On computing of arbitrary positive integer powers for one type of odd order symmetric circulant matrices—II

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Abstract

This article is an extension of the work [J. Rimas, On computing of arbitrary positive integer powers for one type of symmetric odd order circulant matrices—I, Appl. Math. Comput, in press], in which the general expression of the lth power (l  N) for one type of symmetric odd order circulant matrices is given. In this new paper we present the complete derivation of this general expression. Expressions of eigenvectors and Jordan’s form of the matrix and of the transforming matrix and its inverse are given, too.

Introduction

Solving some difference, differential equations and delay differential equations we meet the necessity to compute the arbitrary positive integer powers of square matrix [1], [2]. In the work [3] the general expression of the lth power (l  N) for one type of symmetric odd order tridiagonal matrices is presented. In this new paper we give the complete derivation of the general expression, presented in [3].

Section snippets

Formulation of the problem

Consider the nth order (n = 2p + 1, p  N) matrix B of the following type:B=011101101101110.

We will derive expression of the lth power (l  N) of the matrix (1) applying the expression Bl = TJlT−1 [5], 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 the matrix B are known. The eigenvalues of B are defined by the characteristic equationB-λE=0.In the paper (3) it is shown that the roots of the characteristic equation

Eigenvectors of matrix B and transforming matrix T

Consider the relation J = T−1BT (BT = TJ); here B is the nth order matrix (1) (n = 2p + 1, p  N), J is the Jordan’s form of B, T is the transforming matrix. We will find the transforming matrix T.

Denoting jth column of T by Tj(j=1,n¯), we have T = (T1T2Tn) and(BT1BT2BTn)=(T1λ1T2λ1T3λ3T4λ3Tn-2λn-2Tn-1λn-2Tnλn).The latter expression gives:BT1=T1λ1,BT2=T2λ1,BT3=T3λ3,BT4=T4λ3,BTn-2=Tn-2λn-2,BTn-1=Tn-1λn-2,BTn=Tnλn.DenotingTj=Tj1,j=2p-1,Tj-12,j=2p(pN,j=1,n-1¯)and solving the set of systems (5), we findTj1=

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