# On a Discrete Inverse Problem for Two Spectra

Authors: Gusein Sh. Guseinov
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

A version of the inverse spectral problem for two spectra of finite-order real Jacobi matrices (tridiagonal symmetric matrices) is investigated. The problem is to reconstruct the matrix using two sets of eigenvalues: one for the original Jacobi matrix and one for the matrix obtained by deleting the last row and last column of the Jacobi matrix.


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## Body

## 1. Introduction

The Jacobi matrices (tridiagonal symmetric matrices) appear in variety of applications. A distinguishing feature of the Jacobi matrices from others is that they are related to certain three-term recursion equations (second-order linear difference equations). Therefore, these matrices can be viewed as the discrete analogue of Sturm-Liouville operators, and their investigation have many similarities with Sturm-Liouville theory [1].AnN×N (real) Jacobi matrix J is a matrix of the form(1.1)J=[b0a00⋯000a0b1a1⋯0000a1b2⋯000⋮⋮⋮⋱⋮⋮⋮000…bN-3aN-30000⋯aN-3bN-2aN-2000⋯0aN-2bN-1],
where for each n,an and bn are arbitrary real numbers such that an is different from zero:(1.2)an,bn∈R,an≠0.
Let J1 be the truncated matrix obtained by deleting from J the last row and last column:(1.3)J1=[b0a00⋯00a0b1a1⋯000a1b2⋯00⋮⋮⋮⋱⋮⋮000…bN-3aN-3000⋯aN-3bN-2].
Denote the eigenvalues of the matrices J and J1 by λ1,…,λN and μ1,…,μN-1, respectively. The (finite) sequences {λk}k=1N and {μk}k=1N-1 are called the two spectra of the matrix J.The subject of the present paper is the solution of the inverse problem consisting of the following parts.(i)
Is the matrixJ determined uniquely by its two spectra?(ii)
To indicate an algorithm for the construction of the matrixJ from its two spectra;(iii)
To find necessary and sufficient conditions for two given sequences of real numbers{λk}k=1N and {μk}k=1N-1 to be the two spectra for some matrix of the form (1.1) with entries from class (1.2).This problem was solved earlier in [2, 3]. In the present paper we offer another and more effective, as it seems to us, method of solution for this problem.Other versions of the inverse problem for two spectra are investigated in [1, 4–9].The paper consists, besides this introductory section, of two sections. Section2 is auxiliary and presents briefly the solution of the inverse problem for finite Jacobi matrices in terms of the eigenvalues and normalizing numbers. A solution to this problem is presented in [1, Section  4.6] and [10]. In Section 3, we solve our main problem formulated above. At the basis of this solution is the formula(1.4)βk=a∏j=1,j≠kN(λj-λk)∏j=1N-1(μj-λk),
where(1.5)1a=∑m=1N1∏j=1,j≠mN(λj-λm)∏j=1N-1(μj-λm).
These formulae express the normalizing numbers βk of a finite Jacobi matrix in terms of two of its spectra. The formulae (1.4) and (1.5) give a conditional solution (i.e., assuming that there exists a matrix of the form (1.1) which has the sequences {λk}k=1N and {μk}k=1N-1 as two of its spectra) of the inverse problem in terms of two spectra because once we know the numbers {λk}k=1N and {βk}k=1N, we can form the matrix J by the prescription given in Section 2. Next, we give necessary and sufficient conditions for two sequences of real numbers {λk}k=1N and {μk}k=1N-1 to be two spectra of a Jacobi matrix of the form (1.1) with entries in the class (1.2), that is, we solve the main problem of this paper. The conditions consist of the following single and simple condition:(1.6)λ1<μ1<λ2<μ2<λ3⋯<λN-1<μN-1<λN,
that is, the numbers λk and μk interlace.


## 2. Preliminaries on the Inverse Spectral Problem

In this section, we follow the author’s paper [10]. Given a Jacobi matrix J of the form (1.1) with the entries (1.2), consider the eigenvalue problem Jy=λy for a column vector y={yn}n=0N-1, that is equivalent to the second-order linear difference equation(2.1)an-1yn-1+bnyn+anyn+1=λyn,n∈{0,1,…,N-1},a-1=aN-1=1,
for {yn}n=-1N, with the boundary conditions:(2.2)y-1=yN=0.
Denote by {Pn(λ)}n=-1N and {Qn(λ)}n=-1N the solutions of (2.1) satisfying the initial conditions:(2.3)P-1(λ)=0,P0(λ)=1,(2.4)Q-1(λ)=-1,Q0(λ)=0.
For each n≥0,Pn(λ) is a polynomial of degree n and is called a polynomial of first kind and Qn(λ) is a polynomial of degree n-1 and is known as a polynomial of second kind. The equality(2.5)det⁡(J-λI)=(-1)Na0a1⋯aN-2PN(λ)
holds so that the eigenvalues of the matrix J coincide with the zeros of the polynomial PN(λ). If PN(λ0)=0, then {Pn(λ0)}n=0N-1 is an eigenvector of J corresponding to the eigenvalue λ0. Any eigenvector of J corresponding to the eigenvalue λ0 is a constant multiple of {Pn(λ0)}n=0N-1.As shown in [10, Section  8], the equations(2.6)PN-1(λ)QN(λ)-PN(λ)QN-1(λ)=1,(2.7)PN-1(λ)PN′(λ)-PN(λ)PN-1′(λ)=∑n=0N-1Pn2(λ)
hold, where the prime denotes the derivative with respect to λ.Since the real Jacobi matrixJ of the form (1.1), (1.2) is self-adjoint, its eigenvalues are real. Let λ0 be a zero of the polynomial PN(λ). The zero λ0 is an eigenvalue of the matrix J by (2.5), and hence it is real. Putting λ=λ0 in (2.7) and using PN(λ0)=0, we get(2.8)PN-1(λ0)PN′(λ0)=∑n=0N-1Pn2(λ0).
The right-hand side of (2.8) is different from zero because the polynomials Pn(λ) have real coefficients and hence are real for real values of λ, and besides P0(λ)=1. Therefore, PN′(λ0)≠0, that is, the zero λ0 of the polynomial PN(λ) is simple. Hence the PN(λ), as a polynomial of degree N, has N distinct zeros. Thus, any real Jacobi matrix J of the form (1.1), (1.2) has precisely N real and distinct eigenvalues.LetR(λ)=(J-λI)-1 be the resolvent of the matrix J (by I we denote the identity matrix of needed dimension) and e0 the N-dimensional column vector with the components 1,0,…,0. The rational function(2.9)w(λ)=-〈R(λ)e0,e0〉=〈(λI-J)-1e0,e0〉,
we call the resolvent function of the matrix J, where 〈·,·〉 stands for the standard inner product in ℂN. This function is known also as the Weyl-Titchmarsh function of J.In [10, Section  5] it is shown that the entries Rnm(λ) of the matrix R(λ)=(J-λI)-1 (resolvent of J) are of the form(2.10)Rnm(λ)={Pn(λ)[Qm(λ)+M(λ)Pm(λ)],0≤n≤m≤N-1,Pm(λ)[Qn(λ)+M(λ)Pn(λ)],0≤m≤n≤N-1,
where(2.11)M(λ)=-QN(λ)PN(λ).
Therefore, according to (2.9) and using initial conditions (2.3) and (2.4), we get(2.12)w(λ)=-R00(λ)=-M(λ)=QN(λ)PN(λ).We often will use the following well-known simple useful lemma. We bring it here for easy reference.Lemma 2.1.
LetA(λ) and B(λ) be polynomials with complex coefficients and degA<degB. Next, suppose that B(λ)=b(λ-z1)⋯(λ-zN), where z1,…,zN are distinct complex numbers and b is a nonzero complex number. Then, there exist uniquely determined complex numbers a1,…,aN such that
(2.13)A(λ)B(λ)=∑k=1Nakλ-zk,
for all values of λ different from z1,…,zN. The numbers ak are given by the equation
(2.14)ak=lim⁡λ→zk(λ-zk)A(λ)B(λ)=A(zk)B′(zk),k∈{1,…,N}.Proof.
For eachk∈{1,…,N}, define the polynomial
(2.15)Lk(λ)=b∏j=1,j≠kN(λ-zj)=B(λ)λ-zk,
of degree N-1 and set
(2.16)F(λ)=A(λ)-∑k=1NakLk(λ),
where ak is defined by (2.14). Obviously F(λ) is a polynomial and degF≤N-1 (recall that degA<degB=N). Since
(2.17)Lk(zj)=0forj≠k,Lk(zk)=B′(zk)≠0,
we have
(2.18)F(zj)=A(zj)-∑k=1NakLk(zj)=A(zj)-ajLj(zj)=A(zj)-A(zj)B′(zj)B′(zj)=0,
for all j∈{1,…,N}. Thus, the polynomial F(λ) of degree ≤N-1 has N distinct zeros z1,…,zN. Then F(λ)≡0 and we get
(2.19)A(λ)=∑k=1NakLk(λ)=∑k=1NakB(λ)λ-zk=B(λ)∑k=1Nakλ-zk.
This proves (2.13). Note that the decomposition (2.13) is unique as for the ak in this decomposition (2.14) necessarily holds.Denote byλ1,…,λN all the zeros of the polynomial PN(λ) (which coincide by (2.5) with the eigenvalues of the matrix J and which are real and distinct):(2.20)PN(λ)=c(λ-λ1)⋯(λ-λN),
where c is a nonzero constant. Therefore applying Lemma 2.1 to (2.12), we can get for the resolvent function w(λ) the following decomposition:(2.21)w(λ)=∑k=1Nβkλ-λk,
where(2.22)βk=QN(λk)PN′(λk).
Further, putting λ=λk in (2.6) and (2.7) and taking into account that PN(λk)=0, we get(2.23)PN-1(λk)QN(λk)=1,(2.24)PN-1(λk)PN′(λk)=∑n=0N-1Pn2(λk),
respectively. It follows from (2.23) that QN(λk)≠0 and therefore βk≠0. Comparing (2.22), (2.23), and (2.24), we find that(2.25)βk={∑n=0N-1Pn2(λk)}-1,
whence we obtain, in particular, that βk>0.Since{Pn(λk)}n=0N-1 is an eigenvector of the matrix J corresponding to the eigenvalue λk, it is natural, according to the formula (2.25), to call βk the normalizing number of the matrix J corresponding to the eigenvalue λk.The collection of the eigenvalues and normalizing numbers:(2.26){λk,βk(k=1,…,N)},
of the matrix J of the form (1.1), (1.2) is called the spectral data of this matrix.Determination of the spectral data of a given Jacobi matrix is called thedirect spectral problem for this matrix.Thus, the spectral data consist of the eigenvalues and associated normalizing numbers derived by decomposing the resolvent function (Weyl-Titchmarsh function) into partial fractions using the eigenvalues. The resolvent functionw(λ) of the matrix J can be constructed by using (2.12). Another convenient formula for computing the resolvent function is (see [10, Section  5])(2.27)w(λ)=-det⁡(J(1)-λI)det⁡(J-λI),
where J(1) is the matrix obtained from J by deleting the first row and first column of J.It follows from (2.27) that λw(λ) tends to 1 as λ→∞. Therefore multiplying (2.21) by λ and passing then to the limit as λ→∞, we find(2.28)∑k=1Nβk=1.Theinverse spectral problem is stated as follows.(i)
To see if it is possible to reconstruct the matrixJ, given its spectral data (2.26). If it is possible, to describe the reconstruction procedure;(ii)
To find the necessary and sufficient conditions for a given collection (2.26) to be spectral data for some matrix J of the form (1.1) with entries belonging to the class (1.2).The solution of this problem is well known (see [1, Section  4.6] and [10]) and let us bring here the final result.Given a collection (2.26), where λ1,…,λN are real and distinct and β1,…,βN are positive, define the numbers:(2.29)sl=∑k=1Nβkλkl,l=0,1,2,…,
and using these numbers introduce the determinants:(2.30)Dn=|s0s1⋯sns1s2⋯sn+1⋮⋮⋱⋮snsn+1⋯s2n|,n=0,1,2,….Lemma 2.2.
For the determinantsDn defined by (2.30) and (2.29), we have Dn>0 for n∈{0,1,…,N-1} and Dn=0 for n≥N.Proof.
Denote byA the (n+1)×(n+1) matrix corresponding to the determinant Dn given by (2.30). Then for arbitrary real column vector x=(x0,x1,…,xn)T, we have
(2.31)〈Ax,x〉=∑j,m=0nsj+mxmxj=∑j,m=0n(∑k=1Nβkλkj+m)xmxj=∑k=1Nβk(∑j=0nxjλkj)2=∑k=1Nβk[G(λk)]2≥0,
where
(2.32)G(λ)=∑j=0nxjλj.
Further, it follows that if 〈Ax,x〉=0, then
(2.33)G(λk)=0,k=1,…,N.
If n≤N-1, then degG≤N-1 and (2.33) is possible only if G(λ)≡0 (recall that λ1,…,λN are distinct). But then x0=x1=⋯=xn=0. Therefore,
(2.34)〈Ax,x〉>0,
for all nonzero real vectors x=(x0,x1,…,xn)T if n≤N-1. Then as is well known from Linear Algebra, we have detA>0. Thus we have proved that Dn>0 for n≤N-1.
To prove thatDn=0 for n≥N, let us define the linear functional Ω on the linear space of all polynomials in λ with complex coefficients as follows: if G(λ) is a polynomial, then the value 〈Ω,G(λ)〉 of the functional Ω on the element (polynomial) G is
(2.35)〈Ω,G(λ)〉=∑k=1NβkG(λk).
Let m≥0 be a fixed integer and set
(2.36)T(λ)=λm(λ-λ1)⋯(λ-λN)=tmλm+tm+1λm+1+⋯+tm+N-1λm+N-1+λm+N.
Then, according to (2.35),
(2.37)〈Ω,λlT(λ)〉=0,l=0,1,2,….
Consider (2.37) for l=0,1,2,…,N+m, and substitute (2.36) in it for T(λ). Taking into account that
(2.38)〈Ω,λl〉=∑k=1Nβkλkl=sl,l=0,1,2,…,
we get
(2.39)tmsl+m+tm+1sl+m+1+⋯+tm+N-1sl+m+N-1+sl+m+N=0,l=0,1,2,…,N+m.
Therefore, (0,…,0,tm,tm+1,…,tm+N-1,1) is a nontrivial solution of the homogeneous system of linear algebraic equations:
(2.40)x0sl+x1sl+1+⋯+xmsl+m+xm+1sl+m+1+⋯+xm+N-1sl+m+N-1+xm+Nsl+m+N=0,l=0,1,2,…,N+m,
with the unknowns x0,x1,…,xm,xm+1,…,xm+N-1,xm+N. Therefore, the determinant of this system, which coincides with DN+m, must be equal to zero.Theorem 2.3.
Let an arbitrary collection (2.26) of numbers be given. In order for this collection to be the spectral data for a Jacobi matrix J of the form (1.1) with entries belonging to the class (1.2), it is necessary and sufficient that the following two conditions be satisfied:(i)
The numbersλ1,…,λN are real and distinct.(ii)
The numbersβ1,…,βN are positive and such that β1+⋯+βN=1.
Under the conditions (i) and (ii) we haveDn>0 for  n∈{0,1,…,N-1} and the entries  an and  bn of the matrix  J for which the collection (2.26) is spectral data, are recovered by the formulae
(2.41)an=±Dn-1Dn+1Dn,n∈{0,1,…,N-2},D-1=1,(2.42)bn=ΔnDn-Δn-1Dn-1,n∈{0,1,…,N-1},Δ-1=0,Δ0=s1,
where Dn is defined by (2.30) and (2.29), and  Δn is the determinant obtained from the determinant  Dn by replacing in  Dn the last column by the column with the components  sn+1,sn+2,…,s2n+1.It follows from the above solution of the inverse problem that the matrix (1.1) is not uniquely restored from the spectral data. This is linked with the fact that the an are determined from (2.41) uniquely up to a sign. To ensure that the inverse problem is uniquely solvable, we have to specify additionally a sequence of signs + and −. Namely, let {σ0,σ1,…,σN-2} be a given finite sequence, where for each n∈{0,1,…,N-2} the σn is + or −. We have 2N-1 such different sequences. Now to determine an uniquely from (2.41) for n∈{0,1,…,N-2}, we can choose the sign σn when extracting the square root. In this way, we get precisely 2N-1 distinct Jacobi matrices possessing the same spectral data. The inverse problem is solved uniquely from the data consisting of the spectral data and a sequence {σ0,σ1,…,σN-2} of signs + and −. Thus, we can say that the inverse problem with respect to the spectral data is solved uniquely up to signs of the off-diagonal elements of the recovered Jacobi matrix. In particular, the inverse problem is solvable uniquely in the class of entries an>0,  bn∈ℝ.


## 3. Inverse Problem for Two Spectra

LetJ be an N×N Jacobi matrix of the form (1.1) with entries satisfying (1.2). Define J1 to be the truncated Jacobi matrix given by (1.3). We denote the eigenvalues of the matrices J and J1 by λ1<⋯<λN and μ1<⋯<μN-1, respectively. We call the collections {λk(k=1,…,N)} and {μk(k=1,…,N-1)} the two spectra of the matrix J.The inverse problem for two spectra consists in the reconstruction of the matrixJ by two of its spectra.We will reduce the inverse problem for two spectra to the inverse problem for eigenvalues and normalizing numbers solved above in Section2.First, let us study some necessary properties of the two spectra of the Jacobi matrixJ.LetPn(λ) and Qn(λ) be the polynomials of the first and second kind for the matrix J. By (2.5) we have(3.1)det⁡(J-λI)=(-1)Na0a1⋯aN-2PN(λ),(3.2)det⁡(J1-λI)=(-1)N-1a0a1⋯aN-2PN-1(λ).
Note that we have used the fact that aN-1=1. Therefore, the eigenvalues λ1,…,λN and μ1,…,μN-1 of the matrices J and J1 coincide with the zeros of the polynymials PN(λ) and PN-1(λ), respectively.Dividing both sides of (2.6) by PN-1(λ)PN(λ) gives(3.3)QN(λ)PN(λ)-QN-1(λ)PN-1(λ)=1PN-1(λ)PN(λ).
Therefore, by formula (2.12) for the resolvent function w(λ), we obtain(3.4)w(λ)=QN-1(λ)PN-1(λ)+1PN-1(λ)PN(λ).Lemma 3.1.
The matricesJ and J1 have no common eigenvalues, that is, λk≠μj for all values of k and j.Proof.
Suppose thatλ is an eigenvalue of the matrices J and J1. Then by (3.1) and (3.2) we have PN(λ)=PN-1(λ)=0. But this is impossible by (2.6).Lemma 3.2.
The equality (trace formula)(3.5)λN+∑k=1N-1(λk-μk)=bN-1
holds.Proof.
For any matrixA=[ajk]j,k=1N the spectral trace of A coincides with the matrix trace of A: If ν1,…,νN are the eigenvalues of A, then
(3.6)∑k=1Nνk=∑k=1Nakk.
Therefore, we can write
(3.7)∑k=1Nλk=b0+b1+⋯+bN-2+bN-1,∑k=1N-1μk=b0+b1+⋯+bN-2.
Subtracting the last two equalities side by side, we arrive at (3.5).Lemma 3.3.
The eigenvalues ofJ and J1 interlace:
(3.8)λ1<μ1<λ2<μ2<λ3⋯<λN-1<μN-1<λN.Proof.
Let us set(3.9)ψ(λ)=PN-1(λ)PN(λ),
so that ψ(λ) is a rational function whose poles coincide with the eigenvalues of J and whose zeros coincide with the eigenvalues of J1. Applying Lemma 2.1 to the rational function ψ(λ) we can write
(3.10)ψ(λ)=∑k=1Nγkλ-λk,
where
(3.11)γk=PN-1(λk)PN′(λk).
Next, (2.24) shows that PN-1(λk)PN′(λk)>0, that is, PN-1(λk) and PN′(λk) have the same sign. Then (3.11) implies that γk>0(k=1,…,N).
Differentiating (3.10) we get
(3.12)ψ′(λ)=-∑k=1Nγk(λ-λk)2.
It follows from (3.12) that ψ′(λ)<0 for real values of λ, different from λ1,…,λN. Therefore, ψ(λ) is strictly decreasing continuous function on the intervals (-∞,λ1),  (λ1,λ2),…,(λN-1,λN),  (λN,∞). Besides, it follows from (3.10) that
(3.13)lim⁡|λ|→∞ψ(λ)=0,lim⁡λ→λk-ψ(λ)=-∞,lim⁡λ→λk+ψ(λ)=∞.
Consequently, the function ψ(λ) has no zero in the intervals (-∞,λ1) and (λN,∞), and exactly one zero in each of the intervals (λ1,λ2),…,(λN-1,λN). Since the zeros of the function ψ(λ) coincide with the eigenvalues of J1 by (3.9), the proof is complete.The following lemma gives a formula for calculating the normalizing numbersβ1,…,βN in terms of the two spectra.Lemma 3.4.
For eachk∈{1,…,N} the formula
(3.14)βk=a∏j=1,j≠kN(λj-λk)∏j=1N-1(μj-λk)
holds, where
(3.15)1a=∑m=1N1∏j=1,j≠mN(λj-λm)∏j=1N-1(μj-λm).Proof.
Substituting (2.21) in the left-hand side of (3.4), we can write
(3.16)∑m=1Nβmλ-λm=QN-1(λ)PN-1(λ)+1PN-1(λ)PN(λ).
Multiply both sides of the last equality by λ-λk and pass then to the limit as λ→λk. Taking into account that PN(λk)=0,  PN′(λk)≠0,  PN-1(λk)≠0 (see (2.23) and (2.24)), we get
(3.17)βk=1PN′(λk)PN-1(λk).
Next, by (3.1) and (3.2) we have
(3.18)(-1)Na0a1⋯aN-2PN(λ)=∏j=1N(λj-λ),(-1)N-1a0a1⋯aN-2PN-1(λ)=∏j=1N-1(μj-λ).
Substituting these in the right-hand side of (3.17), we obtain
(3.19)βk=a∏j=1,j≠kN(λj-λk)∏j=1N-1(μj-λk),
where
(3.20)a=(a0a1⋯aN-2)2.
Replacing k by m in (3.19) and then summing this equation over m=1,…,N and taking into account (2.28), we get (3.15). The lemma is proved.Theorem 3.5.
(uniqueness result). The two spectra{λk}k=1N and {μk}k=1N-1 of the Jacobi matrix J of the form (1.1) in the class
(3.21)an>0,bn∈R,
uniquely determine the matrix J.Proof.
Given the two spectra{λk}k=1N and {μk}k=1N-1 of the matrix J, we determine uniquely the normalizing numbers βk(k=1,…,N) of the matrix J by (3.14) and (3.15). Since the collection of the eigenvalues and normalizing numbers {λk,βk(k=1,…,N)} of the matrix J determines J uniquely in the class (3.21), the proof is complete.The following theorem solves the inverse problem in terms of the two spectra. Its proof given below contains an effective procedure for the construction of the Jacobi matrix from its two spectra.Theorem 3.6.
In order for giving two collections of real numbers{λk}k=1N and {μk}k=1N-1 to be the spectra of two matrices J and J1, respectively, of the forms (1.1) and (1.3) with the entries in the class (1.2), it is necessary and sufficient that the following inequalities be satisfied:
(3.22)λ1<μ1<λ2<μ2<λ3⋯<λN-1<μN-1<λN.Proof.
The necessity of the condition (3.22) has been proved above in Lemma 3.3. To prove the sufficiency, suppose that two collections of real numbers {λk}k=1N and {μk}k=1N-1 are given which satisfy the inequalities in (3.22). We construct βk(k=1,…,N) according to these data by (3.14) and (3.15). It follows from (3.22) that
(3.23)∏j=1,sj≠kN(λj-λk)∏j=1N-1(μj-λk)>0,k=1,…,N.
Therefore, the expression on the right-hand side of (3.14) is positive and hence βk>0(k=1,…,N). Next, it follows directly from (3.14) and (3.15) that β1+⋯+βN=1.
Consequently, the collection{λk,βk(k=1,…,N)} satisfies the conditions of Theorem 2.3, and hence there exists a Jacobi matrix J of the form (1.1) with entries from the class (1.2) such that the λk(k=1,…,N) are the eigenvalues and the βk(k=1,…,N) are the corresponding normalizing numbers for J. Having the matrix J, we construct the matrix J1 by (1.3). It remains to show that {μk}k=1N-1 is the spectrum of the constructed matrix J1. Denote the eigenvalues of J1 by μ̃1<⋯<μ̃N-1. By Lemma 3.3,
(3.24)λ1<μ̃1<λ2<μ̃2<⋯<λN-1<μ̃N-1<λN.
We have to show that μ̃k=μk(k=1,…,N-1).
By the direct spectral problem, we have (Lemma3.4):
(3.25)βk=ã∏j=1,j≠kN(λj-λk)∏j=1N-1(μ̃j-λk),
where
(3.26)1ã=∑m=1N1∏j=1,j≠mN(λj-λm)∏j=1N-1(μ̃j-λm).
On the other hand, by our construction of βk, we have (3.14) and (3.15). Equating the right-hand sides of (3.25) and (3.14), we obtain
(3.27)ã∏j=1N-1(μj-λk)=a∏j=1N-1(μ̃j-λk),k=1,…,N.
This means that the polynomial
(3.28)ã∏j=1N-1(λ-μj)-a∏j=1N-1(λ-μ̃j),
of degree ≤N-1 has N distinct zeros λ1,…,λN. Then, this polynomial identically equals zero. Hence, ã=a and μj=μ̃j(j=1,…,N). The proof is complete.



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*Source: 956407-2012-03-18.xml*