JAMJournal of Applied Mathematics1687-00421110-757XHindawi10.1155/2020/72904037290403Research ArticleExplicit Expression for Arbitrary Positive Powers of Special Tridiagonal MatricesBeiranvandMohammadhttps://orcid.org/0000-0002-6248-697XGhasemi KamalvandMojtabaAndrianovIgorDepartment of MathematicsLorestan UniversityKhorramabadIranlu.ac.ir20201292020202018052020230720202707202012920202020Copyright © 2020 Mohammad Beiranvand and Mojtaba Ghasemi Kamalvand.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

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 , numerical solution of PDEs , telecommunication system analysis [3, 4], texture modeling , and image processing and coding . 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=UJmU1 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 , investigated positive integer powers of certain tridiagonal matrices, Oteles and Akbulak [14, 15] and Gutierrez  generalized some papers of Rimas, and Wang  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 the n-th order tridiagonal matrix A of the following type (1)A=α+bcabc0abc0aβ+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 .

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.

Lemma 1 (see [<xref ref-type="bibr" rid="B19">19</xref>]).

Suppose α=0, β=ac0, 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=ρj1sin2kjπ2n+1,j=1,2,,n.

In this paper, we need the following theorem.

Theorem 2.

(canonical Jordan’s form ). 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)U1AU=J=diagJn1λ1,Jn2λ2,Jnlλl,n1+n2++nl=n.which is called the canonical Jordan’s form, λj being the eigenvalues of A and JkCk×k a Jordan block of the form (5)J1λ=λ,J2λ=λ10λ,,Jkλ=λ1λ100λ1.

Since all the eigenvalues λk for k=1,2,,n are distinct (cosθ is a strictly decreasing function of θ on (0,π), and a0c), 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,,ρn1sin2knπ2n+1T,for k=1,2,,n.

Hence, (8)U=u1,u2,u3,,un.

Let (9)D=diag1,ρ1,ρ2,ρ3,,ρn1,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)U1=U~1D1.

So an explicit expression for U~1 will suffice.

Theorem 3.

Suppose U~ is defined as above, then (13)U~1=42n+1U~.

Proof.

We show that (14)u~iTu~j=2n+14,ifi=j,0,ifij.

From (10), equals sinαsinβ=1/2cosαβcosα+β and (15)k=1ncoskθ=sinn+1/2θ2sin1/212,follows (16)u~iTu~j=k=1nsin2kiπ2n+1sin2kjπ2n+1=12k=1ncos2kijπ2n+1cos2ki+jπ2n+1.

If i=j, (17)u~iTu~j=n212sinn+1/24iπ/2n+12sin4iπ/2n+1/212=n212sin2iπ2sin2iπ/2n+112=n212012=2n+14.

If ij, (18)u~iTu~j=12k=1ncos2kijπ2n+1cos2ki+jπ2n+1=12sinn+1/22ijπ/2n+12sin2ijπ/2n+1/2sinn+1/22i+jπ/2n+12sin2i+jπ/2n+1/2=14sinijπsinijπ/2n+1sini+jπsini+jπ/2n+1.

If i and j are even or odd, then ij and i+j are even; therefore, sinijπ=sini+jπ=0, so we have (19)u~iTu~j=0.

If one of i or j is even and the other is odd, then ij and i+j are odd; therefore, we have (20)sinijπijπ2n+1=sinijπ2n+1,(21)sini+jπi+jπ2n+1=sini+jπ2n+1.

From (18), (20), and (21) follows (22)u~iTu~j=0.

For derivation of the formula for the entries of Am, from (4), (6), (10), (12), and (13), we can conclude (23)Am=UJmU1=UJmU~1D1=42n+1Udiadλ1m,λ2m,,λnmU~1diag1,ac1/2,ac2/2,,acn1/2.

By substituting U and U~1 in the latter equation and doing the necessary computation follows (24)Ami,j=42n+1acij/2k=1nλkmsin2ikπ2n+1sin2jkπ2n+1,for i,j=1,2,,n, where λk=b+2accos2kπ/2n+1,k=1,2,,n.

Lemma 4 (see [<xref ref-type="bibr" rid="B19">19</xref>]).

Suppose α=ac0 and β=0 and ρ=ac, then the eigenvalues λ1,λ2,,λn of An are given by (25)λk=b+2accos2k1π2n+1,k=1,2,n.

The corresponding eigenvectors, vk=v1k,v2k,,vnkT,k=1,2,,n, are given by (26)vjk=ρj1cos2k12j122n+1,j=1,2,,n.

From (26) for k=1,2,,n, we can write the columns transforming matrix V as (27)vk=cos2k1π22n+1,ρ1cos32k1π22n+1,,ρn1cos2n12k1π22n+1T

Hence, (28)V=v1,v2,,vn.

Let (29)D=diag1,ρ1,ρ2,,ρn1,V~=v~1,v~2,,v~n,in which (30)v~k=cos2k1π22n+1,cos32k1π22n+1,,cos2n12k1π22n+1T,for k=1,2,,n.

Therefore, we have (31)V=DV~,V1=V~1D1.

So an explicit expression for V~1 will suffice.

Theorem 5.

Suppose V~ is defined as above, then (32)V~1=42n+1V~.

Proof.

We show that (33)v~iTv~j=2n+14,ifi=j,0,ifij.

From (30) equals cosαcosβ=1/2cosαβ+cosα+β and (34)k=1ncos2k1θ=sin2nθ2sinθ,follows (35)v~iTv~j=k=1ncos2k12i1π22n+1cos2k12j1π22n+1=12k=1ncos2k12i1π22n+12k12j1π22n+1+cos2k12i1π22n+1+2k12j1π22n+1=12k=1ncos22k1ijπ22n+1+cos22k1i+j1π22n+1.

If i=j, (36)v~iTv~j=n2+12k=1ncos22k12i1π22n+1=n2+12sin2n2i1π/2n+12sin2i1π/2n+1=n2+12sin2n/2n+12i1π2sin2i1π/2n+1=n2+12sin2i1π2i1π/2n+12sin2i1π/2n+1n2+12sin2i1π/2n+12sin2i1π/2n+1=2n+14.

If ij, (37)v~iTv~j=12k=1ncos2k1ijπ2n+1+cos2k1i+j1π2n+1=12sin2n2k1ijπ/2n+12sin2k1ijπ/2n+1+sin2n2k1i+j1π/2n+12sin2k1i+j1π/2n+1=12sin2nijπ/2n+12sinijπ/2n+1+sin2ni+j1π/2n+12sini+j1π/2n+1=12sin11/2n+1ijπ2sinijπ/2n+1+sin11/2n+1i+j1π2sini+j1π/2n+1.

If i and j are even or odd, then ij is even and i+j1 is odd; therefore, (38)sin112n+1ijπ=sinijπ2n+1,sin112n+1i+j1π=sini+j1π2n+1.

So we have (39)v~iTv~j=0.

If one of i or j is even and the other is odd, then ij and i+j1 are even; therefore, we have (40)sin112n+1ijπ=sinijπ2n+1and (41)sin112n+1i+j1π=sini+j1π2n+1.

Therefore, (42)v~iTv~j=0.

Similar for (24), we can conclude (43)Ami,j=42n+1acij/2k=1nλkmcos2k12i1π22n+1sin2k12j1π22n+1,for i,j=1,2,,n.

Lemma 6 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

Suppose α=β=ac0 and ρ=ac, then the eigenvalues λ1,λ2,,λn of An are given by (44)λk=b+2accos2k1π2n,k=1,2,,n.

The corresponding eigenvectors are given by (45)uk=u1k,u2k,,unkT,k=1,2,,n,ujk=ρj1sin2k12j1π4n,j=1,2,,n.

In the case α=β=ac0, the eigenvalues are given by (44) and the corresponding eigenvectors by (46)vk=v1k,v2k,,vnkT,k=1,2,,n,vjk=ρj1cos2k12j1π4n,j=1,2,,n.

In these cases which are similar to previous cases, we showed that integer powers of the matrix A for i,j=1,2,,n, respectively, (47)Ami,j=2nacij/2k=1nλkmsin2k12i1π4nsin2k12j1π4n,where λk=b+2accos2k1π/2n,k=1,2,,n, and (48)Ami,j=2nacij/2k=1nλkmcos2k12i1π4ncos2k12j1π4n,where λk=b+2accos2k1π/2n,k=1,2,,n.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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