# Also, the upper and lower bounds for the spectral norm of the Hadamard inverse of that matrix are obtained

## Teljes szövegt

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Vol. 19 (2018), No. 1, pp. 505–515 DOI: 10.18514/MMN.2018.1779

ONK-CIRCULANT MATRICES INVOLVING THE FIBONACCI NUMBERS

Abstract. Letkbe a nonzero complex number. In this paper we consider ak-circulant matrix whose first row is.F1; F2; : : : ; Fn/, whereFnis thent hFibonacci number, and investigate the eigenvalues and Euclidean (or Frobenius) norm of that matrix. Also, the upper and lower bounds for the spectral norm of the Hadamard inverse of that matrix are obtained.

2010Mathematics Subject Classification: 15B05; 11B39; 15A18; 15A60

Keywords: k-circulant matrix, Fibonacci numbers, eigenvalues, Euclidean norm, spectral norm

1. INTRODUCTION

In this paper,kis a nonzero complex number andCnndenotes the set of all complex matrices of ordern. Anynt hroot ofkand any primitivent hroot of unity are denoted by and!, respectively. Symbolsj; jD0; n 1, jCj,kCkE, kCk2 andCı 1 stand for the eigenvalues, the determinant, the Euclidean norm, the spectral norm and the Hadamard inverse ofC2Cnn, respectively. Namely, forCD

ci;j

2Cnn, kCkED.

n

X

i;jD1

jci;j j2/12; kCk2Dq

1maxini.CC /; where C is the conjugate transpose ofC, andCı 1Dh

ci;j1i :

Definition 1. A matrix C of order n with the first row .c0; c1; c2; : : : ; cn 1/ is called ak-circulant matrixifC has the following form:

C D 2 6 6 6 6 6 6 6 4

c0 c1 c2 cn 2 cn 1

kcn 1 c0 c1 cn 3 cn 2

kcn 2 kcn 1 c0 cn 4 cn 3

::: ::: ::: : :: ::: ::: kc2 kc3 kc4 c0 c1

kc1 kc2 kc3 kcn 1 c0

3 7 7 7 7 7 7 7 5

: (1.1)

c 2018 Miskolc University Press

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We shall write CDci rcnfk.c0; c1; c2; : : : ; cn 1/gif a matrixC has the form (1.1).

The designation for the order of a matrix can be omitted if the dimension of a matrix is known. Circulant matrices arek-circulant matrices forkD1and skew circulant matrices arek-circulant matrices forkD 1.

The Fibonacci numbersfFngsatisfy the following recursive relation:

FnDFn 2CFn 1; n2; (1.2)

with initial conditionsF0D0andF1D1.

Let˛andˇbe the roots of the equationx2 x 1D0 i.e.

˛D1Cp 5

2 ; ˇD1 p 5

2 ; ˛ˇD 1; ˛CˇD1 and ˛ ˇDp

5: (1.3) Binet’s formulafor the Fibonacci numbers is:

FnD ˛n ˇn

˛ ˇ D 1

p5

1Cp 5 2

!n

1 p

5 2

!n!

: (1.4)

Let us mention that the Lucas numbersfLngsatisfy the same recursive relation (as the Fibonacci numbers) but with initial conditionsL0D2andL1D1and

LnnnD 1Cp 5 2

!n

C 1 p 5 2

!n

(1.5) isBinet’s formulafor the Lucas numbers.

The following identities hold for the Fibonacci numbers:

n

X

iD1

Fi DFnC2 1 and

n

X

iD1

Fi2DFnFnC1

DL2nC1 . 1/n 5

: (1.6) More information about these numbers can be found in [2,3,8–10,12,14,15].

In [4,13] the authors investigated the determinants and inverses of circulant (of skew circulant) matrices whose first rows are.F1; F2; : : : ; Fn/and.L1; L2; : : : ; Ln/. The paper  is devoted tok-circulant matrices with the Fibonacci and Lucas numbers, and an upper bound estimation of the spectral norm for such matrices was given in that paper.

The motivation for this paper is the paper  in which the authors investigatedk- circulant matrices with the generalizedr-Horadam numbersfHr;ngwhich are defined as follows:

Hr;nC2Df .r/Hr;nC1Cg.r/Hr;n; n0;

wherer2RC,Hr;0Da; Hr;1Db,a; b2Randf2.r/C4g.r/ > 0, and presented the upper and lower bounds for the spectral norms of such matrices.

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Theorem 1(Theorem 5. ). LetHDci rcfk.Hr;0; Hr;1; : : : ; Hr;n 1/g. a) Ifjkj 1, then

v u u t

n 1

X

iD0

Hr;i2 kHk2 v u u

t a2.1 jkj2/Cjkj2

n 1

X

iD0

Hr;i2

! 1 a2C

n 1

X

iD0

Hr;i2

! (1.7)

b) Ifjkj< 1, then

jkj v u u t

n 1

X

iD0

Hr;i2 kHk2 v u u tn

n 1

X

iD0

Hr;i2 : (1.8)

Also, in , the formulae for the eigenvalues and determinant of ak-circulant matrix with the generalizedr-Horadam numbers were derived.

Theorem 2(Theorem 7. ). LetHDci rcfk.Hr;0; Hr;1; : : : ; Hr;n 1/g. Then the eigenvalues ofH are:

j D kHr;nC.g.r/kHr;n 1 bCaf .r// ! j Hr;0

g.r/. ! j/2Cf .r/ ! j 1 ; jD0; n 1 : (1.9) Theorem 3(Theorem 8. ). LetHDci rcfk.Hr;0; Hr;1; : : : ; Hr;n 1/g. Then the determinant ofH is:

jHj D.Hr;0 kHr;n/n .g.r/kHr;n 1 bCaf .r//nk

.1 k˛n/.1 kˇn/ ; (1.10)

where˛andˇare the roots of the equationx2 f .r/x g.r/D0.

Let us also mention that, in , the authors considered circulant matrices with the generalizedr-Horadam numbers and obtained the determinants and inverses of such matrices.

In the present paper, we consider the matrix

F Dci rcfk.F1; F2; : : : ; Fn/g (1.11) and improve the result in relation to the eigenvalues of (1.11) which can be obtained from (1.9) because the authors did not consider the case when the denominator is equal to zero. Also, we determine the Euclidean norm of (1.11) and derive the upper and lower bounds for the spectral norm of the Hadamard inverse of (1.11). The results are presented in the next section.

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2. RESULTS

Let us recall that is anynt hroot ofkand!is any primitivent hroot of unity. Also, throughout this section,˛andˇare the roots of the equationx2 x 1D0. In order to obtain the eigenvalues of (1.11), we need the following lemma.

Lemma 1(Lemma 4. ). The eigenvalues ofCDci rcfk.c0; c1; c2; : : : ; cn 1/g are:

jD

n 1

X

iD0

ci. ! j/i; jD0; n 1 : (2.1) Moreover, in this case:

ciD 1 n

n 1

X

jD0

j. ! j/ i; iD0; n 1 : (2.2) Theorem 4. LetF be the matrix as in(1.11). Then the eigenvalues ofF are given by the following formulae:

1) If ! jD˛1, then

j D 1 p5

n˛C1 . 1/nˇ2n p5

; (2.3)

2) If ! jDˇ1, then

j D 1 p5

1 . 1/n˛2n p5 nˇ

; (2.4)

3) If ! j¤˛1 and ! j¤ˇ1, then

j DkFnC1 1CkFn ! j

. ! j/2C ! j 1 : (2.5)

Proof. Based on Lemma1and (1.4), it follows:

1) Suppose that ! jD˛1. Then, j D

n 1

X

iD0

FiC1. ! j/i D 1 p5

n 1

X

iD0

h

˛iC1 ˇiC1i .1

˛/i D 1

p5

"

˛

n 1

X

iD0

1 ˇ

n 1

X

iD0

˛/i

# D 1

p5

"

n˛ ˇ1 .ˇ˛/n 1 ˇ˛

#

D 1 p5

n˛C1 . 1/nˇ2n p5

;

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2) Suppose that ! jDˇ1. Then, j D

n 1

X

iD0

FiC1. ! j/iD 1 p5

n 1

X

iD0

h

˛iC1 ˇiC1 i

.1 ˇ/i D 1

p5

"

˛

n 1

X

iD0

.˛ ˇ/i ˇ

n 1

X

iD0

1

# D 1

p5

"

˛

1 .ˇ˛/n 1 ˛ˇ

#

D 1 p5

1 . 1/n˛2n p5 nˇ

;

3) Suppose that ! j¤˛1 and ! j¤ˇ1. Then,j follows from (1.9).

The previously obtained result will be illustrated by the following example.

Example1. Let

FDci rcf9 4p

5.1; 1; 2; 3; 5; 8/g i.e.

F D 2 6 6 6 6 6 6 6 6 6 6 6 4

1 1 2 3 5 8

8.9 4p

5/ 1 1 2 3 5

5.9 4p

5/ 8.9 4p

5/ 1 1 2 3

3.9 4p

5/ 5.9 4p

5/ 8.9 4p

5/ 1 1 2

2.9 4p

5/ 3.9 4p

5/ 5.9 4p

5/ 8.9 4p

5/ 1 1

9 4p

5 2.9 4p

5/ 3.9 4p

5/ 5.9 4p

5/ 8.9 4p 5/ 1

3 7 7 7 7 7 7 7 7 7 7 7 5 :

SincenD6andkD9 4p

5i.e. D ˇand!D 12Ci

p3

2 , based on Theorem4, it follows that

F: !0D˛1, so0is obtained based on 1) of Theorem4:0D 29C15p 5;

F: ! j¤1˛ and ! j¤ˇ1, forjD1; 5, soj, for jD1; 5, are obtained based on 3) of Theorem4:1;5D 12

h

51 23p 5˙ip

3.29 13p 5/i

,2;4D7 3p 5 ip

3.29 13p

5/,3D72 32p 5 Bearing in mind thatjF jD

n 1

Y

jD0

j, it follows that

jF jD 238 300 041 216 106 571 018 240p 5 :

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Let us remark, in relation to the previous example, that the determinant of F D ci rcf9 4p

5.1; 1; 2; 3; 5; 8/gis not possible to obtain using the result of Theorem3.

The next theorem is devoted to determining the Euclidean norm of (1.11). The fol- lowing formula will be needed.

For allx,

n 1

X

iD1

ixi Dx nxnC.n 1/xnC1

.1 x/2 : (2.6)

Theorem 5. LetF be the matrix as in(1.11). Then the Euclidean norm ofF is:

kFkED s

1 5

n

L2nC1 . 1/n

C.jkj2 1/

L2nC.n 1/L2nC1C5 2C1 2n

2 . 1/n

: (2.7)

Proof. From the definition of the Euclidean norm, using (1.4), (1.5), (1.6) and (2.6), we obtain:

.kFkE/2 D

n

X

i;jD1

jfi;jj2 DnF12C

.n 1/C jkj2 F22C

.n 2/C2jkj2

F32C C

1C.n 1/jkj2 Fn2 D

n 1

X

iD0

.n i /Fi2C1C jkj2

n 1

X

iD1

iFi2C1

Dn

n 1

X

iD0

Fi2C1C.jkj2 1/

n 1

X

iD1

iFi2C1

D1 5

"

n

L2nC1 . 1/n

C.jkj2 1/

n 1

X

iD1

ih

˛2iC2 2.˛ˇ/iC12iC2i

#

Dn 5

L2nC1 . 1/n

Cjkj2 1

5 .˛2˛22nC.n 1/˛2nC2

˛2 C2 1 n. 1/nC.n 1/. 1/nC1

4 Cˇ2ˇ22nC.n 1/ˇ2nC2

ˇ2 /

Dn 5

L2nC1 . 1/n

Cjkj2 1

5 . n˛2nC.n 1/˛2nC2C5 2

n 2. 1/n n 1

2 . 1/n2nC.n 1/ˇ2nC2/ Dn

5

L2nC1 . 1/n

Cjkj2 1 5

nL2nC.n 1/L2nC2C5

2C1 2n 2 . 1/n

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Dn 5

L2nC1 . 1/n

Cjkj2 1 5

L2nC.n 1/L2nC1C5

2C1 2n 2 . 1/n

: Therefore,

kFkEDq

1 5

˚n ŒL2nC1 . 1/nC.jkj2 1/

L2nC.n 1/L2nC1C52C1 2n2 . 1/n : The upper and lower bounds for the spectral norm of the Hadamard inverse of (1.11) will be given by the following theorem. We use the well - known inequalities

kCkE

pn kCk2 kCkE; (2.8)

which hold for any matrixC of ordern, and the following lemma.

Lemma 2(). LetMD mi;j

andND ni;j

be matrices of ordermn. Then

kMıNk2r1.M /ıc1.N /; (2.9) whereMıN DŒmi;jni;jis the Hadamard product (or Schur product),

r1.M /D max

1im

v u u t

n

X

jD1

jmi;j j2 and c1.N /D max

1jn

v u u t

m

X

iD1

jni;j j2:

Theorem 6. LetF be the matrix as in(1.11).

1) Ifjkj 1, then s

5n

L2nC1 . 1/n kFı 1k2 q

n.1C.n 1/jkj2/ ; (2.10)

2) Ifjkj< 1, then

jkj s

5n

L2nC1 . 1/n kFı 1k2n : (2.11) Proof. From the definition of the Euclidean norm, it follows that

kFı 1k2E D

n 1

X

iD0

.n i / 1

Fi2C1C jkj2

n 1

X

iD1

i 1

Fi2C1: (2.12)

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1) Ifjkj 1, then

kFı 1k2E

n 1

X

iD0

.n i / 1 Fi2C1C

n 1

X

iD1

i 1 Fi2C1 Dn

n 1

X

iD0

1 Fi2C1 Dn

n

X

iD1

1 Fi2 n

n

X

iD1

1

Fn2 D. n Fn

/2 n2 FnFnC1

D 5n2 L2nC1 . 1/n: Therefore,

kFı 1kE

pn s

5n

L2nC1 . 1/n:

We conclude from (2.8) that

kFı 1k2 s

5n

L2nC1 . 1/n:

Now, we shall obtain the upper bound for the spectral norm ofFı 1. Let RandS be the following matrices:

RD 2 6 6 6 6 6 6 6 6 4

1 F1

1 F2

1

F3 F1n k F1

1

1

F2 Fn11

k k F1

1 Fn12 ::: ::: ::: : :: ::: k k k F11

3 7 7 7 7 7 7 7 7 5

andSD 2 6 6 6 6 6 6 6 6 4

1 1 1 1

1

Fn 1 1 1

1 Fn 1

1

Fn 1 1 ::: ::: ::: : :: :::

1 F2

1 F3

1

F4 1 3 7 7 7 7 7 7 7 7 5 :

Then,

r1.R/D max

1in

v u u t

n

X

jD1

jri;j j2D q

1C.n 1/jkj2

and

c1.S /D max

1jn

v u u t

n

X

iD1

jsi;j j2Dp n :

SinceFı 1DRıS, based on Lemma2, we can write kFı 1k2r1.R/ıc1.S /D

q

n.1C.n 1/jkj2/ :

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2) Ifjkj< 1, then

kFı 1k2E

n 1

X

iD0

.n i /jkj2 1 Fi2C1C

n 1

X

iD1

ijkj2 1

Fi2C1 Dnjkj2

n 1

X

iD0

1 Fi2C1 Dnjkj2

n

X

iD1

1

Fi2 njkj2

n

X

iD1

1

Fn2 D jkj2. n Fn

/2 jkj2 n2 FnFnC1

D jkj2 5n2 L2nC1 . 1/n: Therefore,

kFı 1kE

pn jkj s

5n L2nC1 . 1/n: We conclude from (2.8) that

kFı 1k2 jkj

s 5n L2nC1 . 1/n:

Now, we shall obtain the upper bound for the spectral norm ofFı 1. LetQandW be the following matrices:

QD 2 6 6 6 6 6 6 6 6 6 4

1

F1 1 1 1

k Fn

1

F1 1 1

k Fn 1

k Fn

1

F1 1 ::: ::: ::: : :: :::

k F2

k F3

k

F4 F11 3 7 7 7 7 7 7 7 7 7 5

andWD 2 6 6 6 6 6 6 6 6 4

1 F1

2

1

F3 F1n 1 1 F1

2 Fn11 1 1 1 Fn12

::: ::: ::: : :: :::

1 1 1 1

3 7 7 7 7 7 7 7 7 5 :

Then,

r1.Q/D max

1in

v u u t

n

X

jD1

jqi;j j2Dp n

and

c1.W /D max

1jn

v u u t

n

X

iD1

jwi;j j2Dp n : SinceFı 1DQıW, based on Lemma2, we can write

kFı 1k2r1.Q/ıc1.W /Dn :

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3. CONCLUSION

In this paper, we investigated the eigenvalues, the Euclidean norm and the upper and lower bounds for the spectral norm of the Hadamard inverse of

FDci rcfk.F1; F2; : : : ; Fn/g;

whereFnis thent hFibonacci number andkis a nonzero complex number.

From the fact that the eigenvalues of an upper triangular matrix are the diagonal entries, the eigenvalues of a semicirculant matrix (i.e. ak-circulant matrix forkD0) with the first row.F1; F2; : : : ; Fn/are:jD1, (jD0; n 1). The Euclidean norm of such (semicirculant) matrix can be obtained from (2.7) i.e. in (2.7)kcan be equal to 0. Semicirculant matrices are not Hadamard invertible.

ACKNOWLEDGEMENT

We would like to thank the anonymous referee for the very useful suggestions which helped us to improve the quality of our paper.

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