# Explicit Expression for Arbitrary Positive Powers of Special Tridiagonal Matrices

Authors: Mohammad Beiranvand; Mojtaba Ghasemi Kamalvand
Journal: Journal of Applied Mathematics
 (2020)

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## Abstract

Tridiagonal matrices appear frequently in mathematical models. In this paper, we derive the positive integer powers of special tridiagonal matrices of arbitrary order.


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## Body

## 1. Introduction

Recently, computing the integer powers of tridiagonal matrices has been a very popular problem. Tridiagonal matrices are used in different areas of science and engineering, for instance, the solution of difference systems [1], numerical solution of PDEs [2], telecommunication system analysis [3, 4], texture modeling [5], and image processing and coding [6]. In these areas, the computation of the powers of these matrices is necessary. Therefore, there are a lot of studies dealing with the powers of these matrices using the well-known expression Am=UJmU−1 where J is Jordan’s form of the matrix A and U is the transforming matrix. We need the eigenvalues and eigenvectors of these matrices to calculate Am.Rimas [7–13], investigated positive integer powers of certain tridiagonal matrices, Oteles and Akbulak [14, 15] and Gutierrez [16] generalized some papers of Rimas, and Wang [17] derived the entries of positive integer powers of complex persymmetric anti-tridiagonal matrices with constant anti-diagonals. Some authors also investigated integer powers of certain tridiagonal matrices.In this paper, we consider then-th order tridiagonal matrix A of the following type
(1)A=−α+bcabc0⋱⋱⋱abc0a−β+b,where a, b, c, α, and β are the numbers in the complex C. There are many mathematical models that are involved in this form [18].


## 2. Main Results

According to the following lemmas, Wen-Chyuan Yueh obtains eigenvalue and corresponding eigenvectors for matrix (1), in special cases. In this section, we firstly find the transforming matrices and their invers for the matrix (1). Secondly, we present a general expression for the entries of Am for m∈ℕ.Lemma 1 (see [19]).
Supposeα=0, β=ac≠0, and ρ=a/c, then the eigenvalues λ1,λ2,⋯,λn of A are given by
(2)λk=b+2accos2kπ2n+1,k=1,2,⋯,n,the corresponding eigenvectors uk=u1k,u2k,⋯,unkT,k=1,2,⋯,n, are given by
(3)ujk=ρj−1sin2kjπ2n+1,j=1,2,⋯,n.In this paper, we need the following theorem.Theorem 2.
(canonical Jordan’s form [20]). Let A be any square matrix. Then, there exists a nonsingular matrix U which transforms A into a block diagonal matrix J such that
(4)U−1AU=J=diagJn1λ1,Jn2λ2,⋯Jnlλl,n1+n2+⋯+nl=n.which is called the canonical Jordan’s form, λj being the eigenvalues of A and Jk∈Ck×k a Jordan block of the form
(5)J1λ=λ,J2λ=λ10λ,⋯,Jkλ=λ1λ10⋱⋱0λ1.Since all the eigenvaluesλk for k=1,2,⋯,n are distinct (cosθ is a strictly decreasing function of θ on (0,π), and a≠0≠c), columns of the transforming matrix U are the eigenvectors of the matrix (1). Also, all eigenvalues λk correspond to the single Jordan cell J1λk in the matrix J, then we write down Jordan’s form of the matrix A as
(6)J=diagλ1,λ2,⋯,λn.From (3), we can write the column transforming matrix U as
(7)uk=sin2kπ2n+1,ρ1sin4kπ2n+1,⋯,ρn−1sin2knπ2n+1T,for k=1,2,⋯,n.Hence,(8)U=u1,u2,u3,⋯,un.Let(9)D=diag1,ρ1,ρ2,ρ3,⋯,ρn−1,U~=u~1,u~2,u~3,⋯,u~n,in which
(10)u~k=sin2kπ2n+1,sin4kπ2n+1,⋯,sin2knπ2n+1T,for k=1,2,⋯,n.Therefore, we have(11)U=DU~,(12)U−1=U~−1D−1.So an explicit expression forU~−1 will suffice.Theorem 3.
SupposeU~ is defined as above, then
(13)U~−1=42n+1U~.Proof.
We show that(14)u~iTu~j=2n+14,ifi=j,0,ifi≠j.From (10), equals sinαsinβ=1/2cosα−β−cosα+β and
(15)∑k=1ncoskθ=sinn+1/2θ2sin1/2−12,follows
(16)u~iTu~j=∑k=1nsin2kiπ2n+1sin2kjπ2n+1=12∑k=1ncos2ki−jπ2n+1−cos2ki+jπ2n+1.Ifi=j,
(17)u~iTu~j=n2−12sinn+1/24iπ/2n+12sin4iπ/2n+1/2−12=n2−12sin2iπ2sin2iπ/2n+1−12=n2−120−12=2n+14.Ifi≠j,
(18)u~iTu~j=12∑k=1ncos2ki−jπ2n+1−cos2ki+jπ2n+1=12sinn+1/22i−jπ/2n+12sin2i−jπ/2n+1/2−sinn+1/22i+jπ/2n+12sin2i+jπ/2n+1/2=14sini−jπsini−jπ/2n+1−sini+jπsini+jπ/2n+1.Ifi and j are even or odd, then i−j and i+j are even; therefore, sini−jπ=sini+jπ=0, so we have
(19)u~iTu~j=0.If one ofi or j is even and the other is odd, then i−j and i+j are odd; therefore, we have
(20)sini−jπ−i−jπ2n+1=sini−jπ2n+1,(21)sini+jπ−i+jπ2n+1=sini+jπ2n+1.From (18), (20), and (21) follows
(22)u~iTu~j=0.For derivation of the formula for the entries ofAm, from (4), (6), (10), (12), and (13), we can conclude
(23)Am=UJmU−1=UJmU~−1D−1=42n+1Udiadλ1m,λ2m,⋯,λnmU~−1diag1,ac−1/2,ac−2/2,⋯,ac−n−1/2.By substitutingU and U~−1 in the latter equation and doing the necessary computation follows
(24)Ami,j=42n+1aci−j/2∑k=1nλkmsin2ikπ2n+1sin2jkπ2n+1,for i,j=1,2,⋯,n, where λk=b+2accos2kπ/2n+1,k=1,2,⋯,n.Lemma 4 (see [19]).
Supposeα=−ac≠0 and β=0 and ρ=ac, then the eigenvalues λ1,λ2,⋯,λn of An are given by
(25)λk=b+2accos2k−1π2n+1,k=1,2,⋯n.The corresponding eigenvectors,vk=v1k,v2k,⋯,vnkT,k=1,2,⋯,n, are given by
(26)vjk=ρj−1cos2k−12j−122n+1,j=1,2,⋯,n.From (26) for k=1,2,⋯,n, we can write the columns transforming matrix V as
(27)vk=cos2k−1π22n+1,ρ1cos32k−1π22n+1,⋯,ρn−1cos2n−12k−1π22n+1THence,(28)V=v1,v2,⋯,vn.Let(29)D=diag1,ρ1,ρ2,⋯,ρn−1,V~=v~1,v~2,⋯,v~n,in which
(30)v~k=cos2k−1π22n+1,cos32k−1π22n+1,⋯,cos2n−12k−1π22n+1T,for k=1,2,⋯,n.Therefore, we have(31)V=DV~,V−1=V~−1D−1.So an explicit expression forV~−1 will suffice.Theorem 5.
SupposeV~ is defined as above, then
(32)V~−1=42n+1V~.Proof.
We show that(33)v~iTv~j=2n+14,ifi=j,0,ifi≠j.From (30) equals cosαcosβ=1/2cosα−β+cosα+β and
(34)∑k=1ncos2k−1θ=sin2nθ2sinθ,follows
(35)v~iTv~j=∑k=1ncos2k−12i−1π22n+1cos2k−12j−1π22n+1=12∑k=1ncos2k−12i−1π22n+1−2k−12j−1π22n+1+cos2k−12i−1π22n+1+2k−12j−1π22n+1=12∑k=1ncos22k−1i−jπ22n+1+cos22k−1i+j−1π22n+1.Ifi=j,
(36)v~iTv~j=n2+12∑k=1ncos22k−12i−1π22n+1=n2+12sin2n2i−1π/2n+12sin2i−1π/2n+1=n2+12sin2n/2n+12i−1π2sin2i−1π/2n+1=n2+12sin2i−1π−2i−1π/2n+12sin2i−1π/2n+1n2+12sin2i−1π/2n+12sin2i−1π/2n+1=2n+14.Ifi≠j,
(37)v~iTv~j=12∑k=1ncos2k−1i−jπ2n+1+cos2k−1i+j−1π2n+1=12sin2n2k−1i−jπ/2n+12sin2k−1i−jπ/2n+1+sin2n2k−1i+j−1π/2n+12sin2k−1i+j−1π/2n+1=12sin2ni−jπ/2n+12sini−jπ/2n+1+sin2ni+j−1π/2n+12sini+j−1π/2n+1=12sin1−1/2n+1i−jπ2sini−jπ/2n+1+sin1−1/2n+1i+j−1π2sini+j−1π/2n+1.Ifi and j are even or odd, then i−j is even and i+j−1 is odd; therefore,
(38)sin1−12n+1i−jπ=−sini−jπ2n+1,sin1−12n+1i+j−1π=−sini+j−1π2n+1.So we have(39)v~iTv~j=0.If one ofi or j is even and the other is odd, then i−j and i+j−1 are even; therefore, we have
(40)sin1−12n+1i−jπ=−sini−jπ2n+1and
(41)sin1−12n+1i+j−1π=−sini+j−1π2n+1.Therefore,(42)v~iTv~j=0.Similar for (24), we can conclude
(43)Ami,j=42n+1aci−j/2∑k=1nλkmcos2k−12i−1π22n+1sin2k−12j−1π22n+1,for i,j=1,2,⋯,n.Lemma 6 (see [18]).
Supposeα=−β=ac≠0 and ρ=ac, then the eigenvalues λ1,λ2,⋯,λn of An are given by
(44)λk=b+2accos2k−1π2n,k=1,2,⋯,n.The corresponding eigenvectors are given by(45)uk=u1k,u2k,⋯,unkT,k=1,2,⋯,n,ujk=ρj−1sin2k−12j−1π4n,j=1,2,⋯,n.In the caseα=−β=−ac≠0, the eigenvalues are given by (44) and the corresponding eigenvectors by
(46)vk=v1k,v2k,⋯,vnkT,k=1,2,⋯,n,vjk=ρj−1cos2k−12j−1π4n,j=1,2,⋯,n.In these cases which are similar to previous cases, we showed that integer powers of the matrixA for i,j=1,2,⋯,n, respectively,
(47)Ami,j=2naci−j/2∑k=1nλkmsin2k−12i−1π4nsin2k−12j−1π4n,where λk=b+2accos2k−1π/2n,k=1,2,⋯,n, and
(48)Ami,j=2naci−j/2∑k=1nλkmcos2k−12i−1π4ncos2k−12j−1π4n,where λk=b+2accos2k−1π/2n,k=1,2,⋯,n.



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