# An Inverse Eigenvalue Problem for Jacobi Matrices

Authors: Zhengsheng Wang; Baojiang Zhong
Journal: Mathematical Problems in Engineering
 (2011)

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## Abstract

A kind of inverse eigenvalue problem is proposed which is the
reconstruction of a Jacobi matrix by given four or five eigenvalues and corresponding
eigenvectors. The solvability of the problem is discussed, and some sufficient conditions for existence of the solution of this problem are proposed. Furthermore, a numerical algorithm and two examples are presented.


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## Body

## 1. Introduction

Ann×n matrix J is called a Jacobi matrix if it is of the following form:(1.1)J=[a1b1b1a2b2b2a3b3⋱⋱⋱bn-2an-1bn-1bn-1an],bi>0.A Jacobi matrix inverse eigenvalue problem, roughly speaking, is how to determine the elements of Jacobi matrix from given eigen data. This kind of problem has great value for many applications, including vibration theory and structural design, for example, the vibrating rod model [1, 2]. In recent years, some new results have been obtained on the construction of a Jacobi matrix [3, 4]. However, the problem of constructing a Jacobi matrix from its four or five eigenpairs has not been considered yet. The problem is as follows.Problem 1.
Given four different real scalarsλ, μ, ξ, and η (supposed λ>μ>ξ>η) and four real orthogonal vectors of size nx=[x1,x2,…,xn]T,y=[y1,y2,…,yn]T,m=[m1,m2,…,mn]T,r=[r1,r2,…,rn]T, finding a Jacobi matrix J of size n such that (λ,x),(μ,y),(ξ,m),and(η,r) are its four eigenpairs.Problem 2.
Given five different real scalarsλ, μ, ν, ξ,and η(supposed λ>μ>ν>ξ>η) and five real orthogonal vectors of size nx=[x1,x2,…,xn]T,y=[y1,y2,…,yn]T,z=[z1,z2,…,zn]T,m=[m1,m2,…,mn]T,r=[r1,r2,…,rn]T, finding a Jacobi matrix J of size nsuch that (λ,x),(μ,y),(ν,z),(ξ,m),and(η,r) are its five eigenpairs.In Sections2 and 3, the sufficient conditions for the existence and uniqueness of the solution of Problems 1 and 2 are derived, respectively. Numerical algorithms and two numerical examples are given in Section 4. We give conclusion and remarks in Section 5.


## 2. The Solvability Conditions of Problem1

Lemma 2.1 (see [5, 6]).
Given two different real scalarsλ, μ(supposedλ>μ) and two real orthognal vectors of size n,x=[x1,x2,…,xn]T,y=[y1,y2,…,yn]T, there is a unique Jacobi matrix J  such that (λ,x),(μ,y) are its two eigenpairs if the following condition is satisfied:
(2.1)dkDk>0,(k=1,2,…,n-1),
where
(2.2)dk=∑i=1kxiyi,(k=1,2,…,n),Dk=|xkxk+1ykyk+1|≠0,(k=1,2,…,n-1).
And the elements of matrix J  are
(2.3)bk=(λ-μ)dkDk,(k=1,2,…,n-1),a1=λ-b1x2x1,an=λ-bn-1xn-1xn,ak={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,μ-(bk-1yk-1+bkyk+1)yk,xk=0,(k=2,3,…,n-1).From Lemma2.1, we can see that under some conditions two eigenpairs can determine a unique Jacobi matrix. Therefore, for Problem 1, we only prove that the Jacobi matrices determined by (λ,x),(μ,y)  and (ξ,m),(η,r) are the same.The following theorem gives a sufficient condition for the uniqueness of the solution of Problem1.Theorem 2.2.
Problem1 has a unique solution if the following conditions are satisfied:(i)
(λ-μ)dk(1)/Dk(1)=(λ-ξ)dk(2)/Dk(2)=(λ-η)dk(3)/Dk(3)>0;(ii)
ifxk=0, then (λ-μ)dj(1)/Dj(1)=(μ-ξ)dj(4)/Dj(4)=(μ-η)dj(5)/Dj(5),j=k,k-1, where(2.4)dk(1)=∑i=1kxiyi,dk(2)=∑i=1kximi,dk(3)=∑i=1kxiri,dk(4)=∑i=1kyimi,dk(5)=∑i=1kyiri,dk(6)=∑i=1kmiri,(k=1,2,…,n),(2.5)Dk(1)=|ykyk+1xkxk+1|,Dk(2)=|mkmk+1xkxk+1|,Dk(3)=|rkrk+1xkxk+1|,Dk(4)=|mkmk+1ykyk+1|,Dk(5)=|rkrk+1ykyk+1|,Dk(6)=|rkrk+1mkmk+1|,(k=1,2,…,n-1).Proof.
According to Lemma2.1, under certain condition, (λ,x) and (μ,y), (λ,x) and (ξ,m), (λ,x) and (η,r) can determine one unique Jacobi matrix, denoted J,J′,J′′, respectively. Their elements are as follows:
(2.6)bk=(λ-μ)dk(1)Dk(1),(k=1,2,…,n-1),a1=λ-b1x2x1,an=λ-bn-1xn-1xn,ak={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,μ-(bk-1yk-1+bkyk+1)yk,xk=0,(k=2,3,…,n-1),(2.7)bk′=(λ-ξ)dk(2)Dk(2),(k=1,2,…,n-1),a1′=λ-b1x2x1,an′=λ-bn-1xn-1xn,ak′={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,ξ-(bk-1mk-1+bkmk+1)mk,xk=0,(k=2,3,…,n-1),(2.8)bk′′=(λ-η)dk(3)Dk(3),(k=1,2,…,n-1),a1′′=λ-b1x2x1,an′′=λ-bn-1xn-1xn,ak′′={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,η-(bk-1rk-1+bkrk+1)rk,xk=0,(k=2,3,…,n-1).
From the conditions, we have
(2.9)bk=bk′=bk′′>0,k=1,2,…,n-1.
If xk≠0, we have ak=ak′=ak′′; if xk=0,
(2.10)(λ-μ)dk(1)Dk(1)=(μ-ξ)dk(4)Dk(4),(λ-μ)dk-1(1)Dk-1(1)=(μ-ξ)dk-1(4)Dk-1(4),(λ-μ)dk(1)Dk(1)=(μ-η)dk(4)Dk(4),(λ-μ)dk-1(1)Dk-1(1)=(μ-η)dk-1(4)Dk-1(4).
Since (2.6), we have
(2.11)bkDk(4)=(μ-ξ)dk(4),bk-1Dk-1(4)=(μ-ξ)dk-1(4).
That is,
(2.12)(μ-ξ)ykmk+bk-1Dk-1(4)-bkDk(4)=0.
Since Dk(i)≠0 and xk=0, we have yk≠0,mk≠0.      Dk-1(4)=mk-1yk-mkyk-1,Dk(4)=mkyk+1-mk+1yk replacing Dk-1(4),Dk(4) in (2.12), then we have
(2.13)μ-(bk-1yk-1+bkyk+1)yk=ξ-(bk-1mk-1+bkmk+1)mk.
Thus, if xk=0, we also have ak=ak′. In the same way, we have ak=ak′′. Then, ak=ak′=ak′′.    Therefore,
(2.14)J=J′=J′′
with four eigenpairs (λ,x),(μ,y),(ξ,m), and (η,r).


## 3. The Solvability Conditions of Problem2

Lemma 3.1 (see [7]).
Given three different real scalarsλ,μ,ν (supposed  λ>μ>ν) and three real orthogonal vectors of size  nx=[x1,x2,…,xn]T,y=[y1,y2,…,yn]T,z=[z1,z2,…,zn]T, there is a unique Jacobi matrix  J such that  (λ,x),(μ,y),(ν,z) are its three eigenpairs if the following conditions are satisfied: (i)
(λ-μ)dk(1)/Dk(1)=(λ-ν)dk(2)/Dk(2)>0;(ii)
ifxk=0, (λ-μ)dj(1)/Dj(1)=(μ-ν)dj(3)/Dj(3),j=k,k-1, where
(3.1)dk(1)=∑i=1kxiyi,dk(2)=∑i=1kxizi,dk(3)=∑i=1kyizi,Dk(1)=|ykyk+1xkxk+1|,Dk(2)=|zkzk+1xkxk+1|,Dk(3)=|zkzk+1ykyk+1|(k=1,2,…,n-1).
And the elements of matrix J are(3.2)bk=(λ-μ)dkDk,(k=1,2,…,n-1),a1=λ-b1x2x1,an=λ-bn-1xn-1xn,ak={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,μ-(bk-1yk-1+bkyk+1)yk,xk=0(k=2,3,…,n-1).From Lemma3.1, we can see that under some conditions three eigenpairs can determine a unique Jacobi matrix. Therefore, for Problem 2, we only prove that the Jacobi matrices determined by (λ,x),(μ,y),(ν,z);(λ,x),(μ,y),(ξ,m),(λ,x),(μ,y),(η,r) are the same.The following theorem gives a sufficient condition for the uniqueness of the solution of Problem2.Theorem 3.2.
Problem2 has a unique solution if the following conditions are satisfied:(i)(λ-μ)dk(1)/Dk(1)=(λ-ν)dk(2)/Dk(2)=(λ-ξ)dk(3)/Dk(3)=(λ-η)dk(4)/Dk(4)>0;(ii) ifxk=0, then (λ-μ)dj(1)/Dj(1)=(μ-ν)dj(5)/Dj(5)=(μ-ξ)dj(6)/Dj(6)=(μ-η)dj(7)/Dj(7),j=k,k-1, where(3.3)dk(1)=∑i=1kxiyi,dk(2)=∑i=1kxizi,dk(3)=∑i=1kximi,dk(4)=∑i=1kxini,dk(5)=∑i=1kyizi,dk(6)=∑i=1kyimi,dk(7)=∑i=1kyini,dk(8)=∑i=1kzimi,dk(9)=∑i=1kzini,dk(10)=∑i=1kmini,(k=1,2,…,n)Dk(1)=|ykyk+1xkxk+1|,Dk(2)=|zkzk+1xkxk+1|,Dk(3)=|mkmk+1xkxk+1|,Dk(4)=|nknk+1ykyk+1|,Dk(5)=|zkzk+1ykyk+1|,Dk(6)=|mkmk+1ykyk+1|,Dk(7)=|nknk+1ykyk+1|,Dk(8)=|mkmk+1zkzk+1|,Dk(9)=|nknk+1zkzk+1|Dk(10)=|nknk+1mkmk+1|,(k=1,2,…,n-1).Proof.
According to Lemma3.1, under certain condition, (λ,x),(μ,y),(ν,z);(λ,x),(μ,y),(ξ,m),(λ,x),(μ,y),(η,r) can determine one unique Jacobi matrix, denoted J,J′,J′′, respectively. Their elements are as follows:(3.4)bk=(λ-μ)dk(1)Dk(1)(k=1,2,…,n-1),a1=λ-b1x2x1,an=λ-bn-1xn-1xn,ak={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,μ-(bk-1yk-1+bkyk+1)yk,xk=0,(k=2,3,…,n-1),bk′=(λ-μ)dk(1)Dk(1),(k=1,2,…,n-1),a1′=λ-b1x2x1,an′=λ-bn-1xn-1xn,ak′={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,μ-(bk-1yk-1+bkyk+1)yk,xk=0,(k=2,3,…,n-1),bk′′=(λ-μ)dk(1)Dk(1),(k=1,2,…,n-1),a1′′=λ-b1x2x1,an′′=λ-bn-1xn-1xn,ak′′={λ-(bk-1xk-1+bkxk+1)xk,xk≠0,μ-(bk-1yk-1+bkyk+1)yk,xk=0,(k=2,3,…,n-1),
From conditions (i) and (ii) we have obviously (3.5)bk=bk′=bk′′>0,k=1,2,…,n-1,ak=ak′=ak′′.Therefore,(3.6)J=J′=J′′with five eigenpairs(λ,x),(μ,y),(ν,z),(ξ,m), and  (η,r).


## 4. Numerical Algorithms and Examples

The process of the proof of the theorem provides us with a recipe for finding the solution of Problem1 if it exists.From Theorem2.2, we propose a numerical algorithm for finding the unique solution of Problem 1 as follows.Algorithm 1.
Input.The real numbers λ>μ>ξ>η and mutually orthogonal vectors x,y,m,r.
Output. The symmetric Jacobi matrix having the eigenpairs (λ,x),(μ,y),(ξ,m),(η,r): (1)
computedk(1),dk(2),dk(3),dk(4),dk(5),dk(6) and Dk(1),Dk(2),Dk(3),Dk(4),Dk(5),Dk(6);(2)
if any one ofDk(1),Dk(2),Dk(3),Dk(4),Dk(5),Dk(6) is zero, the Problem 1 can not be solved by this method;(3)
fork=1,2,…,n-1.
(a)
whenxk=0, if(4.1)(λ-μ)dj(1)Dj(1)=(μ-ξ)dj(4)Dj(4)=(μ-η)dj(5)Dj(5),j=k,k-1,
then
(4.2)bk=(λ-μ)dk(1)Dk(1),ak=μ-(bk-1yk-1+bkyk+1)yk.
Otherwise, Problem 1 has no solution.(b)
Whenxk≠0, if(4.3)(λ-μ)dk(1)Dk(1)=(λ-ξ)dk(2)Dk(2)=(λ-η)dk(3)Dk(3)>0,
then
(4.4)bk=(λ-μ)dk(1)Dk(1),ak=λ-(bk-1xk-1+bkxk+1)xk.
Otherwise, Problem 1 has no solution; (4)
an=λ-bn-1xn-1/xn.Note that we can also propose a numerical algorithm from Theorem3.2. Because of the limitation of space, we don't describe it here in detail.Now we give two numerical examples here to illustrate that the results obtained in this paper are correct.Example 4.1.
Given four real numbersλ=3,μ=2,ξ=1,η=0.2679, and the four vectors x=[1,1,0,-1,-1]T,y=[1,0,-1,0,1]T,m=[1,-1,0,1,-1]T,r=[1,-3,2,-3,1]T, it is easy to verify that these given data satisfy the conditions of the Theorem 2.2. After calculating on the microcomputer through making program of Algorithm 1, we have a unique Jacobi matrix:
(4.5)J=[2112112112112].Example 4.2.
Given five real numbersλ=7.543,μ=-3.543,ν=2,ξ=4.296, and η=-0.296, and the five vectors: x=[0.1913,0.3536,0.4619,0.5000,0.4619,0.3536,0.1913]T, y=[0.1913,-0.3536,0.4619,-0.5000,0.4619,-0.3536,0.1913]T, z=[0.5000,0,-0.5000,0,0.5000,0,-0.5000]T, m=[0.4619,0.3536,-0.1913,0.5000,-0.1913,0.3536,0.4619]T,  and  r=[0.4619,-0.3536,-0.1913,0.5000,-0.1913,-0.3536,0.4619]T, it is easy to verify that these given numbers can not satisfy the conditions of the Theorem 2.2 but Theorem 3.2. After calculating on the microcomputer through making program of Theorem 3.2, we have a Jacobi matrix:
(4.6)J=[2332332332332332332].


## 5. Conclusion and Remarks

As a summary, we have presented some sufficient conditions, as well as simple methods to construct a Jacobi matrix from its four or five eigenpairs. Numerical examples have been given to illustrate the effectiveness of our results and the proposed method. Also, the idea in this paper may provide some insights for other banded matrix inverse eigenvalue problems.



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*Source: 571781-2011-05-30.xml*