894 = 119895 we obtain

119903119894119894= ℎ119894119894ℎ1015840

119894119894= (L1minusL119899+1) sdot

1

L1minusL119899+1

= 1 (42)

For 119894 ge 119895 + 1 we obtain

119903119894119895=

119899minus2

sum

119896=1

ℎ119894119896ℎ1015840

119896119895= ℎ119894119894minus1

ℎ1015840

119894minus1119895+ ℎ119894119894ℎ1015840

119894119895

= (2 minusL119899) sdot

(L119899minus 2)119894minus119895minus1

(L1minusL119899+1)119894minus119895

+ (L1minusL119899+1) sdot

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

= 0

(43)

We verify HHminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyHminus1H = 119868

119899minus2 Thus

the proof is completed

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

+

119895

sum

119894=1

(F119895+3minus119894

minus 3F119895+2minus119894

+F119895+1minus119894

) (minusF119899)119894minus1

(F1minusF119899+1)119895+1

=(minusF119899)119895+1

(F1minusF119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(29)

We obtain

Aminus1

119899= Circ(

1 minus 3119862(119899minus2)

119899+ 119862(119899minus3)

119899

119891119899

119862(119899minus2)

119899minus 3

119891119899

119862(1)

119899

119891119899

119862(2)

119899minus 3119862(1)

119899

119891119899

119862(3)

119899minus 3119862(2)

119899+ 119862(1)

119899

119891119899

119862(119899minus2)

119899minus 3119862(119899minus3)

119899+ 119862(119899minus4)

119899

119891119899

)

=1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)

(30)

3 Determinant and Inverse of a CirculantMatrix with the Sum of the Fibonacci andLucas Numbers

In this section let B119899= Circ(L

1L2 L

119899) be a circulant

matrix Firstly we give an explicit determinant formula ofmatrix B

119899 Afterwards we prove that B

119899is an invertible

matrix for any positive integer 119899 and then we find its inverse

Theorem 12 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

detB119899= 2[(2 minusL

119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(31)

whereL119899is the 119899th 119865

119899+ 119871119899number

Proof ObviouslyB1= 2 satisfies (31) when 119899 gt 1 Let

Σ =

((((((

(

1

minus2 1

minus1 1 minus1

0 0 1 minus1 minus1

c c c0 1 c c0 1 minus1 c 0

0 1 minus1 minus1

))))))

)119899times119899

Ω1=

(((((((((

(

1 0 0 sdot sdot sdot 0 0

0 (L119899minus 2

L1minusL119899+1

)

119899minus2

0 sdot sdot sdot 0 1

0 (L119899minus 2

L1minusL119899+1

)

119899minus3

0 sdot sdot sdot 1 0

d

0L119899minus 2

L1minusL119899+1

1 sdot sdot sdot 0 0

0 1 0 sdot sdot sdot 0 0

)))))))))

)119899times119899

(32)

Then

ΣB119899Ω1

=((

(

L11198971015840

119899L119899minus1

sdot sdot sdot L3

L2

0 119897119899minus2L119899minus1

+L119899sdot sdot sdot sdot sdot sdot minus2L

2+L3

0 0 L1minusL119899+1

0 0 2 minusL119899

d

0 0 2 minusL119899

L1minusL119899+1

))

)

(33)

where

119897119899=L1minus 2L119899+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(34)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(35)

We can obtain

detΣ detB119899detΩ1

=L1[L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

Abstract and Applied Analysis 7

=L1[L1minusL119899+1

+

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(36)

while

detΣ = detΩ1= (minus1)

(119899minus1)(119899minus2)2

(37)

We have

detB119899

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(38)

Theorem 13 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix thenB119899is invertible for any positive integer 119899

Proof SinceL119899= (120572119899

minus120573119899

)(120572minus120573)+ 120572119899

+ 120573119899 where120572+120573 = 1

120572 sdot 120573 = minus1 We have

119891 (120596119896

) =

119899

sum

119895=1

L119895(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573+ 120572119895

+ 120573119895

)(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573) (120596119896

)119895minus1

+

119899

sum

119895=1

(120572119895

+ 120573119895

) (120596119896

)119895minus1

=1

120572 minus 120573[120572 (1 minus 120572

119899

)

1 minus 120572120596119896minus120573 (1 minus 120573

119899

)

1 minus 120573120596119896]

+120572 (1 minus 120572

119899

)

1 minus 120572120596119896+120573 (1 minus 120573

119899

)

1 minus 120573120596119896

=1 minus 119865119899+1

minus 119865119899120596119896

1 minus 120596119896 minus 1205962119896+1 minus 119871119899+1

+ (2 minus 119871119899) 120596119896

1 minus 120596119896 minus 1205962119896

=2 minus (119865

119899+1+ 119871119899+1) minus (119865

119899+ 119871119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896

=2 minusL

119899+1minus (L119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896(119896 = 1 2 119899 minus 1)

(39)

If there exist 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 2 minusL

119899+1minus (L119899minus 2)120596119897

= 0 for 1 minus120596119897 minus1205962119897 = 0 120596119897 =(2minusL

119899+1)(L119899minus2) is a real number while120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899)Hence sin(2119897120587119899) = 0 so we have 120596119897 = minus1 for 0 lt

2119897120587119899 lt 2120587 But 119909 = minus1 is not the root of the equation2 minus L

119899+1minus (L119899minus 2)119909 = 0 for any positive integer 119899 We

obtain 119891(120596119896) = 0 for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) =sum119899

119895=1L119895= L119899+1

+ L119899minus 4 = 0 By Lemma 4 the proof is

completed

Lemma 14 Let matrixH = [ℎ119894119895]119899minus2

119894119895=1

be of the form

ℎ119894119895=

L1minusL119899+1 119894 = 119895

2 minusL119899 119894 = 119895 + 1

0 otherwise(40)

and then inverseHminus1 = [ℎ1015840119894119895]119899minus2

119894119895=1

of the matrixH is equal to

ℎ1015840

119894119895=

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(41)

Proof Let 119903119894119895= sum119899minus2

119896=1ℎ119894119896ℎ1015840

119896119895 Obviously 119903

119894119895= 0 for 119894 lt 119895 In the

case 119894 = 119895 we obtain

119903119894119894= ℎ119894119894ℎ1015840

119894119894= (L1minusL119899+1) sdot

1

L1minusL119899+1

= 1 (42)

For 119894 ge 119895 + 1 we obtain

119903119894119895=

119899minus2

sum

119896=1

ℎ119894119896ℎ1015840

119896119895= ℎ119894119894minus1

ℎ1015840

119894minus1119895+ ℎ119894119894ℎ1015840

119894119895

= (2 minusL119899) sdot

(L119899minus 2)119894minus119895minus1

(L1minusL119899+1)119894minus119895

+ (L1minusL119899+1) sdot

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

= 0

(43)

We verify HHminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyHminus1H = 119868

119899minus2 Thus

the proof is completed

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

=L1[L1minusL119899+1

+

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

]

times (L1minusL119899+1)119899minus2

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(36)

while

detΣ = detΩ1= (minus1)

(119899minus1)(119899minus2)2

(37)

We have

detB119899

= 2[(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(38)

Theorem 13 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix thenB119899is invertible for any positive integer 119899

Proof SinceL119899= (120572119899

minus120573119899

)(120572minus120573)+ 120572119899

+ 120573119899 where120572+120573 = 1

120572 sdot 120573 = minus1 We have

119891 (120596119896

) =

119899

sum

119895=1

L119895(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573+ 120572119895

+ 120573119895

)(120596119896

)119895minus1

=

119899

sum

119895=1

(120572119895

minus 120573119895

120572 minus 120573) (120596119896

)119895minus1

+

119899

sum

119895=1

(120572119895

+ 120573119895

) (120596119896

)119895minus1

=1

120572 minus 120573[120572 (1 minus 120572

119899

)

1 minus 120572120596119896minus120573 (1 minus 120573

119899

)

1 minus 120573120596119896]

+120572 (1 minus 120572

119899

)

1 minus 120572120596119896+120573 (1 minus 120573

119899

)

1 minus 120573120596119896

=1 minus 119865119899+1

minus 119865119899120596119896

1 minus 120596119896 minus 1205962119896+1 minus 119871119899+1

+ (2 minus 119871119899) 120596119896

1 minus 120596119896 minus 1205962119896

=2 minus (119865

119899+1+ 119871119899+1) minus (119865

119899+ 119871119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896

=2 minusL

119899+1minus (L119899minus 2) 120596

119896

1 minus 120596119896 minus 1205962119896(119896 = 1 2 119899 minus 1)

(39)

If there exist 120596119897 (119897 = 1 2 119899 minus 1) such that 119891(120596119897) = 0we obtain 2 minusL

119899+1minus (L119899minus 2)120596119897

= 0 for 1 minus120596119897 minus1205962119897 = 0 120596119897 =(2minusL

119899+1)(L119899minus2) is a real number while120596119897 = exp(2119897120587119894119899) =

cos(2119897120587119899) + 119894 sin(2119897120587119899)Hence sin(2119897120587119899) = 0 so we have 120596119897 = minus1 for 0 lt

2119897120587119899 lt 2120587 But 119909 = minus1 is not the root of the equation2 minus L

119899+1minus (L119899minus 2)119909 = 0 for any positive integer 119899 We

obtain 119891(120596119896) = 0 for any 120596119896 (119896 = 1 2 119899 minus 1) while 119891(1) =sum119899

119895=1L119895= L119899+1

+ L119899minus 4 = 0 By Lemma 4 the proof is

completed

Lemma 14 Let matrixH = [ℎ119894119895]119899minus2

119894119895=1

be of the form

ℎ119894119895=

L1minusL119899+1 119894 = 119895

2 minusL119899 119894 = 119895 + 1

0 otherwise(40)

and then inverseHminus1 = [ℎ1015840119894119895]119899minus2

119894119895=1

of the matrixH is equal to

ℎ1015840

119894119895=

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

119894 ge 119895

0 119894 lt 119895

(41)

Proof Let 119903119894119895= sum119899minus2

119896=1ℎ119894119896ℎ1015840

119896119895 Obviously 119903

119894119895= 0 for 119894 lt 119895 In the

case 119894 = 119895 we obtain

119903119894119894= ℎ119894119894ℎ1015840

119894119894= (L1minusL119899+1) sdot

1

L1minusL119899+1

= 1 (42)

For 119894 ge 119895 + 1 we obtain

119903119894119895=

119899minus2

sum

119896=1

ℎ119894119896ℎ1015840

119896119895= ℎ119894119894minus1

ℎ1015840

119894minus1119895+ ℎ119894119894ℎ1015840

119894119895

= (2 minusL119899) sdot

(L119899minus 2)119894minus119895minus1

(L1minusL119899+1)119894minus119895

+ (L1minusL119899+1) sdot

(L119899minus 2)119894minus119895

(L1minusL119899+1)119894minus119895+1

= 0

(43)

We verify HHminus1 = 119868119899minus2

where 119868119899minus2

is the (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyHminus1H = 119868

119899minus2 Thus

the proof is completed

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

Theorem 15 Let B119899= Circ(L

1L2 L

119899) be a circulant

matrix then one has

Bminus1

119899=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(44)

where119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(45)

Proof Let

Ω2=

(((((

(

1 minus1198971015840

119899

212059613

12059614

sdot sdot sdot 1205961119899

0 1 12059623

12059624

sdot sdot sdot 1205962119899

0 0 1 0 sdot sdot sdot 0

0 0 0 1 sdot sdot sdot 0

d

0 0 0 0 sdot sdot sdot 1

)))))

)

(46)

where

1205961119894=1

2[1198971015840

119899(L119899+3minus119894

minus 2L119899+2minus119894

)

119897119899

minusL119899+2minus119894

]

1205962119894=2L119899+2minus119894

minusL119899+3minus119894

119897119899

119894 = 3 4 119899

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

1198971015840

119899=

119899minus1

sum

119896=1

L119896+1(

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(47)

We have

ΣB119899Ω1Ω2= D2oplusH (48)

whereD2= diag(L

1 119897119899) is a diagonal matrix andD

2oplusH is

the direct sum of D2and H If we denote Ω = Ω

1Ω2 then

we obtain

Bminus1

119899= Ω(D

minus1

2oplusHminus1

) Σ (49)

Since the last row elements of the matrix Ω are

0 12L119899minus1

minusL119899

119897119899

2L119899minus2

minusL119899minus1

119897119899

2L2minusL3

119897119899

(50)

By Lemma 14 if we letBminus1119899= Circ(119910

1 1199102 119910

119899) then its

last row elements are given by the following equations

1199102= minus

2

119897119899

minus1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199103=1

119897119899

2L2minusL3

(L1minusL119899+1)

1199104= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

+1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199105= minus

1

119897119899

2L2minusL3

(L1minusL119899+1)

minus1

119897119899

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+1

119897119899

3

sum

119894=1

(2L5minus119894minusL6minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

119910119899=1

119897119899

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus1

119897119899

119899minus4

sum

119894=1

(2L119899minus119894minus2

minusL119899minus119894minus1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

1199101=1

119897119899

[1 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

]

(51)

Let 119863(119895)

119899= sum

119895

119894=1((2L119895+2minus119894

minus L119895+3minus119894

)(L119899minus 2)119894minus1

(L1minusL119899+1)119894

) (119895 = 1 2 119899 minus 2) we have

119863(2)

119899minus 119863(1)

119899

= minus2L2minusL3

L1minusL119899+1

+

2

sum

119894=1

(2L4minus119894minusL5minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

=2 (L119899minus 2)

(L1minusL119899+1)2

119863(119899minus3)

119899+ 119863(119899minus2)

119899

=

119899minus3

sum

119894=1

(2L119899minus119894minus1

minusL119899minus119894) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

+

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

+

119899minus3

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

=

119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

119863(119895+2)

119899minus 119863(119895+1)

119899minus 119863(119895)

119899

=

119895+2

sum

119894=1

(2L119895minus119894+4

minusL119895minus119894+5

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895+1

sum

119894=1

(2L119895minus119894+3

minusL119895minus119894+4

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus

119895

sum

119894=1

(2L119895minus119894+2

minusL119895minus119894+3

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

=(2L2minusL3) (L119899minus 2)119895+1

(L1minusL119899+1)119895+2

+(2L3minusL4) (L119899minus 2)119895

(L1minusL119899+1)119895+1

minus(2L2minusL3) (L119899minus 2)119895

(L1minusL119899+1)119895+1

=2(L119899minus 2)119895+1

(L1minusL119899+1)119895+2

(119895 = 1 2 119899 minus 4)

(52)

We obtain

Bminus1

119899

= Circ(1 minus 119863

(119899minus2)

119899minus 119863(119899minus3)

119899

119897119899

minus2 minus 119863

(119899minus2)

119899

119897119899

119863(1)

119899

119897119899

119863(2)

119899minus 119863(1)

119899

119897119899

119863(3)

119899minus 119863(2)

119899minus 119863(1)

119899

119897119899

119863(119899minus2)

119899minus 119863(119899minus3)

119899minus 119863(119899minus4)

119899

119897119899

)

=1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)22(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)

(53)

4 Determinant and Inverse of a Left CirculantMatrix with F

119899and L

119899Numbers

In this section let A1015840119899= LCirc(F

1F2 F

119899) and B1015840

119899=

LCirc(L1L2 L

119899) be left circulant matrices By using the

obtained conclusions we give a determinant formula forthe matrix A1015840

119899and B1015840

119899 Afterwards we prove that A1015840

119899is an

invertible matrix for 119899 gt 2 andB1015840119899is an invertible matrix for

any positive integer 119899The inverses of thematricesA1015840119899andB1015840

119899

are also presentedAccording to Lemma 5 andTheorems 8 9 and 11 we can

obtain the following theorems

Theorem 16 Let A1015840119899= LCirc(F

1F2 F

119899) be a left

circulant matrix then one has

detA1015840119899= (minus1)

(119899minus1)(119899minus2)2

times [(1 +F119899+1)119899minus1

+ (minusF119899)119899minus2

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(54)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 17 LetA1015840119899= LCirc(F

1F2 F

119899) be a left circu-

lant matrix if 119899 gt 2 thenA1015840119899is an invertible matrix

Theorem 18 Let A1015840119899= LCirc(F

1F2 F

119899) (119899 gt 2) be a

left circulant matrix then one has

A1015840minus1

119899=1

119891119899

LCirc(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

(minusF119899)2

(F1minusF119899+1)3

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

minusF119899

(F1minusF119899+1)2

1

F1minusF119899+1

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894)

(55)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(56)

By Lemma 5 and Theorems 12 13 and 15 the followingconclusions can be attained

Theorem 19 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one has

detB1015840119899= 2(minus1)

(119899minus1)(119899minus2)2

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(57)

whereL119899is the 119899th 119865

119899+ 119871119899number

Theorem20 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix thenB1015840119899is invertible for any positive integer 119899

Theorem 21 LetB1015840119899= LCirc(L

1L2 L

119899) be a left circu-

lant matrix then one can obtain

B1015840minus1119899

=1

119897119899

LCirc(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

2(L119899minus 2)2

(L1minusL119899+1)3

2 (L119899minus 2)

(L1minusL119899+1)2

2

L1minusL119899+1

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

)

(58)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(59)

5 Determinant and Inverse of 119892-CirculantMatrix with F

119899and L

119899Numbers

In this section let A119892119899

= 119892-Circ(F1F2 F

119899) and

B119892119899

= 119892-Circ(L1L2 L

119899) be 119892-circulant matrices

By using the obtained conclusions we give a determinantformula for thematricesA

119892119899andB

119892119899 Afterwards we prove

that A119892119899

is an invertible matrix for 119899 gt 2 and B119892119899

is aninvertible matrix if (119899 119892) = 1 The inverse of the matricesA119892119899

andB119892119899

are also presentedFrom Lemmas 6 and 7 and Theorems 8 9 and 11 we

deduce the following results

Theorem 22 LetA119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culant matrix then one has

detA119892119899= detQ

119892[(1 +F

119899+1)119899minus1

+ (minusF119899)119899minus2

times

119899minus1

sum

119896=1

(minusF119896) (

1 +F119899+1

minusF119899

)

119896minus1

]

(60)

whereF119899is the 119899th 119865

119899sdot 119871119899number

Theorem 23 Let A119892119899

= 119892-Circ(F1F2 F

119899) be a 119892-cir-

culantmatrix and (119892 119899) = 1 if 119899 gt 2 thenA119892119899

is an invertiblematrix

Theorem 24 LetA119892119899

= 119892-Circ(F1F2 F

119899) (119899 gt 2) be

a 119892-circulant matrix and (119892 119899) = 1 then

Aminus1

119892119899

= [1

119891119899

Circ(1 minus119899minus2

sum

119894=1

F119899minus119894(minusF119899)119894minus1

(F1minusF119899+1)119894

minus 3 +

119899minus2

sum

119894=1

F119899minus1minus119894

(minusF119899)119894minus1

(F1minusF119899+1)119894

1

F1minusF119899+1

minusF119899

(F1minusF119899+1)2

(minusF119899)2

(F1minusF119899+1)3

(minusF119899)119899minus3

(F1minusF119899+1)119899minus2

)]Q119879

119892

(61)

where

119891119899= F1minus 3F119899+

119899minus2

sum

119896=1

(minusF119896) (

minusF119899

F1minusF119899+1

)

119899minus(119896+1)

(62)

Taking Lemmas 6 and 7 andTheorems 12 13 and 15 intoaccount one has the following theorems

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 11

Theorem 25 Let B119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-

circulant matrix then one has

detB119892119899= 2 detQ

119892

times [(2 minusL119899+1)119899minus1

+ (L119899minus 2)119899minus2

times

119899minus1

sum

119896=1

(L119896+2

minus 2L119896+1) (

2 minusL119899+1

L119899minus 2

)

119896minus1

]

(63)

whereL119899is the 119899 119865

119899+ 119871119899number

Theorem 26 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then B119892119899

is invertible for anypositive integer 119899

Theorem 27 LetB119892119899

= 119892-Circ(L1L2 L

119899) be a 119892-cir-

culant matrix and (119892 119899) = 1 then

Bminus1

119892119899

= [1

119897119899

Circ(1 minus119899minus2

sum

119894=1

L119899minus119894minus1

(L119899minus 2)119894minus1

(L1minusL119899+1)119894

minus 2 minus

119899minus2

sum

119894=1

(2L119899minus119894minusL119899minus119894+1

) (L119899minus 2)119894minus1

(L1minusL119899+1)119894

2

L1minusL119899+1

2 (L119899minus 2)

(L1minusL119899+1)2

2(L119899minus 2)2

(L1minusL119899+1)3

2(L119899minus 2)119899minus3

(L1minusL119899+1)119899minus2

)]Q119879

119892

(64)

where

119897119899= L1minus 2L119899

+

119899minus2

sum

119896=1

(L119896+2

minus 2L119896+1) (

L119899minus 2

L1minusL119899+1

)

119899minus(119896+1)

(65)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the Development Projectof Science amp Technology of Shandong Province (Grant no2012GGX10115) and NSFC (Grant no 11301252) and theAMEP of Linyi University China

References

[1] J C Ruiz-Claeyssen and L A dos Santos Leal ldquoDiagonalizationand spectral decomposition of factor block circulant matricesrdquoLinear Algebra and its Applications vol 99 pp 41ndash61 1988

[2] J Wu and X Zou ldquoAsymptotic and periodic boundary valueproblems of mixed FDEs and wave solutions of lattice differ-ential equationsrdquo Journal of Differential Equations vol 135 no2 pp 315ndash357 1997

[3] E Ahmed A M A El-Sayed and H A A El-Saka ldquoOnsomeRouth-Hurwitz conditions for fractional order differentialequations and their applications in Lorenz Rossler Chua andChen systemsrdquo Physics Letters A vol 358 no 1 pp 1ndash4 2006

[4] E Ahmed and A S Elgazzar ldquoOn fractional order differentialequations model for nonlocal epidemicsrdquo Physica A StatisticalMechanics and its Applications vol 379 pp 607ndash614 2007

[5] W F Trench ldquoOn nonautonomous linear systems of differentialand difference equations with 119877-symmetric coefficient matri-cesrdquo Linear Algebra and Its Applications vol 431 no 11 pp2109ndash2117 2009

[6] P E Kloeden A Neuenkirch and R Pavani ldquoMultilevelMonte Carlo for stochastic differential equations with additivefractional noiserdquo Annals of Operations Research vol 189 pp255ndash276 2011

[7] S J Guo Y M Chen and J H Wu ldquoEquivariant normal formsfor parameterized delay differential equations with applicationsto bifurcation theoryrdquo Acta Mathematica Sinica vol 28 no 4pp 825ndash856 2012

[8] F-R Lin and H-X Yang ldquoA fast stationary iterative methodfor a partial integro-differential equation in pricing optionsrdquoCalcolo A Quarterly on Numerical Analysis and Theory ofComputation vol 50 no 4 pp 313ndash327 2013

[9] S T Lee J Liu and H-W Sun ldquoCombined compact differencescheme for linear second-order partial differential equationswith mixed derivativerdquo Journal of Computational and AppliedMathematics vol 264 pp 23ndash37 2014

[10] P J Davis Circulant Matrices John Wiley amp Sons New YorkNY USA 1979

[11] Z L Jiang and Z X Zhou Circulant Matrices Chengdu Tech-nology University Publishing Company Chengdu China 1999

[12] R M Gray ldquoToeplitz and circulant matrices a reviewrdquo Foun-dations and Trends in Communication and Information Theoryvol 2 pp 155ndash239 2006

[13] A Bose R S Hazra and K Saha ldquoPoisson convergence ofeigenvalues of circulant type matricesrdquo Extremes StatisticalTheory and Applications in Science Engineering and Economicsvol 14 no 4 pp 365ndash392 2011

[14] A Bose R S Hazra and K Saha ldquoSpectral norm of circulant-type matricesrdquo Journal of Theoretical Probability vol 24 no 2pp 479ndash516 2011

[15] C Erbas and M M Tanik ldquoGenerating solutions to the 119873-queens problem using 119892-circulantsrdquo Mathematics Magazinevol 68 no 5 pp 343ndash356 1995

[16] Y-K Wu R-Z Jia and Q Li ldquo(0 1)-circulant solutions to the119860119898

= 119869119899matrix equation 119892rdquo Linear Algebra and Its Applications

vol 345 pp 195ndash224 2002[17] E Ngondiep S Serra-Capizzano and D Sesana ldquoSpectral

features and asymptotic properties for (0 1)-circulants and

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Abstract and Applied Analysis

(0 1)-Toeplitz sequencesrdquo SIAM Journal onMatrix Analysis andApplications vol 31 no 4 pp 1663ndash1687 2010

[18] A Basak and A Bose ldquoLimiting spectral distributions of someband matricesrdquo Periodica Mathematica Hungarica vol 63 no1 pp 113ndash150 2011

[19] A Bose and J Mitra ldquoLimiting spectral distribution of a specialcirculantrdquo Statistics amp Probability Letters vol 60 no 1 pp 111ndash120 2002

[20] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquoThe Fibo-nacci Quarterly vol 40 no 3 pp 280ndash286 2002

[21] S-Q Shen J-M Cen and Y Hao ldquoOn the determinantsand inverses of circulant matrices with Fibonacci and Lucasnumbersrdquo Applied Mathematics and Computation vol 217 no23 pp 9790ndash9797 2011

[22] F L Lu and Z L Jiang ldquoThe sum and product of Fibonacci andLucas numbers Pell and Pell-Lucas numbers representation bymatrix methodrdquo WSEAS Transactions on Mathematics vol 12no 4 p 449

[23] DV Jaiswal ldquoOn determinants involving generalized FibonaccinumbersrdquoThe Fibonacci Quarterly vol 7 pp 319ndash330 1969

[24] D A Lind ldquoA Fibonacci circulantrdquoTheFibonacci Quarterly vol8 no 5 pp 449ndash455 1970

[25] D Bozkurt and T-Y Tam ldquoDeterminants and inverses of circu-lant matrices with Jacobsthal and Jacobsthal-Lucas NumbersrdquoApplied Mathematics and Computation vol 219 no 2 pp 544ndash551 2012

[26] H Karner J Schneid and C W Ueberhuber ldquoSpectral decom-position of real circulant matricesrdquo Linear Algebra and ItsApplications vol 367 pp 301ndash311 2003

[27] W T Stallings and T L Boullion ldquoThe pseudoinverse of anr-circulant matrixrdquo Proceedings of the American MathematicalSociety vol 34 pp 385ndash388 1972

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

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