An Inverse Problem for Two Spectra of Complex Finite Jacobi Matrices

Transcription

1 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 An Inverse Problem for Two Sectra of Comlex Finite Jacobi Matrices Gusein Sh. Guseinov Abstract: This aer deals with the inverse sectral roblem for two sectra of finite order comlex Jacobi matrices (tri-diagonal symmetric matrices with comlex entries). The roblem is to reconstruct the matrix using two sets of eigenvalues, one for the original Jacobi matrix and one for the matrix obtained by relacing the first diagonal element of the Jacobi matrix by some another number. The uniueness and existence results for solution of the inverse roblem are established and an exlicit algorithm of reconstruction of the matrix from the two sectra is given. Keywords: Jacobi matrix, difference euation, eigenvalue, normalizing numbers, inverse sectral roblem. Introduction Let J be an N N Jacobi matrix of the form b 0 a a 0 b a a b J = , () b N 3 a N a N 3 b N 2 a N a N 2 b N where for each n, a n and b n are arbitrary comlex numbers such that a n is different from zero: a n,b n C, a n 0. (2) Deartment of Mathematics, Atilim University, Incek, Ankara, Turkey. guseinov@atilim.edu.tr

2 302 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 A distinguishing feature of the Jacobi matrix () from other matrices is that the eigenvalue roblem Jy = λy for a column vector y = {y n } n=0 N is euivalent to the second order linear difference euation a n y n + b n y n + a n y n+ = λy n, (3) n {0,,...,N }, a = a N =, for {y n } N n=, with the boundary conditions y = y N = 0. (4) This allows, using techniues from the theory of three-term linear difference euations Atkinson (964), to develo a thorough analysis of the eigenvalue roblem Jy = λy. Define J to be the Jacobi matrix where all a n and b n are the same as J, excet b 0 is relaced by b 0 C, that is, J = b0 a a 0 b a a b (5) b N 3 a N a N 3 b N 2 a N a N 2 b N We shall assume that b0 b 0. (6) Denote by λ,...,λ all the distinct eigenvalues of the matrix J and by m,...,m their multilicities, resectively, as the roots of the characteristic olynomial det(j λi) so that N, m m = N, and det(λi J) = (λ λ ) m (λ λ ) m. (7) Further, denote by λ,..., λ all the distinct eigenvalues of the matrix J and by n,...,n their multilicities, resectively, as the roots of the characteristic olynomial det( J λi) so that N, n n = N, and det(λi J) = (λ λ ) n (λ λ ) n. (8)

3 An Inverse Problem for Two Sectra 303 The collections {λ k, m k (k =,..., )} and { λ i, n i (i =,...,)} (9) form the sectra (together with their multilicities) of the matrices J and J, resectively. We call these collections the two sectra of the matrix J. The inverse roblem about two sectra consists in reconstruction of the matrix J by its two sectra. This roblem consists of the following arts: (i) Is the matrix J determined uniuely by its two sectra? (ii) To indicate an algorithm for the construction of the matrix J from its two sectra. (iii) To find necessary and sufficient conditions for two collection of numbers in (9) to be the two sectra for some matrix of the form () with entries from class (2). For real finite Jacobi matrices this roblem is comletely solved in Guseinov (202). Note that in case of real entries the finite Jacobi matrix is selfadjoint and its eigenvalues are real and distinct. In the comlex case the Jacobi matrix is, in general, no longer selfadjoint and its eigenvalues may be comlex and multile. In the resent aer we show, by reducing the inverse roblem about two sectra to the inverse roblem about sectral data consisting of the eigenvalues and normalizing numbers of the matrix, that the comlex Jacobi matrix is determined from two sectra given in (9) uniuely u to signs of the off-diagonal elements of the matrix. We indicate also necessary and sufficient conditions for two collections of numbers of the form given in (9) to be two sectra of a Jacobi matrix J of the form () with entries belonging to the class (2). For given collections in (9), assuming that m k λ k i= we construct the numbers n i λi =: a 0, (0) β k j = a(m k j)! lim λ λ k d mk j dλ m k j ( j =,...,m k ; k =,..., ) (λ λ i ) n i i= (λ λ l ) m l l=,l k ()

4 304 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 and then set s l = m k j= ( l j ) l j+ β k j λk, l = 0,,2,..., (2) where ( ) ( l j is a binomial coefficient and we ut l j ) = 0 if j > l. Using these numbers we introduce the determinants s 0 s s n s s 2 s n+ D n =....., n = 0,,2,.... (3). s n s n+ s 2n The main result of this aer is the following theorem. Theorem. Let two collections of numbers in (9) be given, where λ,...,λ are distinct comlex numbers with {,...,N} and m,...,m are ositive integers such that m m = N; the λ,..., λ are distinct comlex numbers with {,...,N} and n,...,n are ositive integers such that n n = N. In order for these collections to be two sectra for a Jacobi matrix J of the form () with entries belonging to the class (2), it is necessary and sufficient that the following two conditions be satisfied: (i) λ k λ i for all k {,..., }, i {,...,}, and (0) holds; (ii) D n 0, for n {,2,...,N }, where D n is the determinant defined by (3), (2), (). Under the conditions (i) and (ii) the entries a n and b n of the matrix J for which the collections in (9) are two sectra, are recovered by the formulae a n = ± D n D n+ D n, n {0,,...,N 2}, D =, (4) b n = n D n n D n, n {0,,...,N }, = 0, 0 = s, (5) where D n is defined by (3), (2), (), and n is the determinant obtained from the determinant D n by relacing in D n the last column by the column with the comonents s n+,s n+2,...,s 2n+. Further, the element b 0 of the matrix J corresonding to the matrix J by (5) is determined by the formula b0 = b 0 + i= n i λi m k λ k.

5 An Inverse Problem for Two Sectra 305 It follows from the above solution of the inverse roblem about two sectra that the matrix () is not uniuely restored from the two sectra. This is linked with the fact that the a n are determined from (4) uniuely u to a sign. To ensure that the inverse roblem is uniuely solvable, we have to secify additionally a seuence of signs + and. Namely, let {σ 0,σ,...,σ N 2 } be a given finite seuence, where for each n {0,,...,N 2} the σ n is + or. We have 2 N such different seuences. Now to determine a n uniuely from (4) for n {0,,...,N 2} we can choose the sign σ n when extracting the suare root. In this way we get recisely 2 N distinct Jacobi matrices ossessing the same two sectra. The inverse roblem is solved uniuely from the data consisting of the two sectra and a seuence {σ 0,σ,...,σ N 2 } of signs + and. Thus, we can say that the inverse roblem with resect to the two sectra is solved uniuely u to signs of the off-diagonal elements of the recovered Jacobi matrix. The aer is organized as follows. Section 2 is auxiliary and resents solution of the inverse sectral roblem for comlex finite Jacobi matrices in terms of the eigenvalues and normalizing numbers. Last Section 3 resents the solution of the inverse roblem for comlex finite Jacobi matrices in terms of the two sectra. 2 Auxiliary facts In this section we follow the auhtor s aer Guseinov (2009). Given a Jacobi matrix J of the form () with the entries (2), consider the eigenvalue roblem Jy = λy for a column vector y = {y n } N n=0, that is euivalent to the roblem (3), (4). Denote by {P n (λ)} N n= and {Q n(λ)} N n= the solutions of E. (3) satisfying the initial conditions P (λ) = 0, P 0 (λ) = ; (6) Q (λ) =, Q 0 (λ) = 0. (7) For each n 0, P n (λ) is a olynomial of degree n and is called a olynomial of first kind and Q n (λ) is a olynomial of degree n and is known as a olynomial of second kind. These olynomials can be found recurrently from E. (3) using initial conditions (6) and (7). The leading terms of the olynomials P n (λ) and Q n (λ) have the forms P n (λ) = The euality λ n λ n +..., n 0; Q n (λ) = +..., n. (8) a 0 a a n a 0 a a n det(j λi) = ( ) N a 0 a a N 2 P N (λ) (9)

6 306 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 holds [see Guseinov (2009)] so that the eigenvalues of the matrix J coincide with the zeros of the olynomial P N (λ). The Wronskian of the solutions P n (λ) and Q n (λ), a n [P n (λ)q n+ (λ) P n+ (λ)q n (λ)], does not deend on n {,0,,...,N }. On the other hand, the value of this exression at n = is eual to by (6), (7), and a =. Therefore a n [P n (λ)q n+ (λ) P n+ (λ)q n (λ)] = for all n {,0,,...,N }. Putting, in articular, n = N, we arrive at P N (λ)q N (λ) P N (λ)q N (λ) =. (20) Let R(λ) = (J λi) be the resolvent of the matrix J (by I we denote the identity matrix of needed dimension) and e 0 be the N-dimensional column vector with the comonents, 0,..., 0. The rational function w(λ) = R(λ)e 0,e 0 = (λi J) e 0,e 0, (2) introduced earlier in Hochstadt (974), we call the resolvent function of the matrix J, where, denotes the standard inner roduct in C N. This function is known also as the Weyl-Titchmarsh function of J. The entries R nm (λ) of the matrix R(λ) = (J λi) (resolvent of J) are of the form R nm (λ) = { Pn (λ)[q m (λ) + M(λ)P m (λ)], 0 n m N, P m (λ)[q n (λ) + M(λ)P n (λ)], 0 m n N, (22) [see Guseinov (2009)] where M(λ) = Q N(λ) P N (λ). (23) According to (2), (22), (23) and using initial conditions (6), (7), we get w(λ) = R 00 (λ) = M(λ) = Q N(λ) P N (λ). (24) We will use the following well-known useful lemma. We bring it here for easy reference. Lemma. Let A(λ) and B(λ) be olynomials with comlex coefficients and dega < degb = N. Next, suose that B(λ) = b(λ z ) m (λ z ) m, where

7 An Inverse Problem for Two Sectra 307 z,...,z are distinct comlex numbers, b is a nonzero comlex number, and m,...,m are ositive integers such that m m = N. Then there exist uniuely determined comlex numbers a k j ( j =,...,m k ; k =,..., ) such that A(λ) B(λ) = m k j= a k j (λ z k ) j (25) for all values of λ different from z,...,z. The numbers a k j are given by the euation a k j = (m k j)! lim d mk [ j λ z k dλ m (λ z k j k ) m A(λ) ] k, (26) B(λ) j =,...,m k ; k =,...,. Proof. For each k {,..., } we have A(λ) B(λ) = C k(λ) (λ z k ) m, k (27) where the function C k (λ) = (λ z k ) m k A(λ) B(λ) A(λ) = b(λ z ) m (λ zk ) m k (λ zk+ ) m k+ (λ z ) m is regular (analytic) at z k. We can exand C k (λ) into a Taylor series about the oint z k, C k (λ) = where s=0 d ks (λ z k ) s, (28) d ks = C(s) k (z k), s = 0,,2,.... s! Substituting (28) in (27) we get that near z k, mk A(λ) B(λ) = d ks (λ z k ) m k s + (a Taylor series about z k). s=0

8 308 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 Consider the function Φ(λ) = A(λ) B(λ) m k s=0 d ks (λ z k ) m k s. This function is analytic everywhere, that is, Φ(λ) is an entire function. Next, since dega < degb, Φ(λ) 0 as λ. Thus the entire function Φ(λ) is bounded and tends to zero as λ. By the well-known Liouville theorem, we conclude that Φ(λ) 0. Thus we have A(λ) B(λ) = and m k s=0 d ks d k,mk j = C(m k j) k (z k ) = (m k j)! (λ z k ) m k s = m k j= (m k j)! lim λ z k d k,mk j (λ z k ) j d mk [ j dλ m (λ z k j k ) m A(λ) ] k. B(λ) These rove (25) and (26). Note that decomosition (25) is uniue as for the a k j in this decomosition E. (26) necessarily holds. By (9) and (7) we have P N (λ) = c(λ λ ) m (λ λ ) m, where c is a nonzero constant. Therefore we can decomose the rational function w(λ) exressed by (24) into artial fractions (Lemma ) to get w(λ) = m k j= β k j (λ λ k ) j, (29) where β k j = (m k j)! lim d mk [ j λ λ k dλ m (λ λ k j k ) m Q ] k N(λ) P N (λ) (30) are called the normalizing numbers of the matrix J. The collection of the eigenvalues and normalizing numbers {λ k, β k j ( j =,...,m k ; k =,..., )}, (3)

9 An Inverse Problem for Two Sectra 309 of the matrix J of the form (), (2) is called the sectral data of this matrix. Determination of the sectral data of a given Jacobi matrix is called the direct sectral roblem for this matrix. Thus, the sectral data consist of the eigenvalues and associated normalizing numbers derived by decomosing the resolvent function (Weyl-Titchmarsh function) w(λ) into artial fractions using the eigenvalues. It follows from (24) by (8) that λw(λ) tends to as λ. Therefore multilying (29) by λ and assing then to the limit as λ, we find that β k =. (32) The inverse sectral roblem consists in reconstruction of the matrix J by its sectral data. This roblem was solved by the author in Guseinov (2009) and we will resent here the final result. Let us set s l = m k j= ( ) l j l j+ β k j λk, l = 0,,2,..., (33) where ( ) ( l j is a binomial coefficient and we ut l j ) = 0 if j > l. Next, using these numbers s l we introduce the determinants s 0 s s n s s 2 s n+ D n =....., n = 0,,2,.... (34). s n s n+ s 2n Let us bring two imortant roerties of the determinants D n in the form of two lemmas. Lemma 2. Given any collection (3), for the determinants D n defined by (34), (33), we have D n = 0 for n N, where N = m m. Proof. Given a collection (3), define a linear functional Ω on the linear sace of all olynomials in λ with comlex coefficients as follows: if G(λ) is a olynomial then the value Ω,G(λ) of the functional Ω on the element (olynomial) G is Ω,G(λ) = m k j= G ( j ) (λ k ) β k j, (35) ( j )!

10 30 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 where G (n) (λ) denotes the n-th order derivative of G(λ) with resect to λ. Let m 0 be a fixed integer and set T (λ) = λ m (λ λ ) m (λ λ ) m = t m λ m +t m+ λ m t m+n λ m+n + λ m+n. (36) Then, according to (35), Ω,λ l T (λ) = 0, l = 0,,2,.... (37) Consider (37) for l = 0,,2,...,N + m, and substitute (36) in it for T (λ). Taking into account that Ω,λ l = s l, l = 0,,2,..., (38) where s l is defined by (33), we get t m s l+m +t m+ s l+m t m+n s l+m+n + s l+m+n = 0, l = 0,,2,...,N + m. Therefore (0,...,0,t m,t m+,...,t m+n,) is a nontrivial solution of the homogeneous system of linear algebraic euations x 0 s l + x s l x m s l+m + x m+ s l+m x m+n s l+m+n +x m+n s l+m+n = 0, l = 0,,2,...,N + m, with the unknowns x 0,x,...,x m,x m+,...,x m+n,x m+n. Therefore the determinant of this system, which coincides with D N+m, must be zero. Lemma 3. If collection (3) is the sectral data of the matrix J of the form () with entries belonging to the class (2), then for the determinants D n defined by (34), (33) we have D n 0 for n {0,,...,N }. Proof. We have D 0 = s 0 = β k = 0 by (32). Consider now D n for n {,...,N }. For any n {,...,N } let us consider the homogeneous system of linear algebraic euations n g k s k+m = 0, m = 0,,...,n, (39) k=0

11 An Inverse Problem for Two Sectra 3 with unknowns g 0,g,...,g n. The determinant of system (39) coincides with the D n. Therefore to rove D n 0, it is sufficient to show that system (39) has only a trivial solution. Assume the contrary: let (39) have a nontrivial solution {g 0,g,...,g n }. For each m {0,,...,n} take an arbitrary comlex number h m. Multily both sides of (39) by h m and sum the resulting euation over m {0,,...,n} to get n n m=0 k=0 h m g k s k+m = 0. Substituting exression (38) for s k+m in this euation and denoting G(λ) = we obtain n k=0 g k λ k, H(λ) = n m=0 h m λ m, Ω,G(λ)H(λ) = 0. (40) Since degg(λ) n, degh(λ) n and the lynomials P 0 (λ),p (λ),...,p n (λ) form a basis (their degrees are different) of the linear sace of olynomials of degree n, we have exansions G(λ) = n k=0 c k P k (λ), H(λ) = n k=0 d k P k (λ). Substituting these in (40) and using the orthogonality relations [see Guseinov (2009)] Ω,P m (λ)p n (λ) = δ mn, m,n {0,,...,N }, where δ mn is the Kronecker delta (at this lace we use the condition that collection (3) is the sectral data for a matrix J of the form (), (2)), we get n c k d k = 0. k=0 Since the olynomial H(λ) is arbitrary, we can take d k = c k in the last euality and get that c 0 = c =... = c n = 0, that is, G(λ) 0. But this is a contradiction and the roof is comlete. The solution of the above inverse roblem is given by the following theorem [see Guseinov (2009)]. Theorem 2. Let an arbitrary collection (3) of numbers be given, where N, m,...,m are ositive integers with m m = N, λ,...,λ are distinct

12 32 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 comlex numbers. In order for this collection to be the sectral data for a Jacobi matrix J of the form () with entries belonging to the class (2), it is necessary and sufficient that the following two conditions be satisfied: (i) β k = ; (ii) D n 0, for n {,2,...,N }, where D n is the determinant defined by (34), (33). Under the conditions (i) and (ii) the entries a n and b n of the matrix J for which the collection (3) is sectral data, are recovered by the formulae a n = ± D n D n+ D n, n {0,,...,N 2}, D =, (4) b n = n D n n D n, n {0,,...,N }, = 0, 0 = s, (42) where D n is defined by (34), (33), and n is the determinant obtained from the determinant D n by relacing in D n the last column by the column with the comonents s n+,s n+2,...,s 2n+. It follows from the above solution of the inverse roblem that the matrix () is not uniuely restored from the sectral data. This is linked with the fact that the a n are determined from (4) uniuely u to a sign. To ensure that the inverse roblem is uniuely solvable, we have to secify additionally a seuence of signs + and. Namely, let {σ 0,σ,...,σ N 2 } be a given finite seuence, where for each n {0,,...,N 2} the σ n is + or. We have 2 N such different seuences. Now to determine a n uniuely from (4) for n {0,,...,N 2} we can choose the sign σ n when extracting the suare root. In this way we get recisely 2 N distinct Jacobi matrices ossessing the same sectral data. The inverse roblem is solved uniuely from the data consisting of the sectral data and a seuence {σ 0,σ,...,σ N 2 } of signs + and. Thus, we can say that the inverse roblem with resect to the sectral data is solved uniuely u to signs of the off-diagonal elements of the recovered Jacobi matrix. Note that in the case of arbitrary real distinct numbers λ,...,λ N and ositive numbers β,...,β N the condition (ii) of Theorem 2 is satisfied automatically and in this case we have D n > 0, for n {,2,...,N }; see Guseinov (2009). However, in the comlex case the condition (ii) of Theorem 2 need not be satisfied automatically. Indeed, let N = 3 and as the collection (3) we take {λ, λ 2, λ 3, β, β 2, β 3 }, where λ, λ 2, λ 3, β, β 2, β 3 are arbitrary comlex numbers such that λ λ 2, λ λ 3, λ 2 λ 3,

13 An Inverse Problem for Two Sectra 33 β 0, β 2 0, β 3 0, β + β 2 + β 3 =. We have s l = β λ l + β 2 λ l 2 + β 3 λ l 3, l = 0,,2,..., and it is not difficult to show that D = s 0 s s s 2 = β β 2 (λ λ 2 ) 2 + β β 3 (λ λ 3 ) 2 + β 2 β 3 (λ 2 λ 3 ) 2, s 0 s s 2 D 2 = s s 2 s 3 s 2 s 3 s 4 = β β 2 β 3 (λ λ 2 ) 2 (λ λ 3 ) 2 (λ 2 λ 3 ) 2, 0 = s = β λ + β 2 λ 2 + β 3 λ 3, = s 0 s 2 s s 3 = β β 2 (λ + λ 2 )(λ λ 2 ) 2 +β β 3 (λ + λ 3 )(λ λ 3 ) 2 + β 2 β 3 (λ 2 + λ 3 )(λ 2 λ 3 ) 2, s 0 s s 3 2 = s s 2 s 4 s 2 s 3 s 5 = β β 2 β 3 λ λ 2 λ 3 λ λ 2 λ 3 λ 2 λ2 2 λ3 2 λ 3 λ2 3 λ3 3 We see that the condition D 0 is not satisfied automatically and therefore one must reuire D 0 as a condition. For examle, if take β = β 2 = β 3 = 3, λ = ± i 3, λ 2 =, λ 3 = 0, 2 then we get D = 0. 3 Construction of a comlex Jacobi matrix from two of its sectra Let J be an N N Jacobi matrix of the form () with entries satisfying (2). Define J to be the Jacobi matrix given by (5), where the number b 0 satisfies (6). We denote all the distinct eigenvalues of the matrices J and J by λ,...,λ and λ,..., λ, resectively. Let m k be the multilicity of λ k as the root of the characteristic olynomial det(λi J) and n i the multilicity of λ i as the root of the characteristic olynomial det(λi J). We call the collections {λ k, m k (k =,..., )} and { λ i, n i (i =,...,)} the two sectra of the matrix J. Note that m m = N and n n = N..

14 34 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 The inverse roblem for two sectra consists in the reconstruction of the matrix J by two of its sectra. We will reduce the inverse roblem for two sectra to the inverse roblem for eigenvalues and normalizing numbers solved above in Section 2. First let us study some necessary roerties of the two sectra of the Jacobi matrix J. Let P n (λ) and Q n (λ) be the olynomials of the first and second kind for the matrix J. The similar olynomials for the matrix J we denote by P n (λ) and Q n (λ). By (9) we have det(j λi) = ( ) N a 0 a a N 2 P N (λ), (43) ) det( J λi = ( ) N a 0 a a N 2 P N (λ), (44) so that the eigenvalues λ,...,λ and λ,..., λ of the matrices J and J and their multilicities coincide with the zeros and their multilicities of the olynomials P N (λ) and P N (λ), resectively. The P n (λ) and P n (λ) satisfy the same euation a n y n + b n y n + a n y n+ = λy n, n {,...,N }, a N =, (45) subject to the initial conditions P 0 (λ) =, P (λ) = λ b 0 a 0 ; (46) P 0 (λ) =, P (λ) = λ b 0 a 0. (47) The Q n (λ) also satisfies E. (45); besides Q 0 (λ) = 0, Q (λ) = a 0. (48) Since P n (λ) and P n (λ) form, for b 0 b 0, linearly indeendent solutions of E. (45), the solution Q n (λ) will be a linear combination of the solutions P n (λ) and P n (λ). Using initial conditions (46), (47), and (48) we find that Q n (λ) = [ ] P n (λ) P n (λ), n {0,,...N}. (49) b0 b 0 Relacing Q N (λ) and Q N (λ) in (20) by their exressions from (49), we get P N (λ) P N (λ) P N (λ) P N (λ) = b 0 b 0. (50)

15 An Inverse Problem for Two Sectra 35 Lemma 4. The matrices J and J have no common eigenvalues, that is, λ k λ i for all values of k and i. Proof. Suose that λ is an eigenvalue of the matrices J and J. Then by (43) and (44) we have P N (λ) = P N (λ) = 0. But this is imossible by (50) and the condition b0 b 0. The following lemma allows us to calculate the difference b 0 b 0 in terms of the two sectra. Lemma 5. The euality holds. m k λ k i= n i λi = b 0 b 0 (5) Proof. For any matrix A = [a jk ] N j, the sectral trace of A coincides with the matrix trace of A: if µ,..., µ are the distinct eigenvalues of A of multilicities m,...,m as the roots of the characteristic olynomial det(j λi), then m k µ k = N a kk. Indeed, this follows from det(λi A) = (λ µ ) m...(λ µ ) m by comarison of the coefficients of λ N on the two sides. Therefore we can write m k λ k = b 0 + b b N and i= n i λi = b 0 + b b N. Subtracting the last two eualities side by side we arrive at (5). Corollary. It follows from (5) that, under the conditon (6), m k λ k i= n i λi 0. The following lemma (together with Lemma 5) gives a formula for calculating the normalizing numbers β k j ( j =,...,m k ; k =,..., ) in terms of the two sectra. Lemma 6. For each k {,..., } and j {,...,m k } the formula β k j = (b 0 b 0 )(m k j)! lim λ λ k d mk j dλ m k j (λ λ i ) n i i= (λ λ l ) m l l=,l k (52)

16 36 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 holds. Proof. Relacing Q N (λ) in (24) by its exression from (49), we get [ w(λ) = P ] N (λ). (53) b0 b 0 P N (λ) Substituting (29) in the left-hand side we can write [ β k j (λ λ k ) j = P ] N (λ). b0 b 0 P N (λ) m k j= Hence, we get, taking into account (43) and (44), { [ β k j = (m k j)! lim d mk j (λ λ k ) m k λ λ k dλ m P ]} N (λ) k j b0 b 0 P N (λ) [ ] = (b 0 b 0 )(m k j)! lim d mk j P λ λ k dλ m (λ λ k j k ) m k N (λ) P N (λ) ( ) = (b 0 b 0 )(m k j)! lim d mk j det λi J) (λ λ λ k dλ m λ k j k ) m k. det(λi J) Substituting here (7) and (8) we arrive at (52). The lemma is roved. Theorem 3 (Uniueness result). The two sectra in (9) determine the Jacobi matrix J of the form () in the class (2) and the number b 0 C in the matrix J defined by (5) uniuely u to signs of the off-diagonal elements of J. Proof. Given the two sectra in (9) of the matrix J we determine uniuely the number (difference) b 0 b 0 by (5) and then the normalizing numbers β k j ( j =,...,m k ; k =,..., ) of the matrix J by (52). Since the collection of the eigenvalues and normalizing numbers of the matrix J determines J uniuely u to signs of the off-diagonal elements of J and the number b 0 is determined uniuely by the euation (Lemma 5) b0 = b 0 + i= n i λi the roof is comlete. m k λ k, Let us now rove Theorem stated above in the Introduction. The necessity of the conditions of Theorem has been roved above. To rove sufficiency suose that two collections of numbers in (9) are given which satisfy the

17 An Inverse Problem for Two Sectra 37 conditions of Theorem. We construct β k j ( j =,...,m k ; k =,..., ) according to Es. () and (0). Let us show that β k =. (54) Indeed, we have, for sufficiently large ositive number R such that λ,...,λ are inside the sircle {λ C : λ = R}, = a β k = Res λ=λk = 2πia = 2πia = 2πia = a ( λ =R λ =R λ =R a(m k )! lim λ λ k d mk dλ m k (λ λ i ) n i i= (λ λ ) n (λ λ ) n (λ λ ) m (λ λ ) m (λ λ ) n (λ λ ) n (λ λ ) m (λ λ ) m dλ (λ λ l ) m l l=,l k λ N (n λ n λ )λ N +... λ N (m λ m λ )λ N +... dλ [ + λ m k λ k i= ( n i λi ) m k λ k + 2πia i= ) ( )] n i λi + O λ 2 dλ λ =R ( ) O λ 2 dλ. Passing here to the limit as R and noting the definition (0) of a and that ( ) lim O R λ 2 dλ = 0, λ =R we arrive at (54). Thus, the collection {λ k, β k j ( j =,...,m k ; k =,..., )} satisfies all the conditions of Theorem 2 and hence there exist a Jacobi matrix J of the form () with entries from the class (2) such that λ k are the eigenvalues of the multilicity m k

18 38 Coyright 202 Tech Science Press CMES, vol.86, no.4,.30-39, 202 and β k j are the corresonding normalizing numbers for J. Having the matrix J, in articular, its entry b 0, we construct the number b 0 by b0 = b 0 + i= n i λi m k λ k (55) and then the matrix J by (5) according to the matrix J and (55). It remains to show that λ i are the eigenvalues of J of multilicity n i. To do this we denote the eigenvalues of J by µ,..., µ s and their multilicities by ñ,...,ñ s. We have to show that s =, µ i = λ i, ñ i = n i (i =,...,). Let us set f (λ) = Then l= g(λ) f (λ) = g (λ) f (λ), where (λ λ l ) m l, g(λ) = i= h(λ) f (λ) = h (λ) f (λ), (λ λ i ) n i, h(λ) = s i= (λ µ i )ñi. g (λ) = f (λ) g(λ), h (λ) = f (λ) h(λ) are olynomials and degg < deg f, degh < deg f. Next, by the direct roblem we have (Lemma 6) β k j = = = (b 0 b 0 )(m k j)! lim λ λ k (b 0 b 0 )(m k j)! lim λ λ k (b 0 b 0 )(m k j)! lim λ λ k d mk j dλ m k j d mk j dλ m k j d mk j dλ m k j s (λ µ i )ñi i= (λ λ l ) m l l=,l k { (λ λ k ) m k [ h(λ) f (λ) [ (λ λ k ) m k h (λ) f (λ) On the other hand, by our construction of β k j we have () which can be written in the form β k j = a(m k j)! lim d mk { [ j λ λ k dλ m (λ λ k j k ) m k g(λ) ]} f (λ) = a(m k j)! lim d mk [ j λ λ k dλ m (λ λ k j k ) m g ] k (λ). (57) f (λ) ] ]} (56)

19 An Inverse Problem for Two Sectra 39 Therefore, by Lemma it follows from (56) and (57), taking into account a = b 0 b 0, that h (λ) f (λ) = g (λ) f (λ). Hence h (λ) g (λ), that is, h(λ) g(λ) and conseuently s =, µ i = λ i, ñ i = n i (i =,...,). The roof is comlete. References Atkinson, F. V. (964): Discrete and Continuous Boundary Problems. Academic Press. Guseinov, G. Sh. (2009): Inverse sectral roblems for tridiagonal N by N comlex Hamiltonians. Symmetry, Integrability and Geometry: Methods and Alications (SIGMA), vol. 5, aer 08, 28 ages. Guseinov, G. Sh. (202): On determination of a finite Jacobi matrix from two sectra. CMES: Comuter Modeling in Engineering & Sciences, vol. 84, Hochstadt, H. (974): On construction of a Jacobi matrix from sectral data. Linear Algebra Al., vol. 8,

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