# Positive Integer Powers of Certain Tridiagonal Matrices and Corresponding Anti-Tridiagonal Matrices

Authors: Mohammad Beiranvand; Mojtaba Ghasemi Kamalvand
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this paper, we firstly derive a general expression for the entries of themth (m∈ℕ) power for two certain types of tridiagonal matrices of arbitrary order. Secondly, we present a method for computing the positive integer powers of the anti-tridiagonal matrix corresponding to these matrices. Also, we give Maple 18 procedures in order to verify our calculations.


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## Body

## 1. Introduction

In recent years, computing the arbitrary positive integer powers of tridiagonal matrices has been a very popular problem for researchers. Tridiagonal matrices are used in different areas of science and engineering. Solution of difference systems [1], the numerical solution of PDE’s [2], telecommunication system analysis [3, 4], texture modeling [5], image processing, and coding [6] are examples for applications of these matrices. In these areas, the computation of the powers of these matrices is necessary. There have been several papers on computing the integer powers of various kinds on tridiagonal matrices [7–16].In this paper, we derive a general expression for the entries of themth (m∈ℕ) power for two certain types of tridiagonal matrices of arbitrary order as follows:
(1)A=a±bbbab0b⋱⋱⋱b0babba+b,where b≠0 and a and b are in the complex numbers. This model matrix has been known for a long time. For example, we have the influential works by D.E. Rutherford [3, 4]. The matrices that calculate their powers in [7, 8] are special cases of matrix (1), for (a=0, b=1), in cases that the 1,1th entry is (a+b) and (a−b), respectively.


## 2. Main Results

In this section, we firstly derive two formulas for calculating themth power for matrix (1) in two cases, where m∈ℕ and ℕ denotes the set of natural numbers; then, we present a method for computing the positive integer powers for one type of the anti-tridiagonal matrix corresponding to the matrix (1). According to the following lemmas, we can find the eigenvalues and eigenvectors of the matrix (1).Lemma 1 (see [17]).
The eigenvaluesλ1,λ2,⋯,λn of the matrix (1) are given as follows: when the 1,1th entry is a+b.
(2)λk=a+2bcosk−1πn,k=1,2,⋯,n,the corresponding eigenvectors uk=u1k,u2k,⋯,unkT,k=1,2,⋯,n, are given by
(3)ui1=1n,i=1,2,⋯,n,uik=2ncos2i−1k−1π2n,i=1,2,⋯,n;k=2,3,⋯,n.Lemma 2 (see [17]).
The eigenvaluesλ1,λ2,⋯,λn of the matrix (1) are given as follows: when the 1,1th entry is a−b.
(4)λk=a+2bcos2k−1π2n,k=1,2,⋯,n,the corresponding eigenvectors vk=v1k,v2k,⋯,vnkT,k=1,2,⋯,n, are given by
(5)vik=2nsin2i−12k−1π4n,i=1,2,⋯,n;k=1,2,⋯,n.Theorem 3 (see [5]).
IfA∈Mn has n distinct eigenvalues, then A is diagonalizable.Sinceθ°cosθ is strictly decreasing on [0,π], the eigenvalues of matrix (1), i.e., of A, are all distinct and the matrix is hence diagonalizable by [5], Theorem 1.3.9. The proof of [5], Theorem 1.3.7 shows that if we define the matrix U=u0,u1,⋯,un−1, then U−1AU=diagλ0,λ1,⋯,λn−1=J will be a diagonal. Conversely, we will have Am=Udiagλ0m,λ1m,⋯,λn−1mU−1, for any integer m. We will use this to give explicit formulas for Ami,j. From Lemma 1, we can write the columns of the matrix U as
(6)u1=1n11⋮1,uk=2ncosk−1π2ncos3k−1π2n⋮cos2n−1k−1π2n,k=2,3,⋯,n.Hence,(7)U=u1,u2,u3,⋯,un.Since matrix (1) is symmetric, the associated eigenvectors are orthogonal. Therefore, we can write
(8)U−1=UT=u1,u2,u3,⋯,unT.By using following corollaries, we can calculate themth power of the matrix (1).Corollary 4.
LetA be a tridiagonal matrix defined in (1) in case a+b and m is a positive integer; then,
(9)Ami,j=1nλ1m+2n∑k=2nλkmcos2i−1k−1π2ncos2j−1k−1π2n,for i,j=1,2,⋯,n, where λk=a+2bcosk−1π/n,k=1,2,⋯,n.Corollary 5.
LetA be a tridiagonal matrix defined in (1) in case a−b and m is a positive integer. Then,
(10)Ami,j=2n∑k=1nλkmsin2k−12i−1π4nsin2k−12j−1π4n,for i,j=1,2,⋯,n, where λk=a+2bcos2k−1π/2n,k=1,2,⋯,n.In the continuation, we present a method for computing the positive integer powers of the one type of the anti-tridiagonal matrices corresponding to the matrix (1), as follows:
(11)B=ba+b0babb⋰⋰bbab⋰0a+bb.Let(12)J=101⋰101.Lemma 6.
If matricesA, B, and J have the form (1), (11), and (12), respectively, then
(13)B=AJ.Proof.
According to the definition of the multiplication of matrices, we can conclude(14)AJi,j=∑k=1nAi,kJk,j=Ai,n+1−j=a+bifi=1,j=n,a+bifi=n,j=1,aifn+1−i+j=0,bifn+1−i+j=±1,0ifotherwise.=Bi,j;i,j=1,2,⋯,n,Lemma 7.
If the matricesA, B, and J have the form (1), (11), and (12), respectively, then the mth power m∈ℕ of matrix B is computed as follows:
Fork=1,2,3,⋯,
(15)Bm=Amforr=2k,AmJforr=2k−1.Proof.
We prove this lemma by induction onk. The base case of k=1 is true, because from Lemma 6and from that, AJ=JA and J=J−1 follow
(16)B=AJ,B2=AJ2=AJAJ=AJ2A=AIA=A2.From (16), we have B2=A2 so B2k=A2k and from B2=A2 and B=AJ, we can conclude B3=A3J. By repeating this method, we have B2k−1=A2k−1J.Suppose that the result is true fork>1 and consider case k+1.By the induction hypothesis, we have(17)B2k=A2k,B2k−1=A2k−1J.We show that casek+1 also is true. By using (17), we can write
(18)B2k+2=B2kB2=A2kA2=A2k+2,and also,
(19)B2k+1=B2kB=A2kAJ=A2k+1J.Thus, the formulas also hold fork+1 and the induction arguments are completed.We can compute thei,jth entry of the mth power for the matrix B in (11) by using formula (9) and Lemma 7.We leave the calculation of the positive powers of the following antisymmetric matrix to the reader.(20)C=ba−b0babb⋱⋱bbab⋱0a+bb.



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