On computing of positive integer powers of some special types of matrices

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OF

SELÇUK UNIVERSITY

ON COMPUTING OF POSITIVE INTEGER POWERS OF SOME SPECIAL TYPES OF

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iv ABSTRACT Ph.D THESIS

ON COMPUTING OF POSITIVE INTEGER POWERS OF SOME SPECIAL TYPES OF MATRICES

Ahmet ÖTELEŞ

THE GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCE OF SELÇUK UNIVERSITY

THE DEGREE OF DOCTOR OF PHILOSOPHY IN MATHEMATICS

Advisor: Assoc. Prof. Dr. Mehmet AKBULAK 2015, 66 Pages

Jury

Assoc. Prof. Dr. Mehmet AKBULAK Prof. Dr. Durmuş BOZKURT Prof. Dr. Ahmet Sinan ÇEVİK

Prof. Dr. Süleyman SOLAK Assoc. Prof. Dr. Ramazan TÜRKMEN

In this dissertation, we obtain formulas for the entries of positive integer powers of some certain tridiagonal and circulant matrices, in terms of the Chebyshev polynomials. Then we present some examples concerned with these formulas. Finally, we give some Maple 13 procedures that calculate these formulas.

Keywords: Chebyshev polynomials, Circulant matrices, Eigenvalues, Eigenvectors, Jordan's

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v ÖZET

DOKTORA TEZİ

BAZI ÖZEL TİPTEKİ MATRİSLERİN POZİTİF TAMSAYI KUVVETLERİNİN HESAPLANMASI

Ahmet ÖTELEŞ

Selçuk Üniversitesi Fen Bilimleri Enstitüsü Matematik Anabilim Dalı

Danışman: Doç. Dr. Mehmet AKBULAK 2015, 66 Sayfa

Jüri

Doç. Dr. Mehmet AKBULAK Prof. Dr. Durmuş BOZKURT Prof. Dr. Ahmet Sinan ÇEVİK

Prof. Dr. Süleyman SOLAK Doç. Dr. Ramazan TÜRKMEN

Bu tezde, bazı tridiagonal ve circulant matrislerin pozitif tamsayı kuvvetlerinin elemanlarını Chebyshev polinomlarına bağlı olarak veren formüller elde ettik. Daha sonra bu formüllerle ilgili bazı örnekler sunduk. Son olarak bu formülleri hesaplayan Maple 13 prosedürleri verdik.

Anahtar Kelimeler: Chebyshev polinomları, Circulant matrisler, Jordan formu, Özdeğer,

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vi

ACKNOWLEDGMENTS

I would like to thank my previous advisor Prof. Dr. Durmuş Bozkurt, has supported and guided me wonderfully during my academic life. I am very glad to meet with such an excellent person.

I would like to express my gratitude to Prof. Dr. Ahmet Sinan Çevik for his valuable suggestions.

I would like to thank the other members of my committee: Prof. Dr. Süleyman Solak and Prof. Dr. Ramazan Türkmen for their contributions during the preparation of this dissertation.

I am very thankful for the financial support of the Council of Higher Education of Turkey (YÖK).

I would like to thank the Chair of the Department of Mathematics, Prof. Dr. Tin-Yau Tam and Assoc. Prof. Dr. Erkan Nane for their great support during my studies at the Auburn University.

I would like to acknowledge my family for their love, support, attention and encouragement during my entire life.

Assoc. Prof. Dr. Mehmet Akbulak, my advisor, has supported, encouraged and guided me excellently throughout my studies. I am very grateful to him for teaching me the clear ways of doing research, providing me significant documentation for my research, listening me for hours, accompanying me during my studies. I am so pleased to meet such a wonderful person and great mathematician in my life. This dissertation has become valuable with his contributions.

Ahmet ÖTELEŞ KONYA-2015

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vii

TABLE OF CONTENTS

ABSTRACT ... iv

ÖZET ... v

ACKNOWLEDGMENTS ... vi

TABLE OF CONTENTS ... vii

LIST OF SYMBOLS ... ix

1. INTRODUCTION ... 1

2. PRELIMINARIES ... 12

2.1. Eigenvalues, Eigenvectors and Similarity ... 12

2.2. Chebyshev Polynomials ... 13

2.2.1. The first-kind polynomial T ... 13_n 2.2.2. The second-kind polynomial U ... 15_n 2.3. Homogeneous Linear Systems ... 16

2.4. Jordan Blocks and Jordan Form ... 16

3. ON COMPUTING OF POSITIVE INTEGER POWERS FOR SOME CERTAIN TYPES OF COMPLEX TRIDIAGONAL MATRICES ... 18

3.1. Positive Integer Powers of One Type of Complex Tridiagonal Matrices ... 18

3.1.1. Eigenvalues and eigenvectors ofXn ... 19

3.1.2. Derivation of the formula for the entries of X ... 22_nr 3.1.3. Numerical Considerations ... 24

3.2. Positive Integer Powers of One Type of Complex Tridiagonal Matrices with Eigenvalues on Imaginary Axis ... 27

3.2.1. Eigenvalues and eigenvectors of Y ... 27_n 3.2.2. Derivation of the formula for the entries of Ynr ... 31

3.2.3. Numerical Considerations ... 33

3.3. Positive Integer Powers of A Certain Type of Complex Tridiagonal Matrix Family ... 35

3.3.1. Eigenvalues and eigenvectors of B ... 36n 3.3.2. Derivation of the formula for the entries of B_nr ... 38

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viii

4. ON COMPUTING OF POSITIVE INTEGER POWERS FOR CERTAIN

REAL CIRCULANT MATRIX ... 46

4.1. General Expression for the Entries of A ... 46_nr 4.2. Numerical Considerations ... 51

5. CONCLUSION ... 56

REFERENCES ... 57

APPENDICES ... 61

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ix

LIST OF SYMBOLS

ij

A   a : matrix Awith entries a_ij

T

A : transpose of matrix A

A : conjugate of matrix A

*

A : conjugate transpose of matrix A, i.e., * T

A  A

1

A : inverse of matrix A



0 1 1



circ c c, , ,c_n_ : circulant matrix with c c₀, ,₁ ,c_n_₁ on the first row

det A : determinant of matrix A



1, 2, , n



diag

 



: diagonal matrix with

 

₁, ₂, ,



_n on the diagonal

I or E : identity matrix

n

I or E _n : the n n identity matrix

n

M : n n matrices with complex number entries

z : conjugate of the complex number z

 : the set of natural numbers

 : the set of complex numbers

n

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1

INTRODUCTION

A certain type of transformation of a set of numbers can be represented as the

multiplication of a vector by a square matrix. Repetition of the operation is

equivalent to multiplying the originial vector by a power of the matrix. Solving some

di¤erence equations, di¤erential and delay di¤erential equations and

boundary value problems, we need to compute the arbitrary integer powers of a square matrix. Properties of powers of matrices are thus of considerable importance (Cobb, 1958; Rimas, 1977; Rimas and Leonaite, 2006).

One can …nd the r th power (r 2 N) of an n n matrix A using the well-known expression

Ar = P JrP 1; (1.1)

where J is the Jordan’s form of A and P is the transforming matrix. Matrices J and P can be found, provided eigenvalues and eigenvectors of A are known (Horn and Johnson, 1985).

An n n tridiagonal matrix is one having the form

Tn := 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a1 b1 0 0 c1 a2 b2 . .. ... 0 c2 a3 . .. . .. ... .. . . .. ... ... bn 2 0 .. . . .. cn 2 an 1 bn 1 0 0 cn 1 an 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

That is, Tn = [ti;j]_{1 i;j n} is tridiagonal if ti;j = 0 for ji jj > 1 (Horn and

Johnson, 1985). These types of matrices often arise in many areas of mathematics

and engineering such as boundary value problems, parallel computing,

telecommunication system analysis, image processing and so on. In these areas, the computation of powers of tridiagonal matrices are necessary (Cobb, 1958; Rimas, 1977; Rimas and Leonaite, 2006). Therefore, there are a lot of studies dealing with the powers of tridiagonal matrices by using equation (1.1). Now we speak to those.

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n n tridiagonal matrix B of the form B = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 1 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 1 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (1.2)

of order n = 2p (p 2 N) ; in terms of Chebyshev polynomials.

The author …nds the rth power (r 2 N) of this matrix in (1.2) applying the

expression Br _{= T J}r_T 1 _{given by (1.1), where J is the Jordan’s form of B, T is}

the transforming matrix. Matrices J and T can be found provided eigenvalues and eigenvectors of B are known. The eigenvalues of B are de…ned by the characteristic equation

jB E_{j = 0:} (1.3)

He …rstly de…nes n( ) as the following

n( ) = 1 0 0 1 1 . .. ... 0 1 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 1 0 0 1 ; (1.4)

here _{2 R. Taking (1.3) and (1.4) into account he writes}

jB E_{j =} n( ) : (1.5)

From (1.4) follows:

n = n 1 n 2 2 = 2 1, 1 = , 0 = 1 ; (1.6)

here n = n( ). Solving di¤erence equation (1.6) he obtains

n( ) = Un

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All the roots of the polynomial Un(x) are included in the interval [ 1; 1] and can

be found using the relation (Fox and Parke, 1968) xnk = cos

k

n + 1, k = 1; 2; 3; : : : ; n. (1.8)

Taking (1.5), (1.7) and (1.8) into account he …nds the roots of the

characteristic equation (1.3) (eigenvalues of B):

k = 2 cos

k

n + 1, k = 1; 2; 3; : : : ; n. Let n be an even number (n = 2p, p 2 N).

Since all the eigenvalues k (k = 1; 2; : : : ; n, n = 2p, p 2 N) are simple

eigenvalues (lk = 1; lk denotes the multiplicity of the eigenvalue k), for each

eigenvalue corresponds single Jordan cell J1( k) in the matrix J . Taking this into

account and applying the relation

k = n k+1 k = 1; 2; : : : ;

n

2 ,

he writes down the Jordan’s form of B:

J = diag n n 1 n 2 n₂+1 n₂+1 n 2 n 1 n .

Using the equality J = T 1_BT _{he …nds the matrices T , T} 1 _{and derives the}

expression of the rth power (r 2 N) of B:

Br = T JrT 1 = 1 2n + 2Q (r) = 1 2n + 2(qij(r)) here qij(r) = h 1 + ( 1)r+i+ji n=2 X k=1 4 2_{n k+1} r_{n k+1}Ui 1 n k+1 2 Uj 1 n k+1 2 ;

k (k = 1; n) are the eigenvalues of B and Uk(x) be the k th degree Chebyshev

polynomial of the second kind

Uk(x) =

sin ((k + 1) arccos x)

sin arccos x ; 1 x 1

(Rimas, 2005e).

There are a lot of studies on computing powers of tridiagonal matrices by using the method mentioned above.

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Rimas (2005f) derives the general expression of the rth power (r 2 N) of the

n n _{matrix B given by (1.2) for n = 2p + 1 (p 2 N).}

Rimas (2005g, 2005h) derives the general expression of the rth power

(r_{2 N, r > 2) for one type of tridiagonal matrices of the form}

2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 2 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5

of order n = 2p (p 2 N, p > 2) and n = 2p + 1 (p 2 N) with zeros in the …rst row. Rimas (2006c) derives the general expression of the rth power (r 2 N) for one type of skew-symmetric tridiagonal matrices of the form

2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 1 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 1 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 (1.9) of order n = 2p + 1, (p 2 N).

Rimas (2007c) derives the general expression of the rth power (r 2 N) for one type of tridiagonal matrices of the form

2 6 6 6 6 6 6 6 6 6 6 6 6 4 1 1 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 1 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 of arbitrary order.

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(r_{2 N; r > 2) for one type of tridiagonal matrices of the form} 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 of order n = 2p + 1, (p 2 N, p > 2).

Rimas (2007d) derives the general expression of the rth power (r 2 N) for one type of tridiagonal matrices of the form

2 6 6 6 6 6 6 6 6 6 6 6 6 4 1 1 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 1 1 3 7 7 7 7 7 7 7 7 7 7 7 7 5 of order n, (n 2 N, n > 2).

Rimas (2007e) derives the general expression of the rth power (r 2 N) for one type of tridiagonal matrices of the form

2 6 6 6 6 6 6 6 6 6 6 6 6 4 1 1 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 1 1 3 7 7 7 7 7 7 7 7 7 7 7 7 5 of order n (n 2 N).

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(r_{2 N, r > 2) for one type of tridiagonal matrices of the form} 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5

of order n = 2p + 1 (p 2 N) with zeros in the …rst and nth rows.

Kiyak et al (2010) derive the general expression of the rth power (r 2 N) for one type of tridiagonal matrix of the form

2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 1 0 0 1 0 1 . .. ... 0 1 0 1 . .. ... .. . . .. 1 . .. . .. 0 .. . . .. ... . .. ( 1)n 1 0 0 ( 1)n 1 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 in terms of the Chebyshev polynomials of the second kind.

Rimas (2012) gives the eigenvalue decomposition for odd order tridiagonal 2-Toeplitz matrix of the form

2 6 6 6 6 6 6 6 6 6 6 6 6 4 a1 b1 0 0 c1 a2 b2 . .. ... 0 c2 a1 b1 . .. ... .. . . .. c1 a2 b2 0 .. . . .. c2 a1 . .. 0 0 . .. ... 3 7 7 7 7 7 7 7 7 7 7 7 7 5

and derive the explicit expression for integer powers of this matrix.

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power (r 2 N) of the n n tridiagonal matrix Un of the form Un:= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 2 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 2 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (1.10) for all n 2 N:

Rimas (2007a, 2007b) also obtain a general expression for the entries of the

r th power (r 2 N) of the n n tridiagonal matrix Vn of the form

Vn := 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 2 0 0 1 0 1 . .. ... 0 1 0 . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 0 1 0 0 2 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (1.11) for all n 2 N:

Gutiérrez studies powers of tridiagonal matrices di¤erently from the method introduced by Rimas (2005e). For example; he consider the n n complex tridiagonal matrix A of the form

A = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a0 a 1 0 0 a1 a0 a 1 . .. ... 0 a1 a0 . .. ... ... .. . . .. ... ... ... 0 .. . . .. ... ... a 1 0 0 a1 a0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 (1.12)

where a1a 1 6= 0: He …rstly give the following theorem for eigenvalue decomposition

of the matrix A in (1.12) (Gutiérrez, 2008c).

Theorem 1 Let a1; a0; a 1 2 C; a1a 1 6= 0 and n 2 N: Then W DW 1 is an

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matrix W are Wjk = r 2 n + 1 r _a 1 a 1 j sin jk n + 1; 1 j; k n;

and D = diag ( 1; 2; : : : n) with j

j = 0+ 2a 1 r _a 1 a 1 cos j n + 1; 1 j n (Gutiérrez, 2008c).

After that, based on this decomposition the author presents the following result that provides a general expression for the entries of the qth power (q 2 N) of

A; in terms of Chebyshev polynomials of the second kind.

Theorem 2 Consider a1; a0; a 1 2 C; a1a 1 6= 0 and n 2 N: Let A be the

n n complex tridiagonal matrix in (1.12), = q a1

a 1 and h = 2 cos h n+1 for every 1 h n: Then [Aq]_jk = j k 2n + 2 2 6 42 1 + ( 1)n+1 aq₀Uj 1(0) Uk 1(0) + bn 2c X h=1 4 2_{n h+1} Uj 1 n h+1 2 Uk 1 n h+1 2 h (a0+ a 1 n h+1) q + ( 1)j+k (a0 a 1 n h+1) q ] 3 5

for all q 2 N and 1 j; k n; _{where bxc denotes the largest integer less than or}

equal to x (Gutiérrez, 2008c).

Since the matrix A in (1.12) is a matrix family of the matrices given by (1.2) and (1.9), Gutiérrez deduce the expression given by Rimas for the entries of the powers of these matrices by using Theorem 2.

By the similar way mentioned above, Gutiérrez (2008d) derives the entries of

powers of an n n Hermitian tridiagonal matrix of the form

2 6 6 6 6 6 6 6 6 6 6 6 6 4 a0 a1 0 0 a1 a0 a1 . .. ... 0 a1 a0 . .. ... ... .. . . .. ... ... ... 0 .. . . .. ... ... a1 0 0 a1 a0 3 7 7 7 7 7 7 7 7 7 7 7 7 5

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where a0 2 R and a1 6= 0.

An n n circulant matrix Cn := circn(c0; c1; : : : ; cn 1) is a square matrix

having the form

Cn:= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 c0 c1 c2 : : : cn 2 cn 1 cn 1 c0 c1 . .. cn 2 cn 2 cn 1 c0 . .. . .. ... .. . . .. ... ... . .. c2 c2 . .. ... . .. c1 c1 c2 : : : cn 2 cn 1 c0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ;

where each row is a cyclic shift of the row above it. We may also write a circulant matrix in the form

Cn = cij = c(j i) mod n:

Circulant matrices arise; for example, in applications involving the discrete Fourier transform (DFT) and the study of cyclic codes for error correction (Gray, 2006).

Circulant matrices have a wide range of applications in mathematics and engineering such as graph theory, signal processing, coding theory and image processing, etc. Numerical solutions of certain types of elliptic and parabolic partial di¤erential equations with periodic boundary conditions often involve linear systems associated with circulant matrices (Tsitsas et al, 2007; Zhao, 2009).

The well-known eigenvalue decomposition of an n n circulant matrix (Davis, 1994) is that

circn(c0; c1; : : : ; cn 1) = FnDnFn; (1.13)

where * denotes conjugate transpose (i.e F_n = FT_n), Fn is the n n Fourier matrix:

[Fn]_u;v = 1 p ne 2 (u 1)(v 1) n i; 1 u; v n;

and Dn= diag ( 1; 2; : : : ; n) with

k = n X r=1 cr 1e 2 (k 1)(r 1) n i; 1 k n:

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It can be given the entries of the r th power (r 2 N) of an n n circulant matrix by using (1.13) as the following:

[C_nr]_i;j = [(FnDnFn) r ]_i;j = [FnDrnFn]i;j = n X k=1 [Fn]_i;k[DrnFn]k;j = n X k=1 [Fn]i;k r k[Fn]j;k = 1 n n X k=1 r ke 2 (i 1)(k 1) n ie 2 (j 1)(k 1) n i;

according to above the last equation, we get [C_nr]_i;j = 1 n n X k=1 r ke 2 (k 1)(j i) n i: (1.14)

We can easily calculate the entries of the r th power (r 2 N) of an

n n circulant matrix by using (1.14). But in recent years, computing the integer

powers of certain types of circulant matrices depending on Chebyshev polynomials has been a very popular problem. For instance, Rimas obtain a general expression for the entries of the r th power (r 2 N) of the n n real symmetric circulant matrix of the form circn(0; 1; 0; : : : ; 0; 1) := 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 1 0 0 1 1 0 1 . .. 0 0 1 0 . .. ... ... .. . . .. ... ... 1 0 0 . .. 1 0 1 1 0 0 1 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ;

in terms of the Chebyshev polynomials (see Rimas (2005c, 2005d) for the odd case and Rimas (2006a, 2006b) for the even case).

Gutiérrez (2008b) derive a single formula by generalizing the results of Rimas

(2005c, 2005d) for the entries of the positive integer powers of the n n complex

symmetric circulant matrix of odd and even order given as circn b0; b1; : : : ; bn 1 2 ; b n 1 2 ; : : : ; b1 T if n is odd, circn b0; b1; : : : ; bn₂ 1; bn₂; bn₂ 1; : : : ; b1 T if n is even, (1.15)

in terms of the Chebyshev polynomials of the …rst kind and the author also derives two separate formulas for the entries of the positive integer powers of the

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n n complex skew-symmetric circulant matrix of odd and even order given as circn 0; b1; : : : ; bn 1 2 ; b n 1 2 ; : : : ; b1 T if n is odd, circn 0; b1; : : : ; bn₂ 1; 0; bn₂ 1; : : : ; b1 T if n is even,

in terms of the Chebyshev polynomials of the …rst and second kinds (Gutiérrez 2008a).

Köken and Bozkurt (2011) consider the n n non-symmetric complex

circulant matrix of the form

circn(0; a; 0; : : : ; 0; b) := 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 a 0 0 b b 0 a . .. 0 0 b 0 . .. ... ... .. . . .. ... ... a 0 0 . .. b 0 a a 0 0 b 0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (1.16)

di¤erently from complex symmetric circulant matrix given by (1.15). They derive a general expression for the entries of the r th power (r 2 N) of the matrix of odd order in (1.16), in terms of the Chebyshev polynomials.

The dissertation is organized as follows.

In Chapter 2, we present some basic notations, de…nitions and fundamental properties that we will need for the main results during this dissertation.

In Chapter 3, we present formulas for the entries of positive integer powers of some certain types of tridiagonal matrices, in terms of Chebyshev polynomials of the …rst kind.. Then we give some numerical examples.

In Chapter 4, we present a single formula for the entries of the r th power

(r_{2 N) of a certain type of circulant matrix of odd and even order, in terms of the}

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2

PRELIMINARIES

In this chapter, we present preliminary de…nitions and the necessary theorems which will be referred throughout the presentation of the main results of this dissertation.

2.1 Eigenvalues, Eigenvectors and Similarity

De…nition 3 _{If A 2 M}n and x 2 Cn; we consider the equation

Ax = x; x_{6= 0;}

where is a scalar. If a scalar and a nonzero vector x happen to satisfy this

equation, then is called an eigenvalue of A and x is called an eigenvector of

A associated with (Horn and Johnson,1985).

De…nition 4 The set of all _{2 C that are eigenvalues of A 2 M}n is called the

spectrum of A and is denoted by (A) (Horn and Johnson,1985).

Example 5 Consider the matrix

A = 2 4 7 2 4 1 3 5 2 M2:

Then we have 3 2 (A) with 2

4 1

2 3

5 as an associated eigenvector since

A 2 4 1 2 3 5 = 2 4 3 6 3 5 = 3 2 4 1 2 3 5 :

De…nition 6 _{The characteristic polynomial of A 2 M}n, denoted by pA(t), is the

polynomial de…ned by

pA(t) = det (tI A) ;

where I denotes the identity matrix (Horn and Johnson,1985).

De…nition 7 _{A matrix B 2 M}n is said to be similar to a matrix A 2 Mn if there

exists a nonsingular matrix S 2 Mn such that

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The transformation A ! S 1_AS _{is called a similarity transformation by the}

similarty matrix S: The relation "B is similar to A" sometimes abbreviated as

B A (Horn and Johnson,1985).

Theorem 8 _{Let A; B 2 M}n:If B is similar to A; then the characteristic polynomial

of B is the same as that of A (Horn and Johnson,1985).

Corollary 9 _{If A; B 2 M}n and if A and B are similar, then they have the same

eigenvalues (Horn and Johnson,1985).

De…nition 10 _{If the matrix A 2 M}n is similar to a diagonal matrix, then A is said

to be diagonalizable (Horn and Johnson,1985).

Theorem 11 _{Let A 2 M}n: Then A is diagonalizable if and only if there is set

of n linearly independent vectors, each of which is an eigenvector of A (Horn and Johnson,1985).

Theorem 12 _{If A 2 M}n has n distinct eigenvalues, then A is diagonalizable (Horn

and Johnson,1985).

Theorem 13 _{If A 2 M}n is diagonalizable, then for k 2 N

Ak = S kS 1

(Horn and Johnson,1985).

2.2 Chebyshev Polynomials

2.2.1 The …rst-kind polynomial Tn

De…nition 14 The Chebyshev polynomial Tn(x) of the …rst kind is a polynomial in

x of degree n, de…ned by the relation

Tn(x) = cos n when x = cos : (2.1)

If the range of the variable x is the interval [ 1; 1] ; then the range of the

corresponding variable can be taken as [0; ] : These ranges are traversed in

opposite directions, since x = 1 corresponds to = and x = 1 corresponds

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It is well known (as a consequence of de Moivre’s Theorem) that cos n is a polynomial of degree n in cos , and indeed we are familiar with the elementary formula

cos 0 = 1; cos 1 = cos ; cos 2 = 2 cos2 1;

cos 3 = 4 cos3 3 cos ; cos 4 = 8 cos4 8 cos2 + 1; : : : :

We may immediately deduce from (2.1), that the …rst few Chebyshev polynomials are

T0(x) = 1; T1(x) = x; T2(x) = 2x2 1;

T3(x) = 4x3 3x; T4(x) = 8x4 8x2+ 1:

In practice it is neither convenient nor e¢ cient to work out each Tn(x)from

…rst principles. Rather by combining the trigonometric identity

cos n + cos (n 2) = 2 cos cos (n 1) ;

with De…nition 14, we obtain the fundamental recurrence relation Tn(x) = 2xTn 1(x) Tn 2(x) ; n = 2; 3; : : : ;

which together with the initial conditions

T0(x) = 1; T1(x) = x;

recursively generates all the polynomials fTn(x)g very e¢ ciently.

Tn(x) can also be satis…ed the determinant equation as the following

Tn(x) = x 1 0 0 1 2x 1 . .. ... 0 1 2x . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 2x 1 0 0 1 2x

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2.2.2 The second-kind polynomial Un

De…nition 15 The Chebyshev polynomial Un(x) of the second kind is a polynomial

of degree n in x de…ned by

Un(x) = sin (n + 1) = sin when x = cos : (2.2)

The ranges of x and are the same as for Tn(x).

sin 1 = sin ; sin 2 = 2 sin cos ; sin 3 = sin 4 cos2 1 ;

sin 4 = sin 8 cos3 4 cos ; : : : ;

so that we see that the ratio of sine functions (2.2) is indeed a polynomial in

cos ; and we may immediately deduce that

U0(x) = 1; U1(x) = 2x; U3(x) = 4x2 1;

U4(x) = 8x3 4x; : : : :

By combining the trigonometric identity

sin (n + 1) + sin (n 1) = 2 cos sin n ;

with De…nition 15, we …nd that Un(x) satis…es the recurrence relation

Un(x) = 2xUn 1(x) Un 2(x) ; n = 2; 3; : : : ;

which together with the initial conditions

U0(x) = 1; U1(x) = 2x;

provides an e¢ cient procedure for generating the polynomials. A similar trigonometric identity

sin (n + 1) sin (n 1) = 2 sin cos ;

leads us to a relationship

Un(x) Un 2(x) = 2Tn(x) ; n = 2; 3; : : : ;

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Un(x) can also be satis…ed the determinant equation as the following Un(x) = 2x 1 0 0 1 2x 1 . .. ... 0 1 2x . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 2x 1 0 0 1 2x

(Mason and Handscomb, 2003).

2.3 Homogeneous Linear Systems

De…nition 16 A system of linear equations is said to be homogeneous if the

constant terms are all zero; that is, the system has the form a11x1+ a12x2+ : : : + a1nxn= 0

a21x1+ a22x2+ : : : + a2nxn= 0

.. .

am1x1+ am2x2+ : : : + amnxn= 0:

Every homogeneous system of linear equations is consistent, since all such systems

have x1 = 0; x2 = 0; : : : ; xn = 0 as a solution. This solution is called the trivial

solution; if there are other solutions, they are called nontrivial solutions.

Because a homogeneous linear system always has the trivial solution, there are only two possibilities for its solutions:

i. The system has only the trivial solution,

ii. The system has in…nitely many solutions in addition to the trivial solution (Anton and Rosses, 2005).

2.4 Jordan Blocks and Jordan Form

A Jordan Block of size m and value is a matrix Jm( )having the value repeated

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For example: J2(4) = 2 4 4 1 0 4 3 5 ; J4( 2) = 2 6 6 6 6 6 6 4 2 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 3 7 7 7 7 7 7 5 :

A Jordan Form is a block diagonal matrix consisting of several Jordan blocks.

For example, we can form a Jordan Form from two copies of J2(4) and one copy of

J4( 2) as follows. J = 2 6 6 6 4 J2(4) 0 0 0 J2(4) 0 0 0 J4( 2) 3 7 7 7 5= 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 4 1 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 4 1 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 2 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 (Klavins, 2008).

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3

ON COMPUTING OF POSITIVE INTEGER POWERS

FOR

SOME

CERTAIN

TYPES

OF

COMPLEX

TRIDIAGONAL MATRICES

There are three sections in this chapter. In the …rst section, we consider one type of complex tridiagonal matrix. Then, in the second section we consider another type of complex tridiagonal matrix with eigenvalues on imaginary axis. In the last section, we consider a matrix family that is the general type of the matrices given in Section 1 and Section 2. We study on powers of these matrices. Hence, we …rstly give theoretical background required for …nding the eigenvalues and eigenvectors of these matrices. Secondly we obtain the eigenvalues and eigenvectors of these matrices. Based on the eigenvalues and eigenvectors we introduce the formulas that provides a general expression for the entries of the r th power (r 2 N) of these matrices, depending on the Chebyshev polynomials of the …rst kind. Finally we present some numerical examples related to the formulas.

3.1 Positive Integer Powers of One Type of Complex

Tridiagonal Matrices

In this section, we study the entries of positive integer powers of an n n complex

tridiagonal matrix having the form

Xn:= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a 2b 0 0 b a b . .. ... 0 b a . .. ... ... .. . . .. ... ... b 0 .. . . .. b a b 0 0 2b a 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (3.1)

where b 6= 0: This matrix is a more general type of Un given by (1.10). For the

solution of this problem we shall apply the similarity transformation

X_nr = NnJnrNn1; where Jn is the Jordan’s form of Xn and Nn is the

transforming matrix. Since Matrices Jn and Nn can be found by providing

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3.1.1 Eigenvalues and eigenvectors of Xn

We …rstly give some preliminaries required for …nding the eigenvalues and eigenvectors of Xn:

Lemma 17 _{Let fH}n; n = 1; 2; : : :g be sequence of tridiagonal matrices of the form

Hn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 h1;1 h1;2 0 0 h2;1 h2;2 h2;3 . .. ... 0 h3;2 h3;3 . .. . .. ... .. . . .. ... ... . .. 0 .. . . .. ... . .. hn 1;n 0 0 hn;n 1 hn;n 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

Then the succesive determinants of Hn are given by the recursive formula:

jH1j = h1;1;

jH2j = h1;1h2;2 h1;2h2;1;

jHnj = hn;njH(n 1)j hn 1;nhn;n 1jH(n 2)j

(Cahill et al, 2003).

Let Hy

n; n = 1; 2; : : : be sequence of tridiagonal matrices of the form

H_ny = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 h1;1 h1;2 0 0 h2;1 h2;2 h2;3 . .. ... 0 h3;2 h3;3 . .. . .. ... .. . . .. . .. . .. . .. 0 .. . . .. . .. . .. hn 1;n 0 0 hn;n 1 hn;n 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

Since the determinant of matrices Hn and Hny have the same recursive formula, it

can be written that

jHnj = Hny (3.2)

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Let Un be the n n tridiagonal matrix given by (1.10). By using (3.2), we

can write the characteristic polynomial of Un as the following:

jtIn Unj = t 2 0 0 1 t 1 . .. ... 0 1 t . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 t 1 0 0 2 t : (3.3)

The eigenvalues of Un are

tk = 2 cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n; (3.4)

where tk denotes k th eigenvalue of Un (Rimas, 2005a):

Lemma 18 Let Qn be the following n n tridiagonal matrix

Qn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a 2 0 0 1 a 1 . .. ... 0 1 a . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 a 1 0 0 2 a 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (3.5)

where a 2 C: Then the eigenvalues of Qn are

k = a + 2 cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n (3.6)

(Ötele¸s and Akbulak, 2013).

Proof. The eigenvalues of Qn are the roots of its characteristic polynomial. From

(3.2) we can write the characteristic polynomial of Qn as the following:

j In Qnj = a 2 0 0 1 a 1 . .. ... 0 1 a . .. . .. ... .. . . .. . .. . .. 1 0 .. . . .. 1 a 1 0 0 2 a :

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Substituting t = aand taking (3.3) and (3.4) into account, we …nd the eigenvalues

of Qn as (3.6). So, the lemma is proved.

Theorem 19 _{Consider a; b 2 C and b 6= 0. Let X}n be the n n complex tridiagonal

matrix in (3.1). Then the eigenvalues of Xn are

k = a + 2b cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n (3.7)

and the eigenvectors of Xn with associated with eigenvalues are given by

&j = 2 6 6 6 6 6 6 6 4 T0 j a 2b T1 j a 2b .. . Tn 1 j a 2b 3 7 7 7 7 7 7 7 5 ; for j = 1; 2; : : : ; n; (3.8)

where Tk(x) is the kth degree Chebyshev polynomial of the …rst kind such that

Tk(x) = cos k (arccos x) ; 1 x 1 (3.9)

(Ötele¸s and Akbulak, 2013).

Proof. In order to prove the theorem we need a relation between the matrices Xn

and Qn given by (3.5). Dividing all entries of Xn by nonzero b; we get a new matrix

Mn with Mn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a=b 2 0 0 1 a=b 1 . .. ... 0 1 a=b . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 a=b 1 0 0 2 a=b 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

Taking (3.5) and (3.6) into account we …nd the eigenvalues of Mn to be

a

b + 2 cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n:

Since the eigenvalues of Xn are just b times the eigenvalues of Mn; we get the

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Let us …nd the eigenvectors corresponding to each eigenvalue of Xn: Each

eigenvector of Xnis the solution of the following homogeneous linear equation system

( jI Xn) x = 0; (3.10)

where j is the jth eigenvalue of Xn (1 j n) : We clearly write the expression

(3.10) as follows: ( j a) x1 2bx2 = 0 bx1+ ( j a) x2 bx3 = 0 bx2+ ( j a) x3 bx4 = 0 .. . bxn 2+ ( j a) xn 1 bxn= 0 2bxn 1+ ( j a) xn= 0 : (3.11)

Dividing all terms of the each equations in (3.11) by b 6= 0, choosing x1 = 1arbitrarily

and solving the set of systems (3.11) according to x1; we get (3.8). So the theorem

is proved.

Let us begin to get the formula for Xr

n.

3.1.2 Derivation of the formula for the entries of X_nr

Theorem 20 _{Consider a; b 2 C; b 6= 0. Let X}n be an n n complex tridiagonal

matrix in (3.1). Then (i,j)th entry of Xr

n is [X_nr]_i;j = j 2n 2 n X k=1 k r kTi 1 k a 2b Tj 1 k a 2b ; (3.12) for i; j = 1; 2; :::; n: Here j = 8 < : 1; if j = 1; n; 2; if 1 < j < n: and _k = 8 < : 1; if k = 1; n; 2; if 1 < k < n; (3.13)

k and Tk(x) are de…ned by (3.7) and (3.9), respectively (Ötele¸s and Akbulak, 2013).

Proof. We shall consider the relation Xr

n = NnJnrNn1. In order to get the formula

for the entries of Xr

n; we …rstly …nd the matrices Jn and Nn:

Since all the eigenvalues k (k = 1; 2; : : : ; n) are simple, each eigenvalue k

corresponds single Jordan cell Ji( k)in Jn:Taking this into account we write down

the Jordan’s form of Xn

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Let us …nd the transforming matrix Nn and its inverse Nn1:

From (3.8) we get the transforming matrix Nn as:

Nn= 2 6 6 6 6 6 6 6 4 T0 1_2ba T0 2_2ba : : : T0 n_2ba T1 1_2ba T1 2_2ba : : : T1 n_2ba .. . ... . .. ... Tn 1 1_2ba Tn 1 2_2ba : : : Tn 1 n_2ba 3 7 7 7 7 7 7 7 5 : (3.15)

Denoting the j th column of the inverse matrix N 1

n by j (Nn1 = ( 1; 2; : : : ; n)) ;

from (Rimas, 2005b) we get

j = j 2 6 6 6 6 6 6 6 4 f1Tj 1 1_2ba f2Tj 1 2_2ba .. . fnTj 1 n_2ba 3 7 7 7 7 7 7 7 5 for j = 1; 2; : : : ; n; (3.16)

where fk = _{2n 2}k and j; k are de…ned by (3.13). Taking into account (3.16), we

write down N 1 n as N_n1 = 1 2n 2 2 6 6 6 6 6 6 6 6 6 4 T0 1_2ba 2T1 1_2ba 2Tn 2 1_2ba Tn 1 1_2ba 2T0 2_2ba 4T1 2_2ba 4Tn 2 2_2ba 2Tn 1 2_2ba .. . ... . .. ... ... 2T0 n_2b1 a 4T1 n_2b1 a 4Tn 2 n_2b1 a 2Tn 1 n_2b1 a T0 n_2ba 2T1 n_2ba 2Tn 2 n_2ba Tn 1 n_2ba 3 7 7 7 7 7 7 7 7 7 5 : (3.17)

Denoting i th row of Nn by Li (i = 1; 2; ::; n) ;from (3.15) we have

LT_i = 2 6 6 6 6 6 6 6 4 Ti 1 1_2ba Ti 1 2_2ba .. . Ti 1 n_2ba 3 7 7 7 7 7 7 7 5 ;

where T means operation of transposition. Using the equalities Xnr = NnJnrNn1 (see

(3.14), (3.15), (3.16) and (3.17)) and Jr n = diag ( r 1; r 2; : : : ; r

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entry of Xr n by [Xnr]i;j; we get [X_nr]_i;j = LiJn jr = Li 2 6 6 6 6 6 6 6 4 jf1 r1Tj 1 1_2ba jf2 r2Tj 1 2_2ba .. . jfn rnTj 1 n_2ba 3 7 7 7 7 7 7 7 5 ; Consequently, [X_nr]_i;j = j 2n 2 n X k=1 k r kTi 1 k a 2b Tj 1 k a 2b ; for i; j = 1; 2; :::; n. It is desirable. 3.1.3 Numerical Considerations

In this section, we give two examples related to the formula of Xnr. One of them is

4 4real tridiagonal matrix and the other is 3 3 complex tridiagonal matrix. We

calculate 3th and 4th powers of these matrices, respectively. These examples can be veri…ed by using Maple 13 procedure given in Appendix 1.

Example 21 Let X4 be a 4 4 real tridiagonal matrix given in (3.1) as in the

following: X4 = 2 6 6 6 6 6 6 4 2 6 0 0 3 2 3 0 0 3 2 3 0 0 6 2 3 7 7 7 7 7 7 5

for a = 2 and b = 3: 3th power of X4 is computed as in the following.

From (3.7) eigenvalues of X4 can be written for k = 1; 2; 3; 4 as:

k= 2 + 6 cos

(k 1)

3 ;

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matrix N4; whose columns consist of eigenvectors of X4; and its inverse as: N4 = 2 6 6 6 6 6 6 6 6 4 T0 1₆ 2 T0 2₆ 2 T0 3₆ 2 T0 4₆ 2 T1 1₆ 2 T1 2₆ 2 T1 3₆ 2 T1 4₆ 2 T2 1₆ 2 T2 2₆ 2 T2 3₆ 2 T2 4₆ 2 T3 1₆ 2 T3 2₆ 2 T3 3₆ 2 T3 4₆ 2 3 7 7 7 7 7 7 7 7 5 = 2 6 6 6 6 6 6 4 1 1 1 1 1 1₂ 1₂ 1 1 1 2 1 2 1 1 1 1 1 3 7 7 7 7 7 7 5 and N₄ 1 = 1 6 2 6 6 6 6 6 6 6 6 4 T0 1₆ 2 2T1 1₆2 2T2 1₆2 T3 1₆ 2 2T0 2₆2 4T1 2₆2 4T2 2₆2 2T3 2₆2 2T0 3₆2 4T1 3₆2 4T2 3₆2 2T3 3₆2 T0 4₆ 2 2T1 4₆2 2T2 4₆2 T3 4₆ 2 3 7 7 7 7 7 7 7 7 5 = 2 6 6 6 6 6 6 6 6 4 1 6 1 3 1 3 1 6 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 6 1 3 1 3 1 6 3 7 7 7 7 7 7 7 7 5 : Then we get X₄3 = N4J43N4 1 = 2 6 6 6 6 6 6 4 116 234 108 54 117 170 171 54 54 171 170 117 54 108 234 116 3 7 7 7 7 7 7 5 :

(i; j)th entry of the X₄3 can be veri…ed by formula given in (3.12).

Example 22 Let X3 be a 3 3 complex tridiagonal matrix given in (3.1) as in the

following: X3 = 2 6 6 6 4 1 i 10 + 4i 0 5 + 2i 1 i 5 + 2i 0 10 + 4i 1 i 3 7 7 7 5:

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for a = 1 i and b = 5 + 2i: 4th power of X3 is computed as in the following.

From (3.7) eigenvalues of X3 can be written for k = 1; 2; 3 as:

k = (1 i) + (10 + 4i) cos

(k 1)

2 ;

namely, 1 = 11 + 3i; 2 = 1 iand 3 = 9 5i. We also write the transforming

matrix N3; whose columns consist of eigenvectors of the X3; and its inverse as:

N3 = 2 6 6 6 6 6 4

T0 1_10+4i(1 i) T0 2_10+4i(1 i) T0 3_10+4i(1 i)

T1 1_10+4i(1 i) T1 2_10+4i(1 i) T1 3_10+4i(1 i)

T2 1 (1 i) 10+4i T2 2 (1 i) 10+4i T2 3 (1 i) 10+4i 3 7 7 7 7 7 5 = 2 6 6 6 4 1 1 1 1 0 1 1 1 1 3 7 7 7 5 and N₃ 1 = 1 4 2 6 6 6 6 6 4 T0 1 (1 i) 10+4i 2T1 1 (1 i) 10+4i T2 1 (1 i) 10+4i

2T0 2_10+4i(1 i) 4T1 2_10+4i(1 i) 2T2 2_10+4i(1 i)

T0 3_10+4i(1 i) 2T1 3_10+4i(1 i) T2 3_10+4i(1 i)

3 7 7 7 7 7 5 = 2 6 6 6 6 6 4 1 4 1 2 1 4 1 2 0 1 2 1 4 1 2 1 4 3 7 7 7 7 7 5 : Then we get X₃4 = N3J34N 1 3 = 2 6 6 6 4

804 + 6216i 6576 + 2352i 808 + 6216i

3288 + 1176i 1612 + 12432i 3288 + 1176i

808 + 6216i 6576 + 2352i 804 + 6216i

3 7 7 7 5: (i; j)th entry of the X4

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3.2 Positive Integer Powers of One Type of Complex

Tridiagonal Matrices with Eigenvalues on Imaginary Axis

Yn := 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a 2b 0 0 b a b . .. ... 0 b a . .. ... ... .. . . .. ... ... b 0 .. . . .. b a b 0 0 2b a 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (3.18)

where b 6= 0: This matrix is a more general type of Vn given by (1.11). For the

solution of this problem we shall apply the similarity transformation

Yr

n = NnJnrNn1; where Jn is the Jordan’s form of Yn and Nn is the

transforming matrix. Since Matrices Jn and Nn can be found by providing

eigenvalues and eigenvectors of Yn, we …nd eigenvalues and eigenvectors of Yn.

3.2.1 Eigenvalues and Eigenvectors of Yn

We …rstly give some preliminaries required for …nding the eigenvalues and eigenvectors of Yn:

Let+Hn and Hn be two n n tridiagonal matrices as the following:

+Hn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 h1;1 h1;2 0 0 h2;1 h2;2 h2;3 . .. ... 0 h3;2 h3;3 . .. . .. ... .. . . .. . .. ... . .. 0 .. . . .. ... . .. hn 1;n 0 0 hn;n 1 hn;n 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ;

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and Hn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 h1;1 h1;2 0 0 h2;1 h2;2 h2;3 . .. ... 0 h3;2 h3;3 . .. . .. ... .. . . .. . .. . .. . .. 0 .. . . .. . .. . .. hn 1;n 0 0 hn;n 1 hn;n 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

From Lemma 17 since the matrices+Hn and Hn have the same recursive formula,

it can be written that

j+Hnj = j Hnj (3.19)

(Ötele¸s and Akbulak, 2014b).

Let Vn be the n n tridiagonal matrix given by (1.11). By using (3.19), we

can write the characteristic polynomial of Vn as the following:

jtIn Vnj = t 2 0 0 1 t 1 . .. ... 0 1 t . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 t 1 0 0 2 t : (3.20)

The eigenvalues of Vn are

tk= 2i cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n (3.21)

where tk denotes k th eigenvalue of Vn and i :=

p

1 (Rimas, 2007a).

Lemma 23 _{Consider a 2 C and i =} p 1: Let Rn be the following n n complex

tridiagonal matrix Rn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a 2 0 0 1 a 1 . .. ... 0 1 a . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 a 1 0 0 2 a 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (3.22)

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where a 2 C: Then the eigenvalues of Rn are

k = a 2i cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n (3.23)

(Ötele¸s and Akbulak, 2014b).

Proof. Since the eigenvalues of Rn are the roots of its characteristic polynomial,

we can write the characteristic polynomial of Rn by using (3.19) as the following:

j I Rnj = a 2 0 0 1 a 1 . .. ... 0 1 a . .. . .. ... .. . . .. . .. . .. 1 0 .. . . .. 1 a 1 0 0 2 a :

Substituting t = a and taking (3.20) and (3.21) into account, we …nd the

eigenvalues of Rn as (3.23). So the theorem is proved.

Theorem 24 _{Consider a; b 2 C ; b 6= 0; and i =}p 1. Let Yn be the n n complex

tridiagonal matrix given by (3.18) and j = ( j a) =b. Then the eigenvalues of Yn

are

k = a 2ib cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n; (3.24)

and the eigenvectors matrix of Yn with associated with eigenvalues are given by,

&j = 2 6 6 6 6 6 6 6 4 e (0) T0 i₂j e (1) T1 i j 2 .. . e (n 1) Tn 1 i j 2 3 7 7 7 7 7 7 7 5 ; j = 1; 2; : : : ; n; (3.25) where e (k) = e ikn for k = 0; 1; 2; 3; : : : ; (3.26)

and Tk(x) is the kth degree Chebyshev polynomial of the …rst kind such that

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Proof. In order to prove the theorem, we need a relation between the matrices Yn

and Rn given by (3.22): Dividing all entries of Yn by nonzero b; we get a new matrix

+Mn with +Mn = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a=b 2 0 0 1 a=b 1 . .. ... 0 1 a=b . .. ... ... .. . . .. ... ... 1 0 .. . . .. 1 a=b 1 0 0 2 a=b 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

Taking (3.22) and (3.23) into account, we …nd the eigenvalues of+Mn to be

a

b 2i cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n:

Since the eigenvalues of Ynare just b times the eigenvalues of+Mn; we get eigenvalues

of Yn as in (3.24)

k= a 2ib cos

(k 1)

n 1 ; for k = 1; 2; : : : ; n;

where k denotes kth eigenvalue of Yn:

Let us …nd the eigenvectors corresponding to each eigenvalue of Yn: Each

eigenvector of Ynis the solution of the following homogeneous linear equation system

( jI Yn) x = 0; (3.28)

where j is the jth eigenvalue of Yn (1 j n) : We clearly write the expression

(3.28) as follows: ( j a) x1 2bx2 = 0 bx1+ ( j a) x2 bx3 = 0 bx2+ ( j a) x3 bx4 = 0 .. . bxn 2+ ( j a) xn 1 bxn = 0 2bxn 1+ ( j a) xn= 0 : (3.29)

Dividing all terms of the each equations in (3.29) by b 6= 0, substituting

j = ( j a) =b, choosing x1 = 1 arbitrarily and solving the set of systems (3.29)

according to x1, we get (3.25). So the theorem is proved.

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3.2.2 Derivation of the formula for the entries of Yr n

Theorem 25 _{Consider a; b 2 C; b 6= 0; and i =}p 1. Let Yn be the n n complex

tridiagonal matrix given by (3.18). Then (i,j)th entry of Yr

n is

[Y_nr]_i;j = je(i 1)e(j 1)

2n 2 n X k=1 k r kTi 1 i( k a) 2b Tj 1 i( k a) 2b ; (3.30) for i; j = 1; 2; :::; n: Here j = 8 < : 1; if j = 1; n; 2; if 1 < j < n: and k = 8 < : 1; if k = 1; n; 2; if 1 < k < n: (1)

k; e (k)and Tk(x) are de…ned by (3.24), (3.26) and (3.27), respectively (Ötele¸s and

Akbulak, 2014b).

Proof. We shall consider the relation Ynr = NnJnrNn1. In order to get the formula

for the entries of Yr

n; we …rstly …nd the matrices Jn and Nn:

the Jordan’s form of Yn

Jn= diag ( 1; 2; : : : ; n) : (3.31)

Nn= 2 6 6 6 6 6 6 6 4 e (0) T0 i₂1 e (0) T0 i₂2 : : : e (0) T0 i₂n e (1) T1 i₂1 e (1) T1 i₂2 : : : e (1) T1 i₂n .. . ... . .. ... e (n 1) Tn 1 i₂1 e (n 1) Tn 1 i₂2 : : : e (n 1) Tn 1 i₂n 3 7 7 7 7 7 7 7 5 : (3.32)

n by j (Nn1 = ( 1; 2; : : : ; n)) ;

j = j 2 6 6 6 6 6 6 6 4 f1e (j 1) Tj 1 i₂1 f2e (j 1) Tj 1 i₂2 .. . fne (j 1) Tj 1 i₂n 3 7 7 7 7 7 7 7 5 ; j = 1; 2; : : : ; n; (3.33)

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where e (j 1) is the complex conjugate of the complex number e (j 1),

fk = _{2n 2}k and j; k are de…ned by (1): Taking (3.33) into account, we write

down N 1 n as N_n1 = 2 6 6 6 6 6 6 6 6 4 1f1e (0) T0 i₂1 2f1e (1) T1 i₂1 nf1e (n 1) Tn 1 i₂1 1f2e (0) T0 i₂2 2f2e (1) T1 i₂2 nf2e (n 1) Tn 1 i₂2 .. . ... . .. ... 1fne (0) T0 i₂n 2fne (1) T1 i₂n nfne (n 1) Tn 1 i₂n 3 7 7 7 7 7 7 7 7 5 : (3.34)

Denoting i th row of Nn by Li (i = 1; 2; ::; n) ;from (3.32) we have

LT_i = 2 6 6 6 6 6 6 6 4 e(i 1)Ti 1 i₂1 e(i 1)Ti 1 i₂2 .. . e(i 1)Ti 1 i₂n 3 7 7 7 7 7 7 7 5 ;

where T means operation of transposition. Using the equalities Ynr = NnJnrNn1 (see

(3.31), (3.32), (3.33) and (3.34)) and Jr n = diag ( r 1; r 2; : : : ; r

n) and denoting (i,j)th

entry of Ynr by [Ynr]i;j; we get

[Y_nr]_i;j = LiJn jr = Li 2 6 6 6 6 6 6 6 4 jf1e (j 1) r1Tj 1 i₂1 jf2e (j 1) r2Tj 1 i₂2 .. . jfne (j 1) rnTj 1 i₂n 3 7 7 7 7 7 7 7 5 : Consequently,

[Y_nr]_i;j = je(i 1)e(j 1) n X k=1 fk( k)rTi 1 i k 2 Tj 1 i k 2 ; for i; j = 1; 2; :::; n.

Substituting k= ( k a)=b and fk= _{2n 2}k ; we get

[Y_nr]_i;j = je(i 1)e(j 1)

2n 2 n X k=1 k r kTi 1 i( k a) 2b Tj 1 i( k a) 2b ; for i; j = 1; 2; :::; n. It is desirable.

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3.2.3 Numerical Considerations

In this section, we give two examples related to the formula of Yr

n. One of them is

4 4real tridiagonal matrix and the other is 3 3 complex tridiagonal matrix. We

calculate 3th and 2th powers of these matrices, respectively. These examples can be veri…ed by using Maple 13 procedure given in Appendix 2.

Example 26 Let Y4 be a 4 4 real tridiagonal matrix given in (3.18) as in the

following: Y4 = 2 6 6 6 6 6 6 4 2 6 0 0 3 2 3 0 0 3 2 3 0 0 6 2 3 7 7 7 7 7 7 5 :

for a = 2 and b = 3: 3th power of Y4 is computed as in the following.

From (3.24) eigenvalues of Y4 can be written for k = 1; 2; 3; 4 as:

k = 2 6i cos

(k 1)

3 ;

namely, 1 = 2 6i; 2 = 2 3i; 3 = 2 + 3i and 4 = 2 + 6i. We also write the

transforming matrix N4;whose columns consist of eigenvectors of Y4;and its inverse

as: N4 = 2 6 6 6 6 6 6 6 6 6 4 e (0) T0 i( 1₆ 2) e (0) T0 i( 2₆ 2) e (0) T0 i( 3₆ 2) e (0) T0 i( 4₆ 2) e (1) T1 i( 1₆ 2) e (1) T1 i( 2₆ 2) e (1) T1 i( 3₆ 2) e (1) T1 i( 4₆ 2) e (2) T2 i( 1₆ 2) e (2) T2 i( 2₆ 2) e (2) T2 i( 3₆ 2) e (2) T2 i( 4₆ 2) e (3) T3 i( 1 2) 6 e (3) T3 i( 2 2) 6 e (3) T3 i( 3 2) 6 e (3) T3 i( 4 2) 6 3 7 7 7 7 7 7 7 7 7 5 = 2 6 6 6 6 6 6 6 4 1 1 1 1 i ₂i ₂i i 1 1₂ 1₂ 1 i i i i 3 7 7 7 7 7 7 7 5

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and N₄ 1 = 2 6 6 6 6 6 6 6 4 1f1e (0) T0 i( 1 2) 6 2f1e (1) T1 i( 1 2) 6 3f1e (2) T2 i( 1 2) 6 1f2e (0) T0 i( 2₆ 2) 2f2e (1) T1 i( 2₆ 2) 3f2e (2) T2 i( 2₆ 2) 1f3e (0) T0 i( 3₆ 2) 2f3e (1) T1 i( 3₆ 2) 3f3e (2) T2 i( 3₆ 2) 1f4e (0) T0 i( 4₆ 2) 2f4e (1) T1 i( 4₆ 2) 3f4e (2) T2 i( 4₆ 2) 4f1e (3) T3 i( 1₆ 2) 4f2e (3) T3 i( 2 2) 6 4f3e (3) T3 i( 3₆ 2) 4f4e (3) T3 i( 4₆ 2) 3 7 7 7 7 7 7 7 7 7 5 = 1 6 2 6 6 6 6 6 6 4 1 2i 2 i 2 2i 2 2i 2 2i 2 2i 1 2i 2 i 3 7 7 7 7 7 7 5 ; Then we get Y₄3 = N1J13N 1 1 = 2 6 6 6 6 6 6 4 100 90 108 54 45 154 99 54 54 99 154 45 54 108 90 100 3 7 7 7 7 7 7 5 :

(i; j)th entry of the Y3

4 can be veri…ed by formula given in (3.30).

Example 27 Let Y3 be a 3 3 complex tridiagonal matrix given in (3.18) as in the

following: Y3 = 2 6 6 6 4 1 i 10 + 4i 0 5 2i 1 i 5 + 2i 0 10 4i 1 i 3 7 7 7 5:

for a = 1 i and b = 5 + 2i: 2th power of Y3 is computed as in the following.

From (3.24) eigenvalues of Y3 can be written for k = 1; 2; 3 as:

k= (1 i) + (4 10i) cos

(k 1)

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namely, 1 = 5 11i; 2 = 1 iand 3 = 3 + 9i. We also write the transforming

matrix N3; whose columns consist of eigenvectors of the Y3; and its inverse as:

N3 = 2 6 6 6 4 e (0) T0 i( 1₆ 2) e (0) T0 i( 2₆ 2) e (0) T0 i( 3₆ 2) e (1) T1 i( 1 2) 6 e (1) T1 i( 2 2) 6 e (1) T1 i( 3 2) 6 e (2) T2 i( 1₆ 2) e (2) T2 i( 2₆ 2) e (2) T2 i( 3₆ 2) 3 7 7 7 5 = 2 6 6 6 4 1 1 1 i 0 i 1 1 1 3 7 7 7 5; and N₃ 1 = 2 6 6 6 4 1f1e (0) T0 i( 1₆ 2) 2f1e (1) T1 i( 1₆ 2) 3f1e (2) T2 i( 1₆ 2) 1f2e (0) T0 i( 2₆ 2) 2f2e (1) T1 i( 2₆ 2) 3f2e (2) T2 i( 2₆ 2) 1f3e (0) T0 i( 3₆ 2) 2f3e (1) T1 i( 3₆ 2) 3f3e (2) T2 i( 3₆ 2) 3 7 7 7 5 = 1 4 2 6 6 6 4 1 2i 1 2 0 2 1 2i 1 3 7 7 7 5: Then we get Y₃2 = N3J32N 1 3 = 2 6 6 6 4

42 42i 28 12i 42 + 40i

14 + 6i 84 82i 14 6i

42 + 40i 28 + 12i 42 42i

3 7 7 7 5: (i; j)th entry of the Y₃2, can be veri…ed by formula given in (3.30).

3.3 Positive Integer Powers of A Certain Type of Complex

Tridiagonal Matrix Family

Bn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a 2b 0 0 c a b . .. ... 0 c a . .. ... ... .. . . .. ... ... b 0 .. . . .. c a b 0 0 2c a 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ; (3.35)

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where bc 6= 0: This matrix is a matrix family of the matrices Un; Vn; Xnand Yngiven

by (1.10), (1.11), (3.1) and (3.18) respectively. For the solution of this problem we

shall apply the similarity transformation Br

n= NnJnrNn1; where Jn is the Jordan’s

form of Bn and Nn is the transforming matrix. Since Matrices Jn and Nn can be

found by providing eigenvalues and eigenvectors of Bn, we shall eigenvalues and

eigenvectors of Bn.

3.3.1 Eigenvalues and Eigenvectors of Bn

Theorem 28 _{Consider a; b; c 2 C; bc 6= 0. Let B}n be the n n complex tridiagonal

matrix given by (3.35). Then the eigenvalues of Bn are

k= a + 2

p

bc cos(k 1)

n 1 ; for k = 1; 2; : : : ; n; (3.36)

and the eigenvectors of Bn associated with eigenvalues are given by

&j = 2 6 6 6 6 6 6 6 4 pc b 0 T0 ₂jp_bca pc b 1 T1 j a 2pbc .. . pc b n 1 Tn 1 j a 2pbc 3 7 7 7 7 7 7 7 5 ; for j = 1; 2; : : : ; n: (3.37)

where Tk(x) is the kth degree Chebyshev polynomial of the …rst kind such that

(Ötele¸s and Akbulak, 2014a).

Proof. In order to prove the theorem, we need a relation between the matrices Bn

and Qn given by (3.5). Now let us consider the n n tridiagonal matrix fMn as the

following: f Mn= 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a=pbc 2 0 0 1 a=pbc 1 . .. ... 0 1 a=pbc . .. . .. ... .. . . .. . .. . .. 1 0 .. . . .. 1 a=pbc 1 0 0 2 a=pbc 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ;

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where bc 6= 0: Taking (3.5) and (3.6) into account, we …nd the eigenvalues of fMnto be a p bc+ 2 cos (k 1) n 1 ; for k = 1; 2; : : : ; n:

Dividing all entries of Bn by

p

bc; we also get a new n n matrix My

n as the following: M_ny = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a=pbc 2b=pbc 0 0 c=pbc a=pbc b=pbc . .. ... 0 c=pbc a=pbc . .. . .. ... .. . . .. . .. . .. b=pbc 0 .. . . .. c=pbc a=pbc b=pbc 0 0 2c=pbc a=pbc 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

Since the multiplication of subdiagonal entries of the matrices fMn and Mny are

the same, the characteristic polynomials of fMn and Mny have the same recursive

formula from Lemma 17. Therefore, the eigenvalues of these matrices are the same.

Moreover, since the eigenvalues of Bn are just

p

bctimes the eigenvalues of My

n; we

get eigenvalues of Bn as in (3.36).

Let us …nd the eigenvectors corresponding to each eigenvalue of Bn: Each

eigenvector of Bnis the solution of the following homogeneous linear equation system

( jIn Bn) x = 0; (3.39)

where j is the j th eigenvalue of Bn (1 j n) : We clearly write the expression

(3.39) as follows: ( j a) x1 2bx2 = 0 cx1+ ( j a) x2 bx3 = 0 cx2+ ( j a) x3 bx4 = 0 .. . cxn 2+ ( j a) xn 1 bxn= 0 2cxn 1+ ( j a) xn= 0 : (3.40)

Dividing all terms of the each equations in (3.40) bypbc, choosing x1 = 1arbitrarily

and solving the set of systems (3.40) according to x1;we get (3.37). So the theorem

is proved.

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3.3.2 Derivation of the formula for the entries of Br n

Theorem 29 _{Consider a; b; c 2 C; bc 6= 0. Let B}n be an n n complex tridiagonal

matrix given by (3.35). Then (i,j)th entry of Br

n is [B_nr]_i;j = j 2n 2 n X k=1 k r k r c b i j Ti 1 k a 2pbc Tj 1 k a 2pbc (3.41) for i; j = 1; 2; :::; n: Here j = 8 < : 1; if j = 1; n; 2; if 1 < j < n: and _k = 8 < : 1; if k = 1; n; 2; if 1 < k < n; (3.42)

k and Tk(x) are de…ned by (3.36) and (3.38).

Proof. We shall consider the relation Br

n = NnJnrNn1. In order to get the general

expression for the entries of Bnr; we …rstly …nd the matrices Jn and Nn:

the Jordan’s form of Bn

Jn= diag ( 1; 2; : : : ; n) : (3.43)

Nn= 2 6 6 6 6 6 6 6 4 pc b 0 T0 ₂1p_bca pc b 0 T0 ₂2p_bca : : : pc b 0 T0 ₂np_bca p_c b 1 T1 ₂1p_bca p_c b 1 T1 ₂2p_bca : : : p_c b 1 T1 ₂np_bca .. . ... . .. ... pc b n 1 Tn 1 ₂1p_bca pc b n 1 Tn 1 ₂2p_bca : : : pc b n 1 Tn 1 ₂np_bca 3 7 7 7 7 7 7 7 5 : (3.44)

n by j (Nn1 = ( 1; 2; : : : ; n)) ;

j = j 2 6 6 6 6 6 6 6 6 4 f1 q b c j 1 Tj 1 ₂1p_bca f2 q b c j 1 Tj 1 ₂2p_bca .. . fn q b c j 1 Tj 1 ₂np_bca 3 7 7 7 7 7 7 7 7 5 for j = 1; 2; : : : ; n; (3.45)

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where fk = _{2n 2}k ; and j; k are de…ned by (3.42). Taking (3.45) into account, we write down N 1 n as N_n1 = 1 2n 2 2 6 6 6 6 6 6 6 6 6 6 6 6 6 4 q b c 0 T0 ₂1p_bca q b c 1 2T1 ₂1p_bca q b c 0 2T0 ₂2p_bca q b c 1 4T1 ₂2p_bca .. . ... . .. q b c 0 2T0 n₂p1_bca q b c 1 4T1 n₂p1_bca q b c 0 T0 ₂np_bca q b c 1 2T1 ₂np_bca q b c n 2 2Tn 2 ₂1p_bca q b c n 1 Tn 1 ₂1p_bca q b c n 2 4Tn 2 ₂2p_bca q b c n 1 2Tn 1 ₂2p_bca .. . ... q b c n 2 4Tn 2 n₂p1_bca q b c n 1 2Tn 1 n₂p1_bca q b c n 2 2Tn 2 ₂np_bca q b c n 1 Tn 1 ₂np_bca 3 7 7 7 7 7 7 7 7 7 7 7 7 7 5 : (3.46)

Denoting i th row of Nn in (3.44) by Li (i = 1; 2; ::; n) ;we have

LT_i = 2 6 6 6 6 6 6 6 4 pc b i 1 Ti 1 ₂1p_bca p_c b i 1 Ti 1 ₂2p_bca .. . p_c b i 1 Ti 1 ₂np_bca 3 7 7 7 7 7 7 7 5 ;

where T means operation of transposition. Using the equalities Br

n= NnJnrNn1 (see (3.43), (3.44), (3.45) and (3.46)) and Jr n = diag ( r 1; r 2; : : : ; r

n) and denoting (i,j)th

entry of Br n by [Bnr]i;j; we get [B_nr]_i;j = LiJn jr = Li 2 6 6 6 6 6 6 6 6 4 q b c j 1 jf1 r1Tj 1 ₂1p_bca q b c j 1 jf2 r2Tj 1 ₂2p_bca .. . q b c j 1 jfn rnTj 1 ₂np_bca 3 7 7 7 7 7 7 7 7 5 : Consequently, [Br_n]_i;j = j 2n 2 n X k=1 k( k) r r c b i j Ti 1 k a 2pbc Tj 1 k a 2pbc ;

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for i; j = 1; 2; :::; n. It is desirable.

Now, if we write c = b in Bn given by (3.35), we get Xn given by (3.1). From

Theorem 29 we can directly obtain the expression given in (3.12) for the entries of the powers of Xn.

Corollary 30 _{Consider a; b; c 2 C; bc 6= 0: Let B}n be an n n complex tridiagonal

matrix given by (3.35). If c = b; then (i,j)th entry of Br

n is [B_nr]_i;j = j 2n 2 n X k=1 k r kTi 1 k a 2b Tj 1 k a 2b ;

for i; j = 1; 2; :::; n: Here k; Tk(x) and j; k are de…ned by (3.36), (3.38) and

(3.42), respectively.

If we write c = b in Bn given by (3.35), we get Yn given by (3.18). From

Theorem 29 we can also obtain the expression given in (3.30) for the entries of the powers of Yn:

Corollary 31 _{Consider a; b; c 2 C; bc 6= 0 and i =} p 1. Let Bn be the

n n complex tridiagonal matrix given by (3.35), If c = b; then (i,j)th entry

of Bnr is

[Br_n]_i;j = je(i 1)e(j 1)

2n 2 n X k=1 k r kTi 1 i( k a) 2b Tj 1 i( k a) 2b ;

for i; j = 1; 2; :::; n: Here e(k); k; Tk(x) and j; k are de…ned by (3.26), (3.36),

(3.38) and (3.42), respectively.

Proof. From Theorem 29 we have

[B_nr]_i;j = j 2n 2 n X k=1 k r k(i) i j Ti 1 i( k a) 2b Tj 1 i( k a) 2b ;

for c = b: Since Tk( x) = ( 1)kTk(x) (Mason and Handscomb, 2003), We get

[B_nr]_i;j = j( 1) i+j 2 (i)i j 2n 2 n X k=1 k r kTi 1 i( k a) 2b Tj 1 i( k a) 2b :

To conclude the proof we need to show that

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Since ( 1)i+j 2 = ( 1)i+j = ( 1)j i = (i)2j 2i; we get ( 1)i+j 2(i)i j = (i)j i:

Since e(i 1)e(j 1) = ei(j i)₂_{and (i)}l

= eil2 (Mason and Handscomb, 2003),

( 1)i+j 2(i)i j = (i)j i = ei(j i)2

= e(i 1)e(j 1):

It is desirable.

3.3.3 Numerical Considerations

In this section, we give three examples related to the formula of Bnr. First one is

4 4real tridiagonal matrix, the other is 3 3complex tridiagonal matrix and the

last one is 6 6 real tridiagonal matrix with eigenvalues on imaginary axis. We

calculate 3th, 4th and 5th powers of these matrices, respectively. These examples can be veri…ed by using Maple 13 procedure given in Appendix 3.

Example 32 Let B4 be a 4 4 real tridiagonal matrix (for a = 3; b = 1 and c = 4,

given in (3.35)) as in the following:

B4 = 2 6 6 6 6 6 6 4 3 2 0 0 4 3 1 0 0 4 3 1 0 0 8 3 3 7 7 7 7 7 7 5 :

3th power of B4 is computed as in the following.

From (3.36) the eigenvalues of B4 can be written for k = 1; 2; 3; 4 as:

k = 3 + 4 cos

(k 1)

3 ;

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matrix N4; whose columns consist of eigenvectors of B4; and its inverse as: N4 = 2 6 6 6 6 6 6 6 6 4 T0 1₄ 3 T0 2₄3 T0 3₄ 3 T0 4₄ 3 2T1 1₄ 3 2T1 2₄ 3 2T1 3₄ 3 2T1 4₄3 (22_{) T} 2 1₄ 3 (22) T2 2₄ 3 (22) T2 3₄3 (22) T2 4₄ 3 (23_{) T} 3 1₄ 3 (23) T3 2₄ 3 (23) T3 3₄3 (23) T3 4₄ 3 3 7 7 7 7 7 7 7 7 5 = 2 6 6 6 6 6 6 4 1 1 1 1 2 1 1 2 4 2 2 4 8 8 8 8 3 7 7 7 7 7 7 5 ; and N₄ 1 = 1 6 2 6 6 6 6 6 6 6 6 4 T0 1₄ 3 1₂ 2T1 1₄3 ₂12 2T2 1₄ 3 ₂13 T3 1₄ 3 2T0 2₄ 3 1₂ 4T1 2₄3 ₂12 4T2 2₄ 3 ₂13 2T3 2₄ 3 2T0 3₄ 3 1₂ 4T1 3₄3 ₂12 4T2 3₄ 3 ₂13 2T3 3₄ 3 T0 4₄ 3 1₂ 2T1 4₄3 ₂12 2T2 4₄ 3 ₂13 T3 4₄ 3 3 7 7 7 7 7 7 7 7 5 = 2 6 6 6 6 6 6 6 6 6 4 1 6 1 6 1 12 1 48 1 3 1 6 1 12 1 24 1 3 1 3 1 12 1 24 1 6 1 3 1 12 1 48 3 7 7 7 7 7 7 7 7 7 5 : Then we get B₄3 = N4J43N 1 4 = 2 6 6 6 6 6 6 4 99 78 18 2 156 135 47 9 144 188 135 39 128 288 312 99 3 7 7 7 7 7 7 5 :

(i,j)th entry of the B43, can be veri…ed by formula given in (3.41).

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b = 3 + 4i and c = 3 4i given in (3.35)) as in the following: B3 = 2 6 6 6 4 1 i 6 + 8i 0 3 4i 1 i 3 + 4i 0 6 8i 1 i 3 7 7 7 5:

From (3.36) the eigenvalues of B3 can be written for k = 1; 2; 3 as:

k= (1 i) + 10 cos

(k 1)

2 ;

namely, 1 = 11 i; 2 = 1 i and 3 = 9 i. We also write the transforming

matrix N3; whose columns consist of eigenvectors of the B3; and its inverse as:

N3 = 2 6 6 6 6 6 4 T0 1 ₁₀(1 i) T0 2 ₁₀(1 i) T0 3 ₁₀(1 i) 3 5 4 5i T1 1 (1 i) 10 3 5 4 5i T1 2 (1 i) 10 3 5 4 5i T1 3 (1 i) 10 3 5 4 5i 2 T2 1 (1 i) 10 3 5 4 5i 2 T2 2 (1 i) 10 3 5 4 5i 2 T2 3 (1 i) 10 3 7 7 7 7 7 5 = 2 6 6 6 6 4 1 1 1 3 5 4 5i 0 3 5 + 4 5i 7 25 24 25i 7 25+ 24 25i 7 25 24 25i 3 7 7 7 7 5 ; and N₃1 = 2 6 6 6 6 6 4 T0 1 ₁₀(1 i) 2 3₅ +4₅i T1 1 ₁₀(1 i) 3₅ + 4₅i 2 T2 1 ₁₀(1 i) 2T0 2 (1 i) 10 4 3 5 + 4 5i T1 2 (1 i) 10 2 3 5 + 4 5i 2 T2 2 (1 i) 10 T0 3 ₁₀(1 i) 2 3₅ +4₅i T1 3 ₁₀(1 i) 3₅ + 4₅i 2 T2 3 ₁₀(1 i) 3 7 7 7 7 7 5 = 1 4 2 6 6 6 6 6 4 1 6₅ +8₅i ₂₅7 + 24₂₅i 2 0 14₂₅ 48₂₅i 1 6₅ 8₅i ₂₅7 + 24₂₅i 3 7 7 7 7 7 5 : Then we get B4₃ = N3J34N3 1 = 2 6 6 6 4

4996 600i 5616 + 688i 824 + 4968i

456 2792i 9996 1200i 2808 + 344i

1976 4632i 912 5584i 4996 600i

3 7 7 7 5: (i,j)th entry of the B34, can be veri…ed by formula given in (3.41).

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Example 34 Let B6 be a 6 6 real tridiagonal matrix (for a = 5; b = 3 and

c = b = 3, given in (3.35)) as in the following:

B6 = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 5 6 0 0 0 0 3 5 3 0 0 0 0 3 5 3 0 0 0 0 3 5 3 0 0 0 0 3 5 3 0 0 0 0 6 5 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

From (3.36) the eigenvalues of B6 can be written for k = 1; 6 as:

k = 5 + 6i cos (k 1) 3 ; namely, 1 = 5 + 6i; 2 = 5 + 4:8541i; 3 = 5 + 1:8541i; 4 = 5 1:8541i; 5 = 5 4:8541i; 6 = 5 6i:

We also write the transforming matrix N6;whose columns consist of eigenvectors of

B6; and its inverse as:

N6 = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 1

i 0:8090i 0:3090i 0:3090i 0:8090i i

1 0:3090 0:8090 0:8090 0:3090 1

i 0:3090i 0:8090i 0:8090i 0:3090i i

1 0:8090 0:3090 0:3090 0:8090 1 i i i i i i 3 7 7 7 7 7 7 7 7 7 7 7 7 5 ;

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and N₆1 = 2 6 6 6 6 6 6 6 6 6 6 6 6 4

0:1 0:2i 0:2 0:2i 0:2 0:1i

0:2 0:3236i 0:1236 0:1236i 0:3236 0:2i

0:2 0:1236i 0:3236 0:3236i 0:1236 0:2i

0:2 0:3236i 0:1236 0:1236i 0:3236 0:2i

0:1 0:2i 0:2 0:2i 0:2 0:1 3 7 7 7 7 7 7 7 7 7 7 7 7 5 : Then we get B₆5 = N6J65N 1 6 = 2 6 6 6 6 6 6 6 6 6 6 6 6 4 7225 16890 6300 11070 4050 486 8445 10375 13980 1125 5049 2025 3150 13980 5200 7959 1125 5535 5535 1125 7959 5200 13980 3150 2025 5049 1125 13980 10375 8445 486 4050 11070 6300 16890 7225 3 7 7 7 7 7 7 7 7 7 7 7 7 5 :

(i,j)th entry of the B5

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4

ON COMPUTING OF POSITIVE INTEGER POWERS

FOR CERTAIN REAL CIRCULANT MATRIX

In this chapter, we present a single formula for the entries of the r th power (r 2 N) of the n n real circulant matrix An := circn(a0; a1; 0; : : : ; 0; a 1) of odd and even

order, in terms of the Chebyshev polynomials. Namely,

An := 2 6 6 6 6 6 6 6 6 6 6 6 6 4 a0 a1 0 : : : 0 a 1 a 1 a0 a1 . .. 0 0 a 1 a0 . .. ... ... .. . . .. ... ... ... 0 0 . .. ... ... a1 a1 0 : : : 0 a 1 a0 3 7 7 7 7 7 7 7 7 7 7 7 7 5 : (4.1)

As we will see, that formula obtained for the powers of the matrix in (4.1) is an extension of the one obtained for the powers of the matrix in (1.16). The chapter …nishes some numerical examples concerned with the formula.

4.1 General Expression for the Entries of

A

r_n

The well-known eigenvalue decomposition of an n ncirculant matrix (Davis, 1994)

is that

circn(c0; c1; : : : ; cn 1) = FnDnFn; (4.2)

where * denotes conjugate transpose (i.e F_n = FT_n), Fn is the n n Fourier matrix:

[Fn]u;v = 1 p ne 2 (u 1)(v 1) n i; 1 u; v n;

and Dn= diag ( 1; 2; : : : ; n) with

k = n X r=1 cr 1e 2 (k 1)(r 1) n i; 1 k n: (4.3)

It is also known that Fn is unitary (Davis, 1994).

Let Uk(x) be the k th degree Chebyshev polynomial of the second kind:

Uk(x) =

sin ((k + 1) arccos x)

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and Tk(x) is the k th degree Chebyshev polynomial of the …rst kind, with

k _{2 N [ f0g :}

Tk(x) = cos (k arccos x) ; 1 x 1 (4.5)

(Mason and Handscomb, 2003).

Based on the eigenvalue decomposition (4.2), we can prove Theorem 35 that

provides a general expression for the entries of Ar

n.

Theorem 35 Let An= circn(a0; a1; 0; : : : ; 0; a 1) be an n n real circulant matrix

and k = cos

2 (k 1)

n ; 1 k n; 3 n 2 N: Then (u,v)th entry of A

r

n is given by:

[Ar_n]_u;v = 1

n(S1+ S2) ;

for all r 2 N and 1 u; v n; where S1 and S2 are respectively that

S1 = bn 2c+1 X k=1 a0+ (a 1+ a1) T1( k) + (a 1 a1) sign n 2 + 1 k q 1 2 kilim_j !kU0( j) r T_{jv uj}( k) + sign (v u) sign n 2 + 1 k q 1 2 kilim j!kUjv uj 1( j) ; and S2 = bn+1 2 c X k=2 a0+ (a 1+ a1) T1( k) (a 1 a1) sign n 2 + 1 k q 1 2 kilim_j !kU0( j) r T_{jv uj}( k) sign (v u) sign n 2 + 1 k q 1 2 kiUjv uj 1( k) :

Here bxc is the largest integer less than or equal to x:

Proof. By using (4.2), [Ar_n]_u;v = [(FnDnFn) r ]_u;v = [FnDrnFn]u;v = n X k=1 [Fn]_u;k[DnrFn]k;v = n X k=1 [Fn]u;k r k[Fn]v;k = 1 n n X k=1 r ke 2 (u 1)(k 1) n ie 2 (v 1)(k 1) n i;

according to above the last equation, we get [Ar_n]_u;v = 1 n n X k=1 r ke 2 (k 1)(v u) n i: (4.6)

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From (4.3) and e 2 (k 1)(rn 1) = e

2 (k 1)(n+2 r 1)

n i for all 2 r n (see (Gutiérrez,

2008a), we can write k as

k = a0+ a1e 2 (k 1) n i+ a ₁e 2 (k 1) n i = a0+ (a 1+ a1) cos 2 (k 1) n + (a 1 a1) i sin 2 (k 1) n :

Observe that from (4.4) and (4.5), we have

Tm cos 2 (k 1) n = cos 2 (k 1) m n ; and Um 1 cos 2 (k 1) n = sin2 (k 1)m_n sin2 (k 1)_n : (4.7)

In (4.7), we have the indeterminate form 0=0 for k = 1 and k = n₂ + 1: So we can

write k = a0+(a 1+ a1) T1 cos 2 (k 1) n +(a 1 a1) sin 2 (k 1) n ijlim!kU0 cos 2 (j 1) n :

Since k = cos2 (k 1)_n and

sin2 (k 1) n = 8 > < > : p 1 2 k if n 2 + 1 k > 0; p 1 2 k if n 2 + 1 k < 0; (4.8) k is obtained as k= a0+ (a 1+ a1) T1( k) + (a 1 a1) sign n 2 + 1 k q 1 2 kilim_j !kU0( j) : (4.9)

From (4.9) we can write k = n+2 k for 2 k n: Namely,

Dn = diag 1; 2; : : : ; n+1 2 ; n+1 2 ; : : : ; 2 if n is odd, Dn = diag 1; 2; : : : ; n₂; n₂+1; n₂; : : : ; 2 if n is even. By using (4.6), [Ar

n]u;v can be expressed as

[Ar_n]_u;v = 8 > > > > > > > < > > > > > > > : 1 n 2 4 r 1+ n+1 2 X k=2 r ke 2 (k 1)(v u) n i+ n X k=n+1₂ +1 r ke 2 (k 1)(v u) n i 3 5 if n is odd, 1 n 2 4 r 1+ n 2+1 X k=2 r ke 2 (k 1)(v u) n i+ n X k=n 2+2 r ke 2 (k 1)(v u) n i 3 5 if n is even,