Research ArticleExact Inverse Matrices of Fermat and MersenneCirculant Matrix

Yanpeng Zheng and Sugoog Shon

Department of Information and Telecommunications Engineering The University of Suwon Wau-ri Bongdam-eupHwaseong-si Gyeonggi-do 445-743 Republic of Korea

Correspondence should be addressed to Sugoog Shon sshonsuwonackr

Received 25 July 2014 Accepted 17 September 2014

Academic Editor Zidong Wang

Copyright copy 2015 Y Zheng and S ShonThis is an open access article distributed under theCreativeCommonsAttributionLicensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The well known circulant matrices are applied to solve networked systems In this paper circulant and left circulant matrices withthe Fermat and Mersenne numbers are considered The nonsingularity of these special matrices is discussed Meanwhile the exactdeterminants and inverse matrices of these special matrices are presented

1 Introduction

Circulant matrices are an important tool in solving net-worked systems In [1] the authors investigated the storageof binary cycles in Hopfield-type and other neural networksinvolving circulant matrix In [2] the authors considered aspecial class of the feedback delay network using circulantmatrices Distributed differential space-time codes that workfor networks with any number of relays using circulantmatrices were proposed by Jing and Jafarkhani in [3] Basic[4] solved the question for when circulant quantum spinnetworks with nearest-neighbor couplings can give perfectstate transfer Wang et al considered two-way transmissionmodel ensured that circular convolution between two fre-quency selective channels in [5] Li et al [6] presented a low-complexity binary framewise network coding encoder designbased on circulant matrix

Circulant matrices have been applied to various disci-plines including image processing communications signalprocessing and encoding Circulant type matrices haveestablished the substantial basis with the work in [7ndash12] andso on

Lately some authors gave the explicit determinant andinverse of the circulant and skew-circulant involving famousnumbers For example Yao and Jiang [13] presented thedeterminants inverses norm and spread of skew circulanttype matrices involving any continuous Lucas numbers

Jiang et al [14] considered circulant type matrices withthe 119896-Fibonacci and 119896-Lucas numbers and presented theexplicit determinant and inverse matrix by constructing thetransformationmatrices Dazheng [15] got the determinant ofthe Fibonacci-Lucas quasi-cyclic matrices Determinants andinverses of circulant matrices with Jacobsthal and Jacobsthal-Lucas numbers were obtained by Bozkurt and Tam in [16]

For any integer119898 ge 0 let 119865119898= 22119898

+1 be the119898th Fermatnumber It is well known that 119865

119898is prime for119898 le 4 but there

is no other m for which 119865119898is known to primeTheMersenne

andFermat sequences are defined by the following recurrencerelations [17 18] respectively

M119899+1= 3M

119899minus 2M119899minus1

F119899+1= 3F119899minus 2F119899minus1

(1)

with the initial conditionM0= 0M

1= 1 F

0= 2 F

1= 3 for

119899 ge 1Let 120572 and120573 be the roots of the characteristic equation 1199092minus

3119909 + 2 = 0 then the Binet formulas of the sequences M119896+119899

and F119896+119899 have the form

M119896+119899=120572119896+119899

minus 120573119896+119899

120572 minus 120573

F119896+119899= 120572119896+119899

+ 120573119896+119899

(2)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2015 Article ID 760823 10 pageshttpdxdoiorg1011552015760823

2 Abstract and Applied Analysis

Lemma 1 LetM119896+119899

be the (119896+119899)th Mersenne number and letF119896+119899

be the (119896 + 119899)th Fermat number then

(1) M119899+1minusM119899= 2119899

M119899+1minus 2M119899= 1

M119899= 2119899

minus 1M2119899minusM119899+1

M119899minus1= 2119899minus1

(2) F119899+1minus F119899= 2119899

F119899+1minus 2F119899= minus1

F119899= 2119899

+ 1F2119899minus F119899+1

F119899minus1= minus2119899minus1

We define a Fermat circulant matrix which is an 119899 times 119899matrix with the following form

Circ (F119896+1 F119896+2 F

119896+119899)

=

[[[[

[

F119896+1

F119896+2

sdot sdot sdot F119896+119899

F119896+119899

F119896+1

sdot sdot sdot F119896+119899minus1

F119896+2

F119896+3

sdot sdot sdot F119896+1

]]]]

]

(3)

A Mersenne circulant matrix which is an 119899 times 119899 matrix isdefined with the following form

Circ (M119896+1M119896+2 M

119896+119899)

=

[[[[

[

M119896+1

M119896+2

sdot sdot sdot M119896+119899

M119896+119899

M119896+1

sdot sdot sdot M119896+119899minus1

M119896+2

M119896+3

sdot sdot sdot M119896+1

]]]]

]

(4)

Besides a Fermat left circulant matrix is given by

LCirc (F119896+1 F119896+2 F

119896+119899)

=

[[[[

[

F119896+1

F119896+2

sdot sdot sdot F119896+119899

F119896+2

F119896+3

sdot sdot sdot F119896+1

F119896+119899

F119896+1

sdot sdot sdot F119896+119899minus1

]]]]

]

(5)

A Mersenne left circulant matrix is given by

LCirc (M119896+1M119896+2 M

119896+119899)

=

[[[[

[

M119896+1

M119896+2

sdot sdot sdot M119896+119899

M119896+2

M119896+3

sdot sdot sdot M119896+1

M119896+119899

M119896+1

sdot sdot sdot M119896+119899minus1

]]]]

]

(6)

The main content of this paper is to obtain the resultsfor the exact determinants and inverses of Fermat andMersenne circulant matrix In this paper let 119896 be a nonneg-ative integer 119860

119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) and 119861

119896119899=

Circ(M119896+1M119896+2 M

119896+119899)

2 Determinant and Inverse ofFermat Circulant Matrix

In this section let 119860119896119899

= Circ(F119896+1 F119896+2 F

119896+119899) be a

Fermat circulant matrix Firstly we obtain the exact formdeterminant of the matrix119860

119896119899 Afterwards we find the exact

form inverse of the matrix 119860119896119899

Theorem 2 Let119860119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) be a Fermat

circulant matrix Then one has

det119860119896119899= F119896+1sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) sdot 119910119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (minus119891)119899minus2

(7)

where 119910 = minus119890119891 119890 = 2(F119896minus F119896+119899) 119891 = F

119896+119899+1minus F119896+1

120591119896=

F119896+2F119896+1

and F119896+119899

is the (119896 + 119899)th Fermat number Moreover119860119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and F

119896+1minus

2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where 120581

119897=

cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Proof It is clear that det1198600119899= F1sdot[sum119899minus2

119895=1(F119895+2minus1205910F119895+1)[2(F119899minus

F0)(F119899+1minus F1)]119899minus119895minus1

+ F1minus 1205910F119899] sdot [F1minus F119899+1]119899minus2 satisfies (7)

In the following let

Σ =

((((((

(

1

minus120591119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Ω1=((

(

1 0 0 sdot sdot sdot 0 0

0 119910119899minus2

0 sdot sdot sdot 0 0

0 119910119899minus3

0 sdot sdot sdot 0 minus1

d

0 119910 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(8)

be two 119899 times 119899matrices we have

Σ119860119896119899Ω1=

((((

(

F119896+1

ℎ1015840

119896119899minusF119896+119899

minusF119896+119899minus1

sdot sdot sdot minusF119896+3

0 ℎ119896119899

1198863

1198864

sdot sdot sdot 119886119899

0 0 119890 119891 0

0 0 0 119890 0

d

0 0 0 0 d 119891

0 0 0 0 119890

))))

)

(9)

Abstract and Applied Analysis 3

where

120591119896=F119896+2

F119896+1

119910 = minus119890

119891

1198863= 120591119896F119896+119899minus F119896+1

119886119895= 120591119896F119896+119899+3minus119895

minus F119896+119899+4minus119895

(119895 = 4 5 119899)

ℎ1015840

119896119899=

119899minus1

sum

119905=1

F119905+119896+1

[2 (F119896+119899minus F119896)

F119896+119899+1

minus F119896+1

]

119899minus119905minus1

ℎ119896119899=

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) 119910119899minus119905minus1

+ F119896+1minus 120591119896F119896+119899

(10)

We obtain

detΣ det119860119896119899

detΩ1

= F119896+1sdot [

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) 119910119899minus119905minus1

+F119896+1minus 120591119896F119896+119899] sdot 119890119899minus2

(11)

while

detΣ = (minus1)(119899minus1)(119899minus2)2

detΩ1= (minus1)

(119899minus1)(119899minus2)2

[2 (F119896minus F119896+119899)

F119896+1minus F119896+119899+1

]

119899minus2

(12)

We have

det119860119896119899= F119896+1

sdot [

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) sdot 119910119899minus119905minus1

+F119896+1minus 120591119896F119896+119899] sdot (minus119891)

119899minus2

(13)

Next we discuss the singularity of the matrix 119860119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos (2119897120587119899) + 119894119904119894119899(2119897120587119899) We have

119891 (120581119897) = F119896+1+ F119896+2120581119897+ sdot sdot sdot + F

119896+119899(120581119897)119899minus1

=F119896+1minus 2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(14)

By Lemma 1 in [14] thematrix119860119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 that is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119860

119896119899is

nonsingular if and only if F119896+1minus 2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899

= 0when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 we obtain 120581

119897= 1120572 or 120581

119897= 1120573

Let 120581119897= 1120572 then the eigenvalue of 119860

119896119899is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1F119899

120572119899minus1 (120572 minus 120573)= 0 (15)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119860

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 3 Let the matrixM = [1198981015840

119894119897]119899minus2

119894119897=1be of the form

119898119894119897=

2 (F119896minus F119896+119899) = 119890 119894 = 119897

F119896+119899+1

minus F119896+1= 119891 119897 = 119894 + 1

0 otherwise(16)

Then the inverseMminus1 = [1198981015840119894119897]119899minus2

119894119897=1of the matrixM is equal to

1198981015840

119894119897=

(F119896+1minus F119896+119899+1

)119897minus119894

[2 (F119896minus F119896+119899)]119897minus119894+1

=(minus119891)119897minus119894

119890119897minus119894+1 119897 ge 119894

0 119897 lt 119894

(17)

Proof Let 119890119894119897= sum119899minus2

119896=11198981198941198961198981015840

119896119897 Distinctly 119888

119894119897= 0 for 119897 lt 119894 In

the case 119894 = 119897 we obtain

119890119894119894= 1198981198941198941198981015840

119894119894

= (F119896+1minus F119896+119899+1

) sdot1

(F119896+1minus F119896+119899+1

)

= 1

(18)

For 119897 ge 119894 + 1 we get

119890119894119897=

119899minus2

sum

119896=1

1198981198941198961198981015840

119896119897

= 1198981198941198941198981015840

119894119897+ 119898119894119894+11198981015840

119894+1119897

= 119890 sdot(minus119891)119897minus119894

119890119897minus119894+1+ 119891 sdot

(minus119891)119897minus119894minus1

119890119897minus119894

= 0

(19)

We check on MMminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyMminus1M = 119868

119899minus2 Thus

the proof is completed

4 Abstract and Applied Analysis

Theorem4 Let119860119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) be a Fermat

circulant matrix Then one acquires 119860minus1119896119899= Circ(V

1 V2

V119899) where

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=

F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=

F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

times(F119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(20)

Proof Let

Ω2=

((((((((

(

1 minusℎ1015840

119896119899

F119896+1

1199091015840

31199091015840

4sdot sdot sdot 1199091015840

119899

0 1 1199101015840

31199101015840

4sdot sdot sdot 1199101015840

119899

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

))))))))

)

(21)

where

120591119896=F119896+2

F119896+1

1199091015840

3=F119896+119899

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot(120591119896F119896+119899minus F119896+1)

F119896+1

1199101015840

3=F119896+1minus 120591119896F119896+119899

ℎ119896119899

1199091015840

119894=F119896+119899+3minus119894

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot120591119896F119896+119899+3minus119894

minus F119896+119899+4minus119894

F119896+1

(119894 = 4 119899)

1199101015840

119894=F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

ℎ119896119899

(119894 = 4 119899)

ℎ1015840

119896119899=

119899minus1

sum

119894=1

F119894+119896+1

[2 (F119896+119899minus F119896)

F119896+119899+1

minus F119896+1

]

119899minus119894minus1

ℎ119896119899=

119899minus2

sum

119894=1

(F119894+119896+2

minus 120591119896F119894+119896+1

) 119910119899minus119894minus1

+ F119896+1minus 120591119896F119896+119899

(22)

We have

Σ119860119896119899Ω1Ω2= D2oplusM (23)

whereD2= diag(F

119896+1 ℎ119896119899) is a diagonal matrix andD

2oplusM

is the direct sum ofD2andM If we denoteΩ = Ω

1Ω2 then

we obtain

119860minus1

119896119899= Ω(D

minus1

2oplusMminus1

) Σ (24)

Let119860minus1119896119899= Circ(V

1 V2 V

119899) Since the last row elements

of the matrix Ω are 0 1 11991010158403minus 1 119910

1015840

4 119910

1015840

119899minus1 1199101015840

119899 according to

Lemma 3 then the last row elements of 119860minus1119896119899

are given by thefollowing equations

V2= minus

120591119896

ℎ119896119899

+1199101015840

3minus 1

F119896minus F119896+119899

V3= (1199101015840

3minus 1)

(minus119891)119899minus3

119890119899minus2+

119899

sum

119894=4

1199101015840

119894sdot(minus119891)119899minus119894

119890119899minus119894+1

V4= (1199101015840

3minus 1) [

(minus119891)119899minus4

(minus119891)119899minus3minus3 (minus119891)

119899minus3

119890119899minus2]

+

119899

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus1

119890119899minus119894minus3 (minus119891)

119899minus119894

119890119899minus119894+1]

(119905 lt 0 (minus119891)119905

= 0)

Abstract and Applied Analysis 5

V119904= (1199101015840

3minus 1) [

(minus119891)119899minus119904

119890119899minus119904+1minus3 (minus119891)

119899minus119904+1

119890119899minus119904+2+2 (minus119891)

119899minus119904+2

119890119899minus119904+3]

+

119899minus119904+5

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus119904+3

119890119899minus119894minus119904+4minus3 (minus119891)

119899minus119894minus119904+4

119890119899minus119894minus119904+5

+2 (minus119891)

119899minus119894minus119904+5

119890119899minus119894minus119904+6]

(119904 = 5 6 119899 119905 lt 0 (minus119891)119905

= 0)

V1=1

ℎ119896119899

+minus2119891 minus 3119890

1198902(1199103minus 1) +

2

1198901199104

(25)

where 119891 = F119896+119899+1

minus F119896+1

119890 = 2(F119896minus F119896+119899) according to

Lemma 1 then we have

(i) 119890 + 119891 = 0

(ii) 119890 + 2119891 = 2119896+119899+1 minus 2119896+1

Hence we obtain

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdot(F119896+1minus F119896+119899+1

)119899minus4

[2 (F119896minus F119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

times(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(26)

Thus the proof is completed

3 Determinant and Inverse ofMersenne Circulant Matrix

In this section let 119861119896119899= Circ(M

119896+1M119896+2 M

119896+119899) be a

Mersenne circulantmatrix Firstly we obtain the determinantof the matrix 119861

119896119899 Afterwards we seek out the inverse of the

matrix 119861119896119899

Theorem 5 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one obtains

det119861119896119899= M119896+1

sdot [

119899minus2

sum

119896=1

(M119896+119896+2

minus 120583119896M119896+119896+1

) 119909119899minus119896minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(27)

where 119909 = minus119888119889 119888 = 2(M119896minus M119896+119899) 119889 = M

119896+119899+1minus M119896+1

120583119896= M119896+2M119896+1

andM119896+119899

is the (119896 + 119899)th Mersenne numberFurthermore119861

119896119899is singular if and only if (1minus120572120581

119897)(1minus120573120581

119897) = 0

andM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+

where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Proof Obviously

det1198600119899= M1

sdot [

119899minus2

sum

119894=1

(M119894+2minus 1205830M119894+1) sdot [

2 (M119899minusM0)

M119899+1minusM1

]

119899minus119896minus1

+M1minus 1205830M119899] sdot [2 (M

0minusM119899)]119899minus2

(28)

6 Abstract and Applied Analysis

satisfies (27) In the following let

Γ =

((((((

(

1

minus120583119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Π1=((

(

1 0 0 sdot sdot sdot 0 0

0 119909119899minus2

0 sdot sdot sdot 0 0

0 119909119899minus3

0 sdot sdot sdot 0 minus1

d

0 119909 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(29)

be two 119899 times 119899matrices then we have

Γ119861119896119899Π1=

((((

(

M119896+1

1198911015840

119896119899minusM119896+119899

sdot sdot sdot minusM119896+3

0 119891119896119899

ℎ3

sdot sdot sdot ℎ119899

0 0 119888 sdot sdot sdot 0

0 0 0 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 119889

0 0 0 sdot sdot sdot 119888

))))

)

(30)

where119888 = 2 (M

119896minusM119896+119899)

119889 = M119896+119899+1

minusM119896+1

119909 = minus119888

119889 120583

119896=M119896+2

M119896+1

ℎ3= 120583119896M119896+119899minusM119896+1

ℎ119895= (120583119896M119896+119899+3minus119895

minusM119896+119899+4minus119895

)

(119895 = 4 5 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(31)

We get

det Γ det119861119896119899

detΠ1

= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot 119888119899minus2

(32)

besides

det Γ = (minus1)(119899minus1)(119899minus2)2

detΠ1= (minus1)

(119899minus1)(119899minus2)2

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus2

(33)

We have

det119861119896119899= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(34)

Now we discuss the singularity of the matrix 119861119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) So we have

119891 (120581119897) = M

119896+1+M119896+2120581119897+ sdot sdot sdot +M

119896+119899(120581119897)119899minus1

=M119896+1minus 2120581119897M119896minusM119896+119899+1

+ 2120581119897M119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(35)

By Lemma 1 in [14] the matrix 119861119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 That is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119861

119896119899is

nonsingular if and only ifM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899

=

0 for 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 When (1minus120572120581

119897)(1minus120573120581

119897) =

0 we obtain 120581119897= 1120572 or 120581

119897= 1120573 Let 120581

119897= 1120572 then the

eigenvalue of 119861119896119899

is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1M119899

120572119899minus1 (120572 minus 120573)= 0 (36)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119861

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 6 Let the matrixG = [119892119894119895]119899minus2

119894119895=1be of the form

119892119894119895=

2 (M119896minusM119896+119899) = 119888 119894 = 119895

M119896+119899+1

minusM119896+1= 119889 119895 = 119894 + 1

0 otherwise(37)

Then the inverseGminus1 = [1198921015840119894119895]119899minus2

119894119895=1of the matrixG is equal to

1198921015840

119894119895=

(M119896+1minusM119896+119899+1

)119895minus119894

[2 (M119896minusM119896+119899)]119895minus119894+1

=(minus119889)119895minus119894

119888119895minus119894+1 119895 ge 119894

0 119895 lt 119894

(38)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Distinctly 119888

119894119895= 0 for 119895 lt 119894

When 119894 = 119895 we obtain

119888119894119894= 1198921198941198941198921015840

119894119894= minus119889 sdot

1

minus119889= 1 (39)

Abstract and Applied Analysis 7

For 119895 ge 119894 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 1198921198941198941198921015840

119894119895+ 119892119894119894+11198921015840

119894+1119895

= 119888 sdot(minus119889)119895minus119894

119888119895minus119894+1+ 119889 sdot

(minus119889)119895minus119894minus1

119888119895minus119894= 0

(40)

We verifyGGminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899minus 2) identitymatrix Similarly we check on Gminus1G = 119868

119899minus2 Thus the proof

is completed

Theorem 7 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one acquires

119861minus1

119896119899= Circ (119906

1 1199062 119906

119899) (41)

where

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899) sdot(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(42)

where

120583119896=M119896+2

M119896+1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(43)

Proof Let

Π2=

(((((((

(

1 minus1198911015840

119896119899

M119896+1

11990931199094sdot sdot sdot 119909119899

0 1 11991031199104sdot sdot sdot 119910119899

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

)))))))

)

(44)

where

120583119896=M119896+2

M119896+1

1199093=M119896+119899

M119896+1

+1198911015840

119896119899

119891119896119899

sdot(120583119896M119896+119899minusM119896+1)

M119896+1

1199103=M119896+1minus 120583119896M119896+119899

119891119896119899

119909119894=M119896+119899+3minus119894

M119896+1

+1198911015840

119896119899

119891119896119899

sdot120583119896M119896+119899+3minus119894

minusM119896+119899+4minus119894

M119896+1

(119894 = 4 119899)

119910119894=M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

119891119896119899

(119894 = 4 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(45)

We have

Γ119861119896119899Π1Π2= D1oplusG (46)

whereD1= diag(M

119896+1 119891119896119899) is a diagonal matrix andD

1oplusG

is the direct sum ofD1andG If we denote Π = Π

1Π2 then

we obtain

119861minus1

119896119899= Π (D

minus1

1oplusGminus1

) Γ (47)

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

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International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

Lemma 1 LetM119896+119899

be the (119896+119899)th Mersenne number and letF119896+119899

be the (119896 + 119899)th Fermat number then

(1) M119899+1minusM119899= 2119899

M119899+1minus 2M119899= 1

M119899= 2119899

minus 1M2119899minusM119899+1

M119899minus1= 2119899minus1

(2) F119899+1minus F119899= 2119899

F119899+1minus 2F119899= minus1

F119899= 2119899

+ 1F2119899minus F119899+1

F119899minus1= minus2119899minus1

We define a Fermat circulant matrix which is an 119899 times 119899matrix with the following form

Circ (F119896+1 F119896+2 F

119896+119899)

=

[[[[

[

F119896+1

F119896+2

sdot sdot sdot F119896+119899

F119896+119899

F119896+1

sdot sdot sdot F119896+119899minus1

F119896+2

F119896+3

sdot sdot sdot F119896+1

]]]]

]

(3)

A Mersenne circulant matrix which is an 119899 times 119899 matrix isdefined with the following form

Circ (M119896+1M119896+2 M

119896+119899)

=

[[[[

[

M119896+1

M119896+2

sdot sdot sdot M119896+119899

M119896+119899

M119896+1

sdot sdot sdot M119896+119899minus1

M119896+2

M119896+3

sdot sdot sdot M119896+1

]]]]

]

(4)

Besides a Fermat left circulant matrix is given by

LCirc (F119896+1 F119896+2 F

119896+119899)

=

[[[[

[

F119896+1

F119896+2

sdot sdot sdot F119896+119899

F119896+2

F119896+3

sdot sdot sdot F119896+1

F119896+119899

F119896+1

sdot sdot sdot F119896+119899minus1

]]]]

]

(5)

A Mersenne left circulant matrix is given by

LCirc (M119896+1M119896+2 M

119896+119899)

=

[[[[

[

M119896+1

M119896+2

sdot sdot sdot M119896+119899

M119896+2

M119896+3

sdot sdot sdot M119896+1

M119896+119899

M119896+1

sdot sdot sdot M119896+119899minus1

]]]]

]

(6)

The main content of this paper is to obtain the resultsfor the exact determinants and inverses of Fermat andMersenne circulant matrix In this paper let 119896 be a nonneg-ative integer 119860

119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) and 119861

119896119899=

Circ(M119896+1M119896+2 M

119896+119899)

2 Determinant and Inverse ofFermat Circulant Matrix

In this section let 119860119896119899

= Circ(F119896+1 F119896+2 F

119896+119899) be a

Fermat circulant matrix Firstly we obtain the exact formdeterminant of the matrix119860

119896119899 Afterwards we find the exact

form inverse of the matrix 119860119896119899

Theorem 2 Let119860119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) be a Fermat

circulant matrix Then one has

det119860119896119899= F119896+1sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) sdot 119910119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (minus119891)119899minus2

(7)

where 119910 = minus119890119891 119890 = 2(F119896minus F119896+119899) 119891 = F

119896+119899+1minus F119896+1

120591119896=

F119896+2F119896+1

and F119896+119899

is the (119896 + 119899)th Fermat number Moreover119860119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and F

119896+1minus

2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where 120581

119897=

cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Proof It is clear that det1198600119899= F1sdot[sum119899minus2

119895=1(F119895+2minus1205910F119895+1)[2(F119899minus

F0)(F119899+1minus F1)]119899minus119895minus1

+ F1minus 1205910F119899] sdot [F1minus F119899+1]119899minus2 satisfies (7)

In the following let

Σ =

((((((

(

1

minus120591119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Ω1=((

(

1 0 0 sdot sdot sdot 0 0

0 119910119899minus2

0 sdot sdot sdot 0 0

0 119910119899minus3

0 sdot sdot sdot 0 minus1

d

0 119910 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(8)

be two 119899 times 119899matrices we have

Σ119860119896119899Ω1=

((((

(

F119896+1

ℎ1015840

119896119899minusF119896+119899

minusF119896+119899minus1

sdot sdot sdot minusF119896+3

0 ℎ119896119899

1198863

1198864

sdot sdot sdot 119886119899

0 0 119890 119891 0

0 0 0 119890 0

d

0 0 0 0 d 119891

0 0 0 0 119890

))))

)

(9)

Abstract and Applied Analysis 3

where

120591119896=F119896+2

F119896+1

119910 = minus119890

119891

1198863= 120591119896F119896+119899minus F119896+1

119886119895= 120591119896F119896+119899+3minus119895

minus F119896+119899+4minus119895

(119895 = 4 5 119899)

ℎ1015840

119896119899=

119899minus1

sum

119905=1

F119905+119896+1

[2 (F119896+119899minus F119896)

F119896+119899+1

minus F119896+1

]

119899minus119905minus1

ℎ119896119899=

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) 119910119899minus119905minus1

+ F119896+1minus 120591119896F119896+119899

(10)

We obtain

detΣ det119860119896119899

detΩ1

= F119896+1sdot [

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) 119910119899minus119905minus1

+F119896+1minus 120591119896F119896+119899] sdot 119890119899minus2

(11)

while

detΣ = (minus1)(119899minus1)(119899minus2)2

detΩ1= (minus1)

(119899minus1)(119899minus2)2

[2 (F119896minus F119896+119899)

F119896+1minus F119896+119899+1

]

119899minus2

(12)

We have

det119860119896119899= F119896+1

sdot [

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) sdot 119910119899minus119905minus1

+F119896+1minus 120591119896F119896+119899] sdot (minus119891)

119899minus2

(13)

Next we discuss the singularity of the matrix 119860119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos (2119897120587119899) + 119894119904119894119899(2119897120587119899) We have

119891 (120581119897) = F119896+1+ F119896+2120581119897+ sdot sdot sdot + F

119896+119899(120581119897)119899minus1

=F119896+1minus 2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(14)

By Lemma 1 in [14] thematrix119860119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 that is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119860

119896119899is

nonsingular if and only if F119896+1minus 2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899

= 0when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 we obtain 120581

119897= 1120572 or 120581

119897= 1120573

Let 120581119897= 1120572 then the eigenvalue of 119860

119896119899is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1F119899

120572119899minus1 (120572 minus 120573)= 0 (15)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119860

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 3 Let the matrixM = [1198981015840

119894119897]119899minus2

119894119897=1be of the form

119898119894119897=

2 (F119896minus F119896+119899) = 119890 119894 = 119897

F119896+119899+1

minus F119896+1= 119891 119897 = 119894 + 1

0 otherwise(16)

Then the inverseMminus1 = [1198981015840119894119897]119899minus2

119894119897=1of the matrixM is equal to

1198981015840

119894119897=

(F119896+1minus F119896+119899+1

)119897minus119894

[2 (F119896minus F119896+119899)]119897minus119894+1

=(minus119891)119897minus119894

119890119897minus119894+1 119897 ge 119894

0 119897 lt 119894

(17)

Proof Let 119890119894119897= sum119899minus2

119896=11198981198941198961198981015840

119896119897 Distinctly 119888

119894119897= 0 for 119897 lt 119894 In

the case 119894 = 119897 we obtain

119890119894119894= 1198981198941198941198981015840

119894119894

= (F119896+1minus F119896+119899+1

) sdot1

(F119896+1minus F119896+119899+1

)

= 1

(18)

For 119897 ge 119894 + 1 we get

119890119894119897=

119899minus2

sum

119896=1

1198981198941198961198981015840

119896119897

= 1198981198941198941198981015840

119894119897+ 119898119894119894+11198981015840

119894+1119897

= 119890 sdot(minus119891)119897minus119894

119890119897minus119894+1+ 119891 sdot

(minus119891)119897minus119894minus1

119890119897minus119894

= 0

(19)

We check on MMminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyMminus1M = 119868

119899minus2 Thus

the proof is completed

4 Abstract and Applied Analysis

Theorem4 Let119860119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) be a Fermat

circulant matrix Then one acquires 119860minus1119896119899= Circ(V

1 V2

V119899) where

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=

F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=

F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

times(F119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(20)

Proof Let

Ω2=

((((((((

(

1 minusℎ1015840

119896119899

F119896+1

1199091015840

31199091015840

4sdot sdot sdot 1199091015840

119899

0 1 1199101015840

31199101015840

4sdot sdot sdot 1199101015840

119899

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

))))))))

)

(21)

where

120591119896=F119896+2

F119896+1

1199091015840

3=F119896+119899

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot(120591119896F119896+119899minus F119896+1)

F119896+1

1199101015840

3=F119896+1minus 120591119896F119896+119899

ℎ119896119899

1199091015840

119894=F119896+119899+3minus119894

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot120591119896F119896+119899+3minus119894

minus F119896+119899+4minus119894

F119896+1

(119894 = 4 119899)

1199101015840

119894=F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

ℎ119896119899

(119894 = 4 119899)

ℎ1015840

119896119899=

119899minus1

sum

119894=1

F119894+119896+1

[2 (F119896+119899minus F119896)

F119896+119899+1

minus F119896+1

]

119899minus119894minus1

ℎ119896119899=

119899minus2

sum

119894=1

(F119894+119896+2

minus 120591119896F119894+119896+1

) 119910119899minus119894minus1

+ F119896+1minus 120591119896F119896+119899

(22)

We have

Σ119860119896119899Ω1Ω2= D2oplusM (23)

whereD2= diag(F

119896+1 ℎ119896119899) is a diagonal matrix andD

2oplusM

is the direct sum ofD2andM If we denoteΩ = Ω

1Ω2 then

we obtain

119860minus1

119896119899= Ω(D

minus1

2oplusMminus1

) Σ (24)

Let119860minus1119896119899= Circ(V

1 V2 V

119899) Since the last row elements

of the matrix Ω are 0 1 11991010158403minus 1 119910

1015840

4 119910

1015840

119899minus1 1199101015840

119899 according to

Lemma 3 then the last row elements of 119860minus1119896119899

are given by thefollowing equations

V2= minus

120591119896

ℎ119896119899

+1199101015840

3minus 1

F119896minus F119896+119899

V3= (1199101015840

3minus 1)

(minus119891)119899minus3

119890119899minus2+

119899

sum

119894=4

1199101015840

119894sdot(minus119891)119899minus119894

119890119899minus119894+1

V4= (1199101015840

3minus 1) [

(minus119891)119899minus4

(minus119891)119899minus3minus3 (minus119891)

119899minus3

119890119899minus2]

+

119899

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus1

119890119899minus119894minus3 (minus119891)

119899minus119894

119890119899minus119894+1]

(119905 lt 0 (minus119891)119905

= 0)

Abstract and Applied Analysis 5

V119904= (1199101015840

3minus 1) [

(minus119891)119899minus119904

119890119899minus119904+1minus3 (minus119891)

119899minus119904+1

119890119899minus119904+2+2 (minus119891)

119899minus119904+2

119890119899minus119904+3]

+

119899minus119904+5

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus119904+3

119890119899minus119894minus119904+4minus3 (minus119891)

119899minus119894minus119904+4

119890119899minus119894minus119904+5

+2 (minus119891)

119899minus119894minus119904+5

119890119899minus119894minus119904+6]

(119904 = 5 6 119899 119905 lt 0 (minus119891)119905

= 0)

V1=1

ℎ119896119899

+minus2119891 minus 3119890

1198902(1199103minus 1) +

2

1198901199104

(25)

where 119891 = F119896+119899+1

minus F119896+1

119890 = 2(F119896minus F119896+119899) according to

Lemma 1 then we have

(i) 119890 + 119891 = 0

(ii) 119890 + 2119891 = 2119896+119899+1 minus 2119896+1

Hence we obtain

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdot(F119896+1minus F119896+119899+1

)119899minus4

[2 (F119896minus F119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

times(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(26)

Thus the proof is completed

3 Determinant and Inverse ofMersenne Circulant Matrix

In this section let 119861119896119899= Circ(M

119896+1M119896+2 M

119896+119899) be a

Mersenne circulantmatrix Firstly we obtain the determinantof the matrix 119861

119896119899 Afterwards we seek out the inverse of the

matrix 119861119896119899

Theorem 5 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one obtains

det119861119896119899= M119896+1

sdot [

119899minus2

sum

119896=1

(M119896+119896+2

minus 120583119896M119896+119896+1

) 119909119899minus119896minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(27)

where 119909 = minus119888119889 119888 = 2(M119896minus M119896+119899) 119889 = M

119896+119899+1minus M119896+1

120583119896= M119896+2M119896+1

andM119896+119899

is the (119896 + 119899)th Mersenne numberFurthermore119861

119896119899is singular if and only if (1minus120572120581

119897)(1minus120573120581

119897) = 0

andM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+

where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Proof Obviously

det1198600119899= M1

sdot [

119899minus2

sum

119894=1

(M119894+2minus 1205830M119894+1) sdot [

2 (M119899minusM0)

M119899+1minusM1

]

119899minus119896minus1

+M1minus 1205830M119899] sdot [2 (M

0minusM119899)]119899minus2

(28)

6 Abstract and Applied Analysis

satisfies (27) In the following let

Γ =

((((((

(

1

minus120583119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Π1=((

(

1 0 0 sdot sdot sdot 0 0

0 119909119899minus2

0 sdot sdot sdot 0 0

0 119909119899minus3

0 sdot sdot sdot 0 minus1

d

0 119909 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(29)

be two 119899 times 119899matrices then we have

Γ119861119896119899Π1=

((((

(

M119896+1

1198911015840

119896119899minusM119896+119899

sdot sdot sdot minusM119896+3

0 119891119896119899

ℎ3

sdot sdot sdot ℎ119899

0 0 119888 sdot sdot sdot 0

0 0 0 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 119889

0 0 0 sdot sdot sdot 119888

))))

)

(30)

where119888 = 2 (M

119896minusM119896+119899)

119889 = M119896+119899+1

minusM119896+1

119909 = minus119888

119889 120583

119896=M119896+2

M119896+1

ℎ3= 120583119896M119896+119899minusM119896+1

ℎ119895= (120583119896M119896+119899+3minus119895

minusM119896+119899+4minus119895

)

(119895 = 4 5 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(31)

We get

det Γ det119861119896119899

detΠ1

= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot 119888119899minus2

(32)

besides

det Γ = (minus1)(119899minus1)(119899minus2)2

detΠ1= (minus1)

(119899minus1)(119899minus2)2

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus2

(33)

We have

det119861119896119899= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(34)

Now we discuss the singularity of the matrix 119861119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) So we have

119891 (120581119897) = M

119896+1+M119896+2120581119897+ sdot sdot sdot +M

119896+119899(120581119897)119899minus1

=M119896+1minus 2120581119897M119896minusM119896+119899+1

+ 2120581119897M119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(35)

By Lemma 1 in [14] the matrix 119861119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 That is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119861

119896119899is

nonsingular if and only ifM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899

=

0 for 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 When (1minus120572120581

119897)(1minus120573120581

119897) =

0 we obtain 120581119897= 1120572 or 120581

119897= 1120573 Let 120581

119897= 1120572 then the

eigenvalue of 119861119896119899

is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1M119899

120572119899minus1 (120572 minus 120573)= 0 (36)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119861

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 6 Let the matrixG = [119892119894119895]119899minus2

119894119895=1be of the form

119892119894119895=

2 (M119896minusM119896+119899) = 119888 119894 = 119895

M119896+119899+1

minusM119896+1= 119889 119895 = 119894 + 1

0 otherwise(37)

Then the inverseGminus1 = [1198921015840119894119895]119899minus2

119894119895=1of the matrixG is equal to

1198921015840

119894119895=

(M119896+1minusM119896+119899+1

)119895minus119894

[2 (M119896minusM119896+119899)]119895minus119894+1

=(minus119889)119895minus119894

119888119895minus119894+1 119895 ge 119894

0 119895 lt 119894

(38)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Distinctly 119888

119894119895= 0 for 119895 lt 119894

When 119894 = 119895 we obtain

119888119894119894= 1198921198941198941198921015840

119894119894= minus119889 sdot

1

minus119889= 1 (39)

Abstract and Applied Analysis 7

For 119895 ge 119894 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 1198921198941198941198921015840

119894119895+ 119892119894119894+11198921015840

119894+1119895

= 119888 sdot(minus119889)119895minus119894

119888119895minus119894+1+ 119889 sdot

(minus119889)119895minus119894minus1

119888119895minus119894= 0

(40)

We verifyGGminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899minus 2) identitymatrix Similarly we check on Gminus1G = 119868

119899minus2 Thus the proof

is completed

Theorem 7 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one acquires

119861minus1

119896119899= Circ (119906

1 1199062 119906

119899) (41)

where

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899) sdot(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(42)

where

120583119896=M119896+2

M119896+1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(43)

Proof Let

Π2=

(((((((

(

1 minus1198911015840

119896119899

M119896+1

11990931199094sdot sdot sdot 119909119899

0 1 11991031199104sdot sdot sdot 119910119899

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

)))))))

)

(44)

where

120583119896=M119896+2

M119896+1

1199093=M119896+119899

M119896+1

+1198911015840

119896119899

119891119896119899

sdot(120583119896M119896+119899minusM119896+1)

M119896+1

1199103=M119896+1minus 120583119896M119896+119899

119891119896119899

119909119894=M119896+119899+3minus119894

M119896+1

+1198911015840

119896119899

119891119896119899

sdot120583119896M119896+119899+3minus119894

minusM119896+119899+4minus119894

M119896+1

(119894 = 4 119899)

119910119894=M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

119891119896119899

(119894 = 4 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(45)

We have

Γ119861119896119899Π1Π2= D1oplusG (46)

whereD1= diag(M

119896+1 119891119896119899) is a diagonal matrix andD

1oplusG

is the direct sum ofD1andG If we denote Π = Π

1Π2 then

we obtain

119861minus1

119896119899= Π (D

minus1

1oplusGminus1

) Γ (47)

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

where

120591119896=F119896+2

F119896+1

119910 = minus119890

119891

1198863= 120591119896F119896+119899minus F119896+1

119886119895= 120591119896F119896+119899+3minus119895

minus F119896+119899+4minus119895

(119895 = 4 5 119899)

ℎ1015840

119896119899=

119899minus1

sum

119905=1

F119905+119896+1

[2 (F119896+119899minus F119896)

F119896+119899+1

minus F119896+1

]

119899minus119905minus1

ℎ119896119899=

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) 119910119899minus119905minus1

+ F119896+1minus 120591119896F119896+119899

(10)

We obtain

detΣ det119860119896119899

detΩ1

= F119896+1sdot [

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) 119910119899minus119905minus1

+F119896+1minus 120591119896F119896+119899] sdot 119890119899minus2

(11)

while

detΣ = (minus1)(119899minus1)(119899minus2)2

detΩ1= (minus1)

(119899minus1)(119899minus2)2

[2 (F119896minus F119896+119899)

F119896+1minus F119896+119899+1

]

119899minus2

(12)

We have

det119860119896119899= F119896+1

sdot [

119899minus2

sum

119905=1

(F119905+119896+2

minus 120591119896F119905+119896+1

) sdot 119910119899minus119905minus1

+F119896+1minus 120591119896F119896+119899] sdot (minus119891)

119899minus2

(13)

Next we discuss the singularity of the matrix 119860119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos (2119897120587119899) + 119894119904119894119899(2119897120587119899) We have

119891 (120581119897) = F119896+1+ F119896+2120581119897+ sdot sdot sdot + F

119896+119899(120581119897)119899minus1

=F119896+1minus 2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(14)

By Lemma 1 in [14] thematrix119860119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 that is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119860

119896119899is

nonsingular if and only if F119896+1minus 2120581119897F119896minus F119896+119899+1

+ 2120581119897F119896+119899

= 0when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 we obtain 120581

119897= 1120572 or 120581

119897= 1120573

Let 120581119897= 1120572 then the eigenvalue of 119860

119896119899is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1F119899

120572119899minus1 (120572 minus 120573)= 0 (15)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119860

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 3 Let the matrixM = [1198981015840

119894119897]119899minus2

119894119897=1be of the form

119898119894119897=

2 (F119896minus F119896+119899) = 119890 119894 = 119897

F119896+119899+1

minus F119896+1= 119891 119897 = 119894 + 1

0 otherwise(16)

Then the inverseMminus1 = [1198981015840119894119897]119899minus2

119894119897=1of the matrixM is equal to

1198981015840

119894119897=

(F119896+1minus F119896+119899+1

)119897minus119894

[2 (F119896minus F119896+119899)]119897minus119894+1

=(minus119891)119897minus119894

119890119897minus119894+1 119897 ge 119894

0 119897 lt 119894

(17)

Proof Let 119890119894119897= sum119899minus2

119896=11198981198941198961198981015840

119896119897 Distinctly 119888

119894119897= 0 for 119897 lt 119894 In

the case 119894 = 119897 we obtain

119890119894119894= 1198981198941198941198981015840

119894119894

= (F119896+1minus F119896+119899+1

) sdot1

(F119896+1minus F119896+119899+1

)

= 1

(18)

For 119897 ge 119894 + 1 we get

119890119894119897=

119899minus2

sum

119896=1

1198981198941198961198981015840

119896119897

= 1198981198941198941198981015840

119894119897+ 119898119894119894+11198981015840

119894+1119897

= 119890 sdot(minus119891)119897minus119894

119890119897minus119894+1+ 119891 sdot

(minus119891)119897minus119894minus1

119890119897minus119894

= 0

(19)

We check on MMminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899 minus 2)identity matrix Similarly we can verifyMminus1M = 119868

119899minus2 Thus

the proof is completed

4 Abstract and Applied Analysis

Theorem4 Let119860119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) be a Fermat

circulant matrix Then one acquires 119860minus1119896119899= Circ(V

1 V2

V119899) where

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=

F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=

F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

times(F119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(20)

Proof Let

Ω2=

((((((((

(

1 minusℎ1015840

119896119899

F119896+1

1199091015840

31199091015840

4sdot sdot sdot 1199091015840

119899

0 1 1199101015840

31199101015840

4sdot sdot sdot 1199101015840

119899

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

))))))))

)

(21)

where

120591119896=F119896+2

F119896+1

1199091015840

3=F119896+119899

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot(120591119896F119896+119899minus F119896+1)

F119896+1

1199101015840

3=F119896+1minus 120591119896F119896+119899

ℎ119896119899

1199091015840

119894=F119896+119899+3minus119894

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot120591119896F119896+119899+3minus119894

minus F119896+119899+4minus119894

F119896+1

(119894 = 4 119899)

1199101015840

119894=F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

ℎ119896119899

(119894 = 4 119899)

ℎ1015840

119896119899=

119899minus1

sum

119894=1

F119894+119896+1

[2 (F119896+119899minus F119896)

F119896+119899+1

minus F119896+1

]

119899minus119894minus1

ℎ119896119899=

119899minus2

sum

119894=1

(F119894+119896+2

minus 120591119896F119894+119896+1

) 119910119899minus119894minus1

+ F119896+1minus 120591119896F119896+119899

(22)

We have

Σ119860119896119899Ω1Ω2= D2oplusM (23)

whereD2= diag(F

119896+1 ℎ119896119899) is a diagonal matrix andD

2oplusM

is the direct sum ofD2andM If we denoteΩ = Ω

1Ω2 then

we obtain

119860minus1

119896119899= Ω(D

minus1

2oplusMminus1

) Σ (24)

Let119860minus1119896119899= Circ(V

1 V2 V

119899) Since the last row elements

of the matrix Ω are 0 1 11991010158403minus 1 119910

1015840

4 119910

1015840

119899minus1 1199101015840

119899 according to

Lemma 3 then the last row elements of 119860minus1119896119899

are given by thefollowing equations

V2= minus

120591119896

ℎ119896119899

+1199101015840

3minus 1

F119896minus F119896+119899

V3= (1199101015840

3minus 1)

(minus119891)119899minus3

119890119899minus2+

119899

sum

119894=4

1199101015840

119894sdot(minus119891)119899minus119894

119890119899minus119894+1

V4= (1199101015840

3minus 1) [

(minus119891)119899minus4

(minus119891)119899minus3minus3 (minus119891)

119899minus3

119890119899minus2]

+

119899

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus1

119890119899minus119894minus3 (minus119891)

119899minus119894

119890119899minus119894+1]

(119905 lt 0 (minus119891)119905

= 0)

Abstract and Applied Analysis 5

V119904= (1199101015840

3minus 1) [

(minus119891)119899minus119904

119890119899minus119904+1minus3 (minus119891)

119899minus119904+1

119890119899minus119904+2+2 (minus119891)

119899minus119904+2

119890119899minus119904+3]

+

119899minus119904+5

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus119904+3

119890119899minus119894minus119904+4minus3 (minus119891)

119899minus119894minus119904+4

119890119899minus119894minus119904+5

+2 (minus119891)

119899minus119894minus119904+5

119890119899minus119894minus119904+6]

(119904 = 5 6 119899 119905 lt 0 (minus119891)119905

= 0)

V1=1

ℎ119896119899

+minus2119891 minus 3119890

1198902(1199103minus 1) +

2

1198901199104

(25)

where 119891 = F119896+119899+1

minus F119896+1

119890 = 2(F119896minus F119896+119899) according to

Lemma 1 then we have

(i) 119890 + 119891 = 0

(ii) 119890 + 2119891 = 2119896+119899+1 minus 2119896+1

Hence we obtain

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdot(F119896+1minus F119896+119899+1

)119899minus4

[2 (F119896minus F119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

times(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(26)

Thus the proof is completed

3 Determinant and Inverse ofMersenne Circulant Matrix

In this section let 119861119896119899= Circ(M

119896+1M119896+2 M

119896+119899) be a

Mersenne circulantmatrix Firstly we obtain the determinantof the matrix 119861

119896119899 Afterwards we seek out the inverse of the

matrix 119861119896119899

Theorem 5 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one obtains

det119861119896119899= M119896+1

sdot [

119899minus2

sum

119896=1

(M119896+119896+2

minus 120583119896M119896+119896+1

) 119909119899minus119896minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(27)

where 119909 = minus119888119889 119888 = 2(M119896minus M119896+119899) 119889 = M

119896+119899+1minus M119896+1

120583119896= M119896+2M119896+1

andM119896+119899

is the (119896 + 119899)th Mersenne numberFurthermore119861

119896119899is singular if and only if (1minus120572120581

119897)(1minus120573120581

119897) = 0

andM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+

where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Proof Obviously

det1198600119899= M1

sdot [

119899minus2

sum

119894=1

(M119894+2minus 1205830M119894+1) sdot [

2 (M119899minusM0)

M119899+1minusM1

]

119899minus119896minus1

+M1minus 1205830M119899] sdot [2 (M

0minusM119899)]119899minus2

(28)

6 Abstract and Applied Analysis

satisfies (27) In the following let

Γ =

((((((

(

1

minus120583119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Π1=((

(

1 0 0 sdot sdot sdot 0 0

0 119909119899minus2

0 sdot sdot sdot 0 0

0 119909119899minus3

0 sdot sdot sdot 0 minus1

d

0 119909 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(29)

be two 119899 times 119899matrices then we have

Γ119861119896119899Π1=

((((

(

M119896+1

1198911015840

119896119899minusM119896+119899

sdot sdot sdot minusM119896+3

0 119891119896119899

ℎ3

sdot sdot sdot ℎ119899

0 0 119888 sdot sdot sdot 0

0 0 0 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 119889

0 0 0 sdot sdot sdot 119888

))))

)

(30)

where119888 = 2 (M

119896minusM119896+119899)

119889 = M119896+119899+1

minusM119896+1

119909 = minus119888

119889 120583

119896=M119896+2

M119896+1

ℎ3= 120583119896M119896+119899minusM119896+1

ℎ119895= (120583119896M119896+119899+3minus119895

minusM119896+119899+4minus119895

)

(119895 = 4 5 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(31)

We get

det Γ det119861119896119899

detΠ1

= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot 119888119899minus2

(32)

besides

det Γ = (minus1)(119899minus1)(119899minus2)2

detΠ1= (minus1)

(119899minus1)(119899minus2)2

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus2

(33)

We have

det119861119896119899= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(34)

Now we discuss the singularity of the matrix 119861119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) So we have

119891 (120581119897) = M

119896+1+M119896+2120581119897+ sdot sdot sdot +M

119896+119899(120581119897)119899minus1

=M119896+1minus 2120581119897M119896minusM119896+119899+1

+ 2120581119897M119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(35)

By Lemma 1 in [14] the matrix 119861119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 That is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119861

119896119899is

nonsingular if and only ifM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899

=

0 for 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 When (1minus120572120581

119897)(1minus120573120581

119897) =

0 we obtain 120581119897= 1120572 or 120581

119897= 1120573 Let 120581

119897= 1120572 then the

eigenvalue of 119861119896119899

is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1M119899

120572119899minus1 (120572 minus 120573)= 0 (36)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119861

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 6 Let the matrixG = [119892119894119895]119899minus2

119894119895=1be of the form

119892119894119895=

2 (M119896minusM119896+119899) = 119888 119894 = 119895

M119896+119899+1

minusM119896+1= 119889 119895 = 119894 + 1

0 otherwise(37)

Then the inverseGminus1 = [1198921015840119894119895]119899minus2

119894119895=1of the matrixG is equal to

1198921015840

119894119895=

(M119896+1minusM119896+119899+1

)119895minus119894

[2 (M119896minusM119896+119899)]119895minus119894+1

=(minus119889)119895minus119894

119888119895minus119894+1 119895 ge 119894

0 119895 lt 119894

(38)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Distinctly 119888

119894119895= 0 for 119895 lt 119894

When 119894 = 119895 we obtain

119888119894119894= 1198921198941198941198921015840

119894119894= minus119889 sdot

1

minus119889= 1 (39)

Abstract and Applied Analysis 7

For 119895 ge 119894 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 1198921198941198941198921015840

119894119895+ 119892119894119894+11198921015840

119894+1119895

= 119888 sdot(minus119889)119895minus119894

119888119895minus119894+1+ 119889 sdot

(minus119889)119895minus119894minus1

119888119895minus119894= 0

(40)

We verifyGGminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899minus 2) identitymatrix Similarly we check on Gminus1G = 119868

119899minus2 Thus the proof

is completed

Theorem 7 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one acquires

119861minus1

119896119899= Circ (119906

1 1199062 119906

119899) (41)

where

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899) sdot(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(42)

where

120583119896=M119896+2

M119896+1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(43)

Proof Let

Π2=

(((((((

(

1 minus1198911015840

119896119899

M119896+1

11990931199094sdot sdot sdot 119909119899

0 1 11991031199104sdot sdot sdot 119910119899

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

)))))))

)

(44)

where

120583119896=M119896+2

M119896+1

1199093=M119896+119899

M119896+1

+1198911015840

119896119899

119891119896119899

sdot(120583119896M119896+119899minusM119896+1)

M119896+1

1199103=M119896+1minus 120583119896M119896+119899

119891119896119899

119909119894=M119896+119899+3minus119894

M119896+1

+1198911015840

119896119899

119891119896119899

sdot120583119896M119896+119899+3minus119894

minusM119896+119899+4minus119894

M119896+1

(119894 = 4 119899)

119910119894=M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

119891119896119899

(119894 = 4 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(45)

We have

Γ119861119896119899Π1Π2= D1oplusG (46)

whereD1= diag(M

119896+1 119891119896119899) is a diagonal matrix andD

1oplusG

is the direct sum ofD1andG If we denote Π = Π

1Π2 then

we obtain

119861minus1

119896119899= Π (D

minus1

1oplusGminus1

) Γ (47)

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Theorem4 Let119860119896119899= Circ(F

119896+1 F119896+2 F

119896+119899) be a Fermat

circulant matrix Then one acquires 119860minus1119896119899= Circ(V

1 V2

V119899) where

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=

F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=

F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

times(F119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(20)

Proof Let

Ω2=

((((((((

(

1 minusℎ1015840

119896119899

F119896+1

1199091015840

31199091015840

4sdot sdot sdot 1199091015840

119899

0 1 1199101015840

31199101015840

4sdot sdot sdot 1199101015840

119899

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

))))))))

)

(21)

where

120591119896=F119896+2

F119896+1

1199091015840

3=F119896+119899

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot(120591119896F119896+119899minus F119896+1)

F119896+1

1199101015840

3=F119896+1minus 120591119896F119896+119899

ℎ119896119899

1199091015840

119894=F119896+119899+3minus119894

F119896+1

+ℎ1015840

119896119899

ℎ119896119899

sdot120591119896F119896+119899+3minus119894

minus F119896+119899+4minus119894

F119896+1

(119894 = 4 119899)

1199101015840

119894=F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

ℎ119896119899

(119894 = 4 119899)

ℎ1015840

119896119899=

119899minus1

sum

119894=1

F119894+119896+1

[2 (F119896+119899minus F119896)

F119896+119899+1

minus F119896+1

]

119899minus119894minus1

ℎ119896119899=

119899minus2

sum

119894=1

(F119894+119896+2

minus 120591119896F119894+119896+1

) 119910119899minus119894minus1

+ F119896+1minus 120591119896F119896+119899

(22)

We have

Σ119860119896119899Ω1Ω2= D2oplusM (23)

whereD2= diag(F

119896+1 ℎ119896119899) is a diagonal matrix andD

2oplusM

is the direct sum ofD2andM If we denoteΩ = Ω

1Ω2 then

we obtain

119860minus1

119896119899= Ω(D

minus1

2oplusMminus1

) Σ (24)

Let119860minus1119896119899= Circ(V

1 V2 V

119899) Since the last row elements

of the matrix Ω are 0 1 11991010158403minus 1 119910

1015840

4 119910

1015840

119899minus1 1199101015840

119899 according to

Lemma 3 then the last row elements of 119860minus1119896119899

are given by thefollowing equations

V2= minus

120591119896

ℎ119896119899

+1199101015840

3minus 1

F119896minus F119896+119899

V3= (1199101015840

3minus 1)

(minus119891)119899minus3

119890119899minus2+

119899

sum

119894=4

1199101015840

119894sdot(minus119891)119899minus119894

119890119899minus119894+1

V4= (1199101015840

3minus 1) [

(minus119891)119899minus4

(minus119891)119899minus3minus3 (minus119891)

119899minus3

119890119899minus2]

+

119899

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus1

119890119899minus119894minus3 (minus119891)

119899minus119894

119890119899minus119894+1]

(119905 lt 0 (minus119891)119905

= 0)

Abstract and Applied Analysis 5

V119904= (1199101015840

3minus 1) [

(minus119891)119899minus119904

119890119899minus119904+1minus3 (minus119891)

119899minus119904+1

119890119899minus119904+2+2 (minus119891)

119899minus119904+2

119890119899minus119904+3]

+

119899minus119904+5

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus119904+3

119890119899minus119894minus119904+4minus3 (minus119891)

119899minus119894minus119904+4

119890119899minus119894minus119904+5

+2 (minus119891)

119899minus119894minus119904+5

119890119899minus119894minus119904+6]

(119904 = 5 6 119899 119905 lt 0 (minus119891)119905

= 0)

V1=1

ℎ119896119899

+minus2119891 minus 3119890

1198902(1199103minus 1) +

2

1198901199104

(25)

where 119891 = F119896+119899+1

minus F119896+1

119890 = 2(F119896minus F119896+119899) according to

Lemma 1 then we have

(i) 119890 + 119891 = 0

(ii) 119890 + 2119891 = 2119896+119899+1 minus 2119896+1

Hence we obtain

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdot(F119896+1minus F119896+119899+1

)119899minus4

[2 (F119896minus F119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

times(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(26)

Thus the proof is completed

3 Determinant and Inverse ofMersenne Circulant Matrix

In this section let 119861119896119899= Circ(M

119896+1M119896+2 M

119896+119899) be a

Mersenne circulantmatrix Firstly we obtain the determinantof the matrix 119861

119896119899 Afterwards we seek out the inverse of the

matrix 119861119896119899

Theorem 5 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one obtains

det119861119896119899= M119896+1

sdot [

119899minus2

sum

119896=1

(M119896+119896+2

minus 120583119896M119896+119896+1

) 119909119899minus119896minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(27)

where 119909 = minus119888119889 119888 = 2(M119896minus M119896+119899) 119889 = M

119896+119899+1minus M119896+1

120583119896= M119896+2M119896+1

andM119896+119899

is the (119896 + 119899)th Mersenne numberFurthermore119861

119896119899is singular if and only if (1minus120572120581

119897)(1minus120573120581

119897) = 0

andM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+

where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Proof Obviously

det1198600119899= M1

sdot [

119899minus2

sum

119894=1

(M119894+2minus 1205830M119894+1) sdot [

2 (M119899minusM0)

M119899+1minusM1

]

119899minus119896minus1

+M1minus 1205830M119899] sdot [2 (M

0minusM119899)]119899minus2

(28)

6 Abstract and Applied Analysis

satisfies (27) In the following let

Γ =

((((((

(

1

minus120583119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Π1=((

(

1 0 0 sdot sdot sdot 0 0

0 119909119899minus2

0 sdot sdot sdot 0 0

0 119909119899minus3

0 sdot sdot sdot 0 minus1

d

0 119909 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(29)

be two 119899 times 119899matrices then we have

Γ119861119896119899Π1=

((((

(

M119896+1

1198911015840

119896119899minusM119896+119899

sdot sdot sdot minusM119896+3

0 119891119896119899

ℎ3

sdot sdot sdot ℎ119899

0 0 119888 sdot sdot sdot 0

0 0 0 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 119889

0 0 0 sdot sdot sdot 119888

))))

)

(30)

where119888 = 2 (M

119896minusM119896+119899)

119889 = M119896+119899+1

minusM119896+1

119909 = minus119888

119889 120583

119896=M119896+2

M119896+1

ℎ3= 120583119896M119896+119899minusM119896+1

ℎ119895= (120583119896M119896+119899+3minus119895

minusM119896+119899+4minus119895

)

(119895 = 4 5 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(31)

We get

det Γ det119861119896119899

detΠ1

= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot 119888119899minus2

(32)

besides

det Γ = (minus1)(119899minus1)(119899minus2)2

detΠ1= (minus1)

(119899minus1)(119899minus2)2

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus2

(33)

We have

det119861119896119899= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(34)

Now we discuss the singularity of the matrix 119861119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) So we have

119891 (120581119897) = M

119896+1+M119896+2120581119897+ sdot sdot sdot +M

119896+119899(120581119897)119899minus1

=M119896+1minus 2120581119897M119896minusM119896+119899+1

+ 2120581119897M119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(35)

By Lemma 1 in [14] the matrix 119861119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 That is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119861

119896119899is

nonsingular if and only ifM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899

=

0 for 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 When (1minus120572120581

119897)(1minus120573120581

119897) =

0 we obtain 120581119897= 1120572 or 120581

119897= 1120573 Let 120581

119897= 1120572 then the

eigenvalue of 119861119896119899

is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1M119899

120572119899minus1 (120572 minus 120573)= 0 (36)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119861

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 6 Let the matrixG = [119892119894119895]119899minus2

119894119895=1be of the form

119892119894119895=

2 (M119896minusM119896+119899) = 119888 119894 = 119895

M119896+119899+1

minusM119896+1= 119889 119895 = 119894 + 1

0 otherwise(37)

Then the inverseGminus1 = [1198921015840119894119895]119899minus2

119894119895=1of the matrixG is equal to

1198921015840

119894119895=

(M119896+1minusM119896+119899+1

)119895minus119894

[2 (M119896minusM119896+119899)]119895minus119894+1

=(minus119889)119895minus119894

119888119895minus119894+1 119895 ge 119894

0 119895 lt 119894

(38)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Distinctly 119888

119894119895= 0 for 119895 lt 119894

When 119894 = 119895 we obtain

119888119894119894= 1198921198941198941198921015840

119894119894= minus119889 sdot

1

minus119889= 1 (39)

Abstract and Applied Analysis 7

For 119895 ge 119894 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 1198921198941198941198921015840

119894119895+ 119892119894119894+11198921015840

119894+1119895

= 119888 sdot(minus119889)119895minus119894

119888119895minus119894+1+ 119889 sdot

(minus119889)119895minus119894minus1

119888119895minus119894= 0

(40)

We verifyGGminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899minus 2) identitymatrix Similarly we check on Gminus1G = 119868

119899minus2 Thus the proof

is completed

Theorem 7 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one acquires

119861minus1

119896119899= Circ (119906

1 1199062 119906

119899) (41)

where

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899) sdot(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(42)

where

120583119896=M119896+2

M119896+1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(43)

Proof Let

Π2=

(((((((

(

1 minus1198911015840

119896119899

M119896+1

11990931199094sdot sdot sdot 119909119899

0 1 11991031199104sdot sdot sdot 119910119899

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

)))))))

)

(44)

where

120583119896=M119896+2

M119896+1

1199093=M119896+119899

M119896+1

+1198911015840

119896119899

119891119896119899

sdot(120583119896M119896+119899minusM119896+1)

M119896+1

1199103=M119896+1minus 120583119896M119896+119899

119891119896119899

119909119894=M119896+119899+3minus119894

M119896+1

+1198911015840

119896119899

119891119896119899

sdot120583119896M119896+119899+3minus119894

minusM119896+119899+4minus119894

M119896+1

(119894 = 4 119899)

119910119894=M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

119891119896119899

(119894 = 4 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(45)

We have

Γ119861119896119899Π1Π2= D1oplusG (46)

whereD1= diag(M

119896+1 119891119896119899) is a diagonal matrix andD

1oplusG

is the direct sum ofD1andG If we denote Π = Π

1Π2 then

we obtain

119861minus1

119896119899= Π (D

minus1

1oplusGminus1

) Γ (47)

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

V119904= (1199101015840

3minus 1) [

(minus119891)119899minus119904

119890119899minus119904+1minus3 (minus119891)

119899minus119904+1

119890119899minus119904+2+2 (minus119891)

119899minus119904+2

119890119899minus119904+3]

+

119899minus119904+5

sum

119894=4

1199101015840

119894sdot [(minus119891)119899minus119894minus119904+3

119890119899minus119894minus119904+4minus3 (minus119891)

119899minus119894minus119904+4

119890119899minus119894minus119904+5

+2 (minus119891)

119899minus119894minus119904+5

119890119899minus119894minus119904+6]

(119904 = 5 6 119899 119905 lt 0 (minus119891)119905

= 0)

V1=1

ℎ119896119899

+minus2119891 minus 3119890

1198902(1199103minus 1) +

2

1198901199104

(25)

where 119891 = F119896+119899+1

minus F119896+1

119890 = 2(F119896minus F119896+119899) according to

Lemma 1 then we have

(i) 119890 + 119891 = 0

(ii) 119890 + 2119891 = 2119896+119899+1 minus 2119896+1

Hence we obtain

V1=1

ℎ119896119899

+ (F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdotminusF119896+119899+1

+ 3F119896+119899+ F119896+1minus 3F119896

2ℎ119896119899(F119896minus F119896+119899)2

+(F119896+119899minus 120591119896F119896+119899minus1

)

ℎ119896119899(F119896minus F119896+119899)

V2=

minus2119896

minus ℎ119896119899F119896+1

F119896+1ℎ119896119899(F119896minus F119896+119899)

V3=F119896+1minus 120591119896F119896+119899minus ℎ119896119899

ℎ119896119899

sdot(F119896+1minus F119896+119899+1

)119899minus3

[2 (F119896minus F119896+119899)]119899minus2+1

ℎ119896119899

sdot

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

sdot(F119896+1minus F119896+119899+1

)119899minus119894

[2 (F119896minus F119896+119899)]119899minus119894+1

V4=F119896+119899+2

minus F119896+2

ℎ119896119899

times [(F119896+1minus 120591119896F119896+119899minus ℎ119896119899)

sdot(F119896+1minus F119896+119899+1

)119899minus4

[2 (F119896minus F119896+119899)]119899minus2

+

119899

sum

119894=4

(F119896+119899+4minus119894

minus 120591119896F119896+119899+3minus119894

)

times(F119896+1minus F119896+119899+1

)119899minus119894minus1

[2 (F119896minus F119896+119899)]119899minus119894+1

]

V119904= 0 (119904 = 5 6 119899)

(26)

Thus the proof is completed

3 Determinant and Inverse ofMersenne Circulant Matrix

In this section let 119861119896119899= Circ(M

119896+1M119896+2 M

119896+119899) be a

Mersenne circulantmatrix Firstly we obtain the determinantof the matrix 119861

119896119899 Afterwards we seek out the inverse of the

matrix 119861119896119899

Theorem 5 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one obtains

det119861119896119899= M119896+1

sdot [

119899minus2

sum

119896=1

(M119896+119896+2

minus 120583119896M119896+119896+1

) 119909119899minus119896minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(27)

where 119909 = minus119888119889 119888 = 2(M119896minus M119896+119899) 119889 = M

119896+119899+1minus M119896+1

120583119896= M119896+2M119896+1

andM119896+119899

is the (119896 + 119899)th Mersenne numberFurthermore119861

119896119899is singular if and only if (1minus120572120581

119897)(1minus120573120581

119897) = 0

andM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+

where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Proof Obviously

det1198600119899= M1

sdot [

119899minus2

sum

119894=1

(M119894+2minus 1205830M119894+1) sdot [

2 (M119899minusM0)

M119899+1minusM1

]

119899minus119896minus1

+M1minus 1205830M119899] sdot [2 (M

0minusM119899)]119899minus2

(28)

6 Abstract and Applied Analysis

satisfies (27) In the following let

Γ =

((((((

(

1

minus120583119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Π1=((

(

1 0 0 sdot sdot sdot 0 0

0 119909119899minus2

0 sdot sdot sdot 0 0

0 119909119899minus3

0 sdot sdot sdot 0 minus1

d

0 119909 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(29)

be two 119899 times 119899matrices then we have

Γ119861119896119899Π1=

((((

(

M119896+1

1198911015840

119896119899minusM119896+119899

sdot sdot sdot minusM119896+3

0 119891119896119899

ℎ3

sdot sdot sdot ℎ119899

0 0 119888 sdot sdot sdot 0

0 0 0 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 119889

0 0 0 sdot sdot sdot 119888

))))

)

(30)

where119888 = 2 (M

119896minusM119896+119899)

119889 = M119896+119899+1

minusM119896+1

119909 = minus119888

119889 120583

119896=M119896+2

M119896+1

ℎ3= 120583119896M119896+119899minusM119896+1

ℎ119895= (120583119896M119896+119899+3minus119895

minusM119896+119899+4minus119895

)

(119895 = 4 5 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(31)

We get

det Γ det119861119896119899

detΠ1

= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot 119888119899minus2

(32)

besides

det Γ = (minus1)(119899minus1)(119899minus2)2

detΠ1= (minus1)

(119899minus1)(119899minus2)2

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus2

(33)

We have

det119861119896119899= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(34)

Now we discuss the singularity of the matrix 119861119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) So we have

119891 (120581119897) = M

119896+1+M119896+2120581119897+ sdot sdot sdot +M

119896+119899(120581119897)119899minus1

=M119896+1minus 2120581119897M119896minusM119896+119899+1

+ 2120581119897M119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(35)

By Lemma 1 in [14] the matrix 119861119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 That is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119861

119896119899is

nonsingular if and only ifM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899

=

0 for 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 When (1minus120572120581

119897)(1minus120573120581

119897) =

0 we obtain 120581119897= 1120572 or 120581

119897= 1120573 Let 120581

119897= 1120572 then the

eigenvalue of 119861119896119899

is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1M119899

120572119899minus1 (120572 minus 120573)= 0 (36)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119861

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 6 Let the matrixG = [119892119894119895]119899minus2

119894119895=1be of the form

119892119894119895=

2 (M119896minusM119896+119899) = 119888 119894 = 119895

M119896+119899+1

minusM119896+1= 119889 119895 = 119894 + 1

0 otherwise(37)

Then the inverseGminus1 = [1198921015840119894119895]119899minus2

119894119895=1of the matrixG is equal to

1198921015840

119894119895=

(M119896+1minusM119896+119899+1

)119895minus119894

[2 (M119896minusM119896+119899)]119895minus119894+1

=(minus119889)119895minus119894

119888119895minus119894+1 119895 ge 119894

0 119895 lt 119894

(38)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Distinctly 119888

119894119895= 0 for 119895 lt 119894

When 119894 = 119895 we obtain

119888119894119894= 1198921198941198941198921015840

119894119894= minus119889 sdot

1

minus119889= 1 (39)

Abstract and Applied Analysis 7

For 119895 ge 119894 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 1198921198941198941198921015840

119894119895+ 119892119894119894+11198921015840

119894+1119895

= 119888 sdot(minus119889)119895minus119894

119888119895minus119894+1+ 119889 sdot

(minus119889)119895minus119894minus1

119888119895minus119894= 0

(40)

We verifyGGminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899minus 2) identitymatrix Similarly we check on Gminus1G = 119868

119899minus2 Thus the proof

is completed

Theorem 7 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one acquires

119861minus1

119896119899= Circ (119906

1 1199062 119906

119899) (41)

where

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899) sdot(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(42)

where

120583119896=M119896+2

M119896+1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(43)

Proof Let

Π2=

(((((((

(

1 minus1198911015840

119896119899

M119896+1

11990931199094sdot sdot sdot 119909119899

0 1 11991031199104sdot sdot sdot 119910119899

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

)))))))

)

(44)

where

120583119896=M119896+2

M119896+1

1199093=M119896+119899

M119896+1

+1198911015840

119896119899

119891119896119899

sdot(120583119896M119896+119899minusM119896+1)

M119896+1

1199103=M119896+1minus 120583119896M119896+119899

119891119896119899

119909119894=M119896+119899+3minus119894

M119896+1

+1198911015840

119896119899

119891119896119899

sdot120583119896M119896+119899+3minus119894

minusM119896+119899+4minus119894

M119896+1

(119894 = 4 119899)

119910119894=M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

119891119896119899

(119894 = 4 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(45)

We have

Γ119861119896119899Π1Π2= D1oplusG (46)

whereD1= diag(M

119896+1 119891119896119899) is a diagonal matrix andD

1oplusG

is the direct sum ofD1andG If we denote Π = Π

1Π2 then

we obtain

119861minus1

119896119899= Π (D

minus1

1oplusGminus1

) Γ (47)

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

satisfies (27) In the following let

Γ =

((((((

(

1

minus120583119896

1

2 1 minus3

0 0 1 minus3 2

c c c0 1 c c0 1 minus3 c 0

0 1 minus3 2

))))))

)

Π1=((

(

1 0 0 sdot sdot sdot 0 0

0 119909119899minus2

0 sdot sdot sdot 0 0

0 119909119899minus3

0 sdot sdot sdot 0 minus1

d

0 119909 0 sdot sdot sdot 0 0

0 1 minus1 sdot sdot sdot 0 0

))

)

(29)

be two 119899 times 119899matrices then we have

Γ119861119896119899Π1=

((((

(

M119896+1

1198911015840

119896119899minusM119896+119899

sdot sdot sdot minusM119896+3

0 119891119896119899

ℎ3

sdot sdot sdot ℎ119899

0 0 119888 sdot sdot sdot 0

0 0 0 sdot sdot sdot 0

d

0 0 0 sdot sdot sdot 119889

0 0 0 sdot sdot sdot 119888

))))

)

(30)

where119888 = 2 (M

119896minusM119896+119899)

119889 = M119896+119899+1

minusM119896+1

119909 = minus119888

119889 120583

119896=M119896+2

M119896+1

ℎ3= 120583119896M119896+119899minusM119896+1

ℎ119895= (120583119896M119896+119899+3minus119895

minusM119896+119899+4minus119895

)

(119895 = 4 5 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(31)

We get

det Γ det119861119896119899

detΠ1

= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot 119888119899minus2

(32)

besides

det Γ = (minus1)(119899minus1)(119899minus2)2

detΠ1= (minus1)

(119899minus1)(119899minus2)2

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus2

(33)

We have

det119861119896119899= M119896+1sdot [

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899] sdot (minus119889)

119899minus2

(34)

Now we discuss the singularity of the matrix 119861119896119899

The roots of polynomial 119892(119909) = 119909119899

minus 1 are 120581119897(119897 =

1 2 119899) where 120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) So we have

119891 (120581119897) = M

119896+1+M119896+2120581119897+ sdot sdot sdot +M

119896+119899(120581119897)119899minus1

=M119896+1minus 2120581119897M119896minusM119896+119899+1

+ 2120581119897M119896+119899

(1 minus 120572120581119897) (1 minus 120573120581

119897)

(35)

By Lemma 1 in [14] the matrix 119861119896119899

is nonsingular if and onlyif 119891(120581

119897) = 0 That is when (1 minus 120572120581

119897)(1 minus 120573120581

119897) = 0 119861

119896119899is

nonsingular if and only ifM119896+1minus2120581119897M119896minusM119896+119899+1

+2120581119897M119896+119899

=

0 for 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 When (1minus120572120581

119897)(1minus120573120581

119897) =

0 we obtain 120581119897= 1120572 or 120581

119897= 1120573 Let 120581

119897= 1120572 then the

eigenvalue of 119861119896119899

is

119891 (120581119897) =119899120572119896+119899

minus 120573119896+1M119899

120572119899minus1 (120572 minus 120573)= 0 (36)

for 120572 = 2 120573 = 1 119896 isin 119873 119899 isin 119873+ 119897 = 1 2 119899 so 119861

119896119899is

nonsingular The arguments for 120581119897= 1120573 are similar Thus

the proof is completed

Lemma 6 Let the matrixG = [119892119894119895]119899minus2

119894119895=1be of the form

119892119894119895=

2 (M119896minusM119896+119899) = 119888 119894 = 119895

M119896+119899+1

minusM119896+1= 119889 119895 = 119894 + 1

0 otherwise(37)

Then the inverseGminus1 = [1198921015840119894119895]119899minus2

119894119895=1of the matrixG is equal to

1198921015840

119894119895=

(M119896+1minusM119896+119899+1

)119895minus119894

[2 (M119896minusM119896+119899)]119895minus119894+1

=(minus119889)119895minus119894

119888119895minus119894+1 119895 ge 119894

0 119895 lt 119894

(38)

Proof Let 119888119894119895= sum119899minus2

119896=11198921198941198961198921015840

119896119895 Distinctly 119888

119894119895= 0 for 119895 lt 119894

When 119894 = 119895 we obtain

119888119894119894= 1198921198941198941198921015840

119894119894= minus119889 sdot

1

minus119889= 1 (39)

Abstract and Applied Analysis 7

For 119895 ge 119894 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 1198921198941198941198921015840

119894119895+ 119892119894119894+11198921015840

119894+1119895

= 119888 sdot(minus119889)119895minus119894

119888119895minus119894+1+ 119889 sdot

(minus119889)119895minus119894minus1

119888119895minus119894= 0

(40)

We verifyGGminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899minus 2) identitymatrix Similarly we check on Gminus1G = 119868

119899minus2 Thus the proof

is completed

Theorem 7 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one acquires

119861minus1

119896119899= Circ (119906

1 1199062 119906

119899) (41)

where

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899) sdot(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(42)

where

120583119896=M119896+2

M119896+1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(43)

Proof Let

Π2=

(((((((

(

1 minus1198911015840

119896119899

M119896+1

11990931199094sdot sdot sdot 119909119899

0 1 11991031199104sdot sdot sdot 119910119899

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

)))))))

)

(44)

where

120583119896=M119896+2

M119896+1

1199093=M119896+119899

M119896+1

+1198911015840

119896119899

119891119896119899

sdot(120583119896M119896+119899minusM119896+1)

M119896+1

1199103=M119896+1minus 120583119896M119896+119899

119891119896119899

119909119894=M119896+119899+3minus119894

M119896+1

+1198911015840

119896119899

119891119896119899

sdot120583119896M119896+119899+3minus119894

minusM119896+119899+4minus119894

M119896+1

(119894 = 4 119899)

119910119894=M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

119891119896119899

(119894 = 4 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(45)

We have

Γ119861119896119899Π1Π2= D1oplusG (46)

whereD1= diag(M

119896+1 119891119896119899) is a diagonal matrix andD

1oplusG

is the direct sum ofD1andG If we denote Π = Π

1Π2 then

we obtain

119861minus1

119896119899= Π (D

minus1

1oplusGminus1

) Γ (47)

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

For 119895 ge 119894 + 1 we obtain

119888119894119895=

119899minus2

sum

119896=1

1198921198941198961198921015840

119896119895= 1198921198941198941198921015840

119894119895+ 119892119894119894+11198921015840

119894+1119895

= 119888 sdot(minus119889)119895minus119894

119888119895minus119894+1+ 119889 sdot

(minus119889)119895minus119894minus1

119888119895minus119894= 0

(40)

We verifyGGminus1 = 119868119899minus2

where 119868119899minus2

is (119899 minus 2) times (119899minus 2) identitymatrix Similarly we check on Gminus1G = 119868

119899minus2 Thus the proof

is completed

Theorem 7 Let 119861119896119899

= Circ(M119896+1M119896+2 M

119896+119899) be a

Mersenne circulant matrix Then one acquires

119861minus1

119896119899= Circ (119906

1 1199062 119906

119899) (41)

where

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899) sdot(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(42)

where

120583119896=M119896+2

M119896+1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(43)

Proof Let

Π2=

(((((((

(

1 minus1198911015840

119896119899

M119896+1

11990931199094sdot sdot sdot 119909119899

0 1 11991031199104sdot sdot sdot 119910119899

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

)))))))

)

(44)

where

120583119896=M119896+2

M119896+1

1199093=M119896+119899

M119896+1

+1198911015840

119896119899

119891119896119899

sdot(120583119896M119896+119899minusM119896+1)

M119896+1

1199103=M119896+1minus 120583119896M119896+119899

119891119896119899

119909119894=M119896+119899+3minus119894

M119896+1

+1198911015840

119896119899

119891119896119899

sdot120583119896M119896+119899+3minus119894

minusM119896+119899+4minus119894

M119896+1

(119894 = 4 119899)

119910119894=M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

119891119896119899

(119894 = 4 119899)

1198911015840

119896119899=

119899minus1

sum

119894=1

M119894+119896+1

[2 (M119896+119899minusM119896)

M119896+119899+1

minusM119896+1

]

119899minus119894minus1

119891119896119899=

119899minus2

sum

119894=1

(M119894+119896+2

minus 120583119896M119894+119896+1

) 119909119899minus119894minus1

+M119896+1minus 120583119896M119896+119899

(45)

We have

Γ119861119896119899Π1Π2= D1oplusG (46)

whereD1= diag(M

119896+1 119891119896119899) is a diagonal matrix andD

1oplusG

is the direct sum ofD1andG If we denote Π = Π

1Π2 then

we obtain

119861minus1

119896119899= Π (D

minus1

1oplusGminus1

) Γ (47)

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

Let119861minus1119896119899= Circ(119906

1 1199062 119906

119899) Since the last row elements

of the matrix Π are 0 1 1199103minus 1 119910

4 119910

119899minus1 119910119899 according to

Lemma 6 then the last row elements of 119861minus1119896119899

are given by thefollowing equations

1199062= minus

120583119896

119891119896119899

+2 (1199103minus 1)

119888

1199063= (1199103minus 1) sdot

(minus119889)119899minus3

119888119899minus2+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894

119888119899minus119894+1

1199064= (1199103minus 1) sdot [

(minus119889)119899minus4

119888119899minus3minus3 (minus119889)

119899minus3

119888119899minus2]

+

119899

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus1

119888119899minus119894minus3 (minus119889)

119899minus119894

119888119899minus119894+1]

= (1199103minus 1) sdot

(minus119889)119899minus4

119888119899minus2(119888 + 3119889)

+

119899

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus1

119888119899minus119894+1(119888 + 3119889) (119905 lt 0 (minus119889)

119905

= 0)

119906119904= (1199103minus 1)

sdot [(minus119889)119899minus119904

119888119899minus119904+1minus3 (minus119889)

119899minus119904+1

119888119899minus119904+2+2 (minus119889)

119899minus119904+2

119888119899minus119904+3]

+

119899minus119904+5

sum

119894=4

119910119894sdot [(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+4

minus3 (minus119889)

119899minus119894minus119904+4

119888119899minus119894minus119904+5+2 (minus119889)

119899minus119894minus119904+5

119888119899minus119894minus119904+6]

= [(1199103minus 1) sdot

(minus119889)119899minus119904

119888119899minus119904+3+

119899minus119904+5

sum

119894=4

119910119894sdot(minus119889)119899minus119894minus119904+3

119888119899minus119894minus119904+6]

times (119888 + 2119889) (119888 + 119889) (119904 = 5 6 119899 119905 lt 0 (minus119889)119905

= 0)

1199061=1

119891119896119899

+minus2119889 minus 3119888

1198882(1199103minus 1) +

2

1198881199104

(48)

where 119889 = M119896+119899+1

minus M119896+1 119888 = 2(M

119896minus M119896+119899) according to

Lemma 1 then we have

(i) 119888 + 119889 = 0(ii) 119888 + 2119889 = 2119896+119899+1 minus 2119896+1

We get

1199061=1

119891119896119899

+ (M119896+1minus 120583119896M119896+119899minus 119891119896119899)

sdotminusM119896+119899+1

+ 3M119896+119899+M119896+1minus 3M119896

2119891119896119899(M119896minusM119896+119899)2

+(M119896+119899minus 120583119896M119896+119899minus1

)

119891119896119899(M119896minusM119896+119899)

1199062=

2119896

minus 119891119896119899M119896+1

M119896+1119891119896119899(M119896minusM119896+119899)

1199063=M119896+1minus 120583119896M119896+119899minus 119891119896119899

119891119896119899

sdot(M119896+1minusM119896+119899+1

)119899minus3

[2 (M119896minusM119896+119899)]119899minus2+1

119891119896119899

sdot

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894

[2 (M119896minusM119896+119899)]119899minus119894+1

1199064=M119896+119899+2

minusM119896+2

119891119896119899

times [(M119896+1minus 120583119896M119896+119899minus 119891119896119899)(M119896+1minusM119896+119899+1

)119899minus4

[2 (M119896minusM119896+119899)]119899minus2

+

119899

sum

119894=4

(M119896+119899+4minus119894

minus 120583119896M119896+119899+3minus119894

)

sdot(M119896+1minusM119896+119899+1

)119899minus119894minus1

[2 (M119896minusM119896+119899)]119899minus119894+1

]

119906119904= 0 (119904 = 5 6 119899)

(49)Thus the proof is completed

4 Determinants and Inverses of Fermat andMersenne Left Circulant Matrix

In this section let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) and

1198611015840

119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) beMersenne and Fermat

left circulant matrices respectively By using the obtainedconclusions we give a determinant formula for the matrix1198601015840

119896119899and 1198611015840

119896119899 In addition the inverse matrices of 1198601015840

119896119899and

1198611015840

119896119899are derivedAccording to Lemma 2 in [14] and Theorems 2 4 5 and

7 we can obtain the following theorems

Theorem 8 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then one has

det1198601015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdot F119896+1

sdot [

[

119899minus2

sum

119895=1

(F119895+119896+2

minus 120591119896F119895+119896+1

) 119901119899minus119895minus1

+ F119896+1minus 120591119896F119896+119899

]

]

sdot (F119896+1minus F119896+119899+1

)119899minus2

(50)where 120591

119896= F119896+2F119896+1

119901 = 2(F119896+119899minusF119896)(F119896+119899+1

minusF119896+1) and F

119896+119899

is the (119896+119899)th Fermat numberMoreover1198601015840119896119899

is singular if and

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

only if (1minus120572120581119897)(1minus120573120581

119897) = 0 and F

119896+1minus2120581119897F119896minusF119896+119899+1

+2120581119897F119896+119899=

0 for 119896 isin 119873 119899 isin 119873+ where 120581

119897= cos(2119897120587119899) + 119894 sin(2119897120587119899)

119897 = 1 2 119899

Theorem 9 Let 1198601015840119896119899

= LCirc(F119896+1 F119896+2 F

119896+119899) be a

Fermat left circulant matrix then

(1198601015840

119896119899)minus1

= Circ minus1 (F119896+1 F119896+2 F

119896+119899) sdot Δ

= Circ (V1 V2 V

119899) sdot Δ

= LCirc (V1 V119899 V

2)

(51)

where V1 V2 V

119899were given by Theorem 4 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

Theorem 10 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

det1198611015840119896119899= (minus1)

(119899minus1)(119899minus2)2

sdotM119896+1

sdot [

[

M119896+1minus 120583119896M119896+119899

+

119899minus2

sum

119895=1

(M119895+119896+2

minus 120583119896M119895+119896+1

) 119911119899minus119895minus1]

]

sdot (M119896+1minusM119896+119899+1

)119899minus2

(52)

where 120583119896= M119896+2M119896+1

119911 = 2(M119896+119899minusM119896)(M119896+119899+1

minusM119896+1)

andM119896+119899

is the (119896+119899)th Mersenne number Furthermore 1198611015840119896119899

is singular if and only if (1 minus 120572120581119897)(1 minus 120573120581

119897) = 0 and M

119896+1minus

2120581119897M119896minus M119896+119899+1

+ 2120581119897M119896+119899= 0 for 119896 isin 119873 119899 isin 119873

+ where

120581119897= cos(2119897120587119899) + 119894 sin(2119897120587119899) 119897 = 1 2 119899

Theorem 11 Let 1198611015840119896119899= LCirc(M

119896+1M119896+2 M

119896+119899) be a

Mersenne left circulant matrix then one has

(1198611015840

119896119899)minus1

= Circ minus1 (M119896+1M119896+2 M

119896+119899) sdot Δ

= Circ (1199061 1199062 119906

119899) sdot Δ

= LCirc (1199061 119906119899 119906

2)

(53)

where 1199061 1199062 119906

119899were given by Theorem 7 and Δ =

LCirc(1 0 0) was given by Lemma 2 in [14]

5 Conclusion

In this paper we present the exact determinants and theinverse matrices of Fermat and Mersenne circulant matrixrespectively Furthermore we give the exact determinantsand the inverse matrices of Fermat and Mersenne left circu-lant matrix Meanwhile the nonsingularity of these specialmatrices is discussed On the basis of circulant matricestechnology we will develop solving the problems in [19ndash22]

Conflict of Interests

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

Acknowledgment

This work was supported by the GRRC Program of GyeonggiProvince ((GRRC SUWON 2014-B4) Development of CloudComputing-Based Intelligent Video Security SurveillanceSystem with Active Tracking Technology) Their support isgratefully acknowledged

References

[1] C Zhang G Dangelmayr and I Oprea ldquoStoring cyclesin Hopfield-type networks with pseudoinverse learning ruleadmissibility and network topologyrdquo Neural Networks vol 46pp 283ndash298 2013

[2] D Rocchesso and J O Smith ldquoCirculant and elliptic feedbackdelay networks for artificial reverberationrdquo IEEE Transactionson Speech and Audio Processing vol 5 no 1 pp 51ndash63 1997

[3] Y Jing and H Jafarkhani ldquoDistributed differential space-timecoding for wireless relay networksrdquo IEEE Transactions onCommunications vol 56 no 7 pp 1092ndash1100 2008

[4] M Basic ldquoCharacterization of quantum circulant networkshaving perfect state transferrdquo Quantum Information Processingvol 12 no 1 pp 345ndash364 2013

[5] G Wang F Gao Y-C Wu and C Tellambura ldquoJoint CFO andchannel estimation for OFDM-based two-way relay networksrdquoIEEE Transactions on Wireless Communications vol 10 no 2pp 456ndash465 2011

[6] J Li J Yuan R Malaney M Xiao and W Chen ldquoFull-diversity binary frame-wise network coding for multiple-source multiple-relay networks over slow-fading channelsrdquoIEEE Transactions on Vehicular Technology vol 61 no 3 pp1346ndash1360 2012

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

[8] Z L Jiang and Z X Zhou Circulant Matrices ChengduTechnology University Publishing Company Chengdu China1999

[9] Z Jiang ldquoOn the minimal polynomials and the inverses of mul-tilevel scaled factor circulant matricesrdquo Abstract and AppliedAnalysis vol 2014 Article ID 521643 10 pages 2014

[10] Z Jiang T Xu and F Lu ldquoIsomorphic operators and functionalequations for the skew-circulant algebrardquo Abstract and AppliedAnalysis vol 2014 Article ID 418194 8 pages 2014

[11] J Li Z L Jiang and F L Lu ldquoDeterminants norms and thespread of circulant matrices with Tribonacci and generalizedLucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 381829 9 pages 2014

[12] Z L Jiang ldquoNonsingularity of two classes of cyclic matricesrdquoMathematics in Practice and Theory no 2 pp 52ndash58 1995

[13] J-J Yao and Z-L Jiang ldquoThe determinants inverses norm andspread of skew circulant typematrices involving any continuousLucas numbersrdquo Journal of Applied Mathematics vol 2014Article ID 239693 10 pages 2014

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

[14] Z Jiang Y Gong and Y Gao ldquoInvertibility and explicitinverses of circulant-type matrices with k-Fibonacci and k-Lucas numbersrdquoAbstract andAppliedAnalysis vol 2014 ArticleID 238953 9 pages 2014

[15] L Dazheng ldquoFibonacci-Lucas quasi-cyclic matricesrdquo TheFibonacci Quarterly vol 40 no 3 pp 280ndash286 2002

[16] D Bozkurt and T-Y Tam ldquoDeterminants and inverses ofcirculant matrices with JACobsthal and JACobsthal-LucasNumbersrdquo Applied Mathematics and Computation vol 219 no2 pp 544ndash551 2012

[17] A F Horadam ldquoFurther appearence of the Fibonacci sequencerdquoThe Fibonacci Quarterly vol 1 no 4 pp 41ndash42 1963

[18] A Ipek and K Arı ldquoOn Hessenberg and pentadiagonal deter-minants related with FIBonacci and FIBonacci-like numbersrdquoApplied Mathematics and Computation vol 229 pp 433ndash4392014

[19] H Dong Z Wang and H Gao ldquoDistributed Hinfin

filteringfor a class of markovian jump nonlinear time-delay systemsover lossy sensor networksrdquo IEEE Transactions on IndustrialElectronics vol 60 no 10 pp 4665ndash4672 2013

[20] Z Wang H Dong B Shen and H Gao ldquoFinite-horizon 119867infin

filtering with missing measurements and quantization effectsrdquoIEEE Transactions on Automatic Control vol 58 no 7 pp 1707ndash1718 2013

[21] D Ding Z Wang J Hu and H Shu ldquoDissipative control forstate-saturated discrete time-varying systems with randomlyoccurring nonlinearities and missing measurementsrdquo Interna-tional Journal of Control vol 86 no 4 pp 674ndash688 2013

[22] J Hu Z Wang B Shen and H Gao ldquoQuantised recursivefiltering for a class of nonlinear systems with multiplicativenoises and missing measurementsrdquo International Journal ofControl vol 86 no 4 pp 650ndash663 2013

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

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

Mathematical PhysicsAdvances in

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

Operations ResearchAdvances in

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

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of