2000 Mathematics Subject Classification: 42C05.We give an easy procedure for solving of the direct and the inverse spectral problems
for the equation. Guseynov used a procedure of the Gelfand-Levitan type
for the case N = 1. We use another procedure and this procedure is more
easy and transparent

Zagorodnyuk, S.

Bulgarian Digital Mathematics Library at IMI-BAS

Serdica Math. J. 30 (2004), 471-482DIRECT AND INVERSE SPECTRAL PROBLEMS FOR(2N + 1)-DIAGONAL, COMPLEX, SYMMETRIC,NON-HERMITIAN MATRICESS. M. ZagorodnyukCommunicated by E. HorozovAbstract. We consider the following linear (2N)-th order difference equa-tion:N∑i=1(αk−i,iyk−i + αk,iyk+i) + αk,0yk = λNyk, k = 0, 1, 2, . . . , whereαi,j ∈ C : αi,N 6= 0, λ is a complex parameter, (y0, y1, . . . , yk, . . .) = ~yT is avector solution, N is a fixed integer. It can be written in the following matrixform: J~y = λN~y, where J is a (2N+1)-diagonal, symmetric matrix. We givean easy procedure for solving of the direct and the inverse spectral problemsfor the equation. Guseynov used a procedure of the Gelfand-Levitan typefor the case N = 1. We use another procedure and this procedure is moreeasy and transparent.1. Introduction. Let us consider the following linear difference equa-tion of the second order:b0y0 + a0y1 = λy0,2000 Mathematics Subject Classification: 42C05.Key words: inverse problems, difference equation.472 S. M. Zagorodnyuk(1) an−1yn−1 + bnyn + anyn+1 = λyn, n = 1, 2, . . . ,where an, bn ∈ C, an 6= 0, λ is a complex parameter, (y0, y1, . . .) is a vectorsolution.Such an equation was considered by Guseynov in [1]. He gave a procedurefor construction of a spectral function. He gave necessary and sufficient conditionsfor a functional σ to be the spectral function. He showed how to reconstruct thedifference equation using σ and a sequence of signs.Our aim here is twofold. First, we shall give a little another definitionof the spectral function and an easy procedure of reconstruction of differenceequation (1). Second, we shall extend our procedure for the following lineardifference equation of the (2N)-th order:(2)N∑i=1(αk−i,iyk−i + αk,iyk+i) + αk,0yk = λNyk, k = 0, 1, 2, . . . ,where αi,j ∈ C : αi,N 6= 0, λ is a complex parameter, (y0, y1, . . . , yk, . . .) = ~yT isa vector solution, N is a fixed integer, and all αi,j and pk with negative indexeshere are equal to zero.Equation (2) can be written in the following matrix form:(3) J~y(λ) = λN~y(λ),where J is a (2N + 1)-diagonal, symmetric, non-Hermitian matrix:(4) J =α0,0 α0,1 α0,2 . . . α0,N 0 0 . . .α0,1 α1,0 α1,1 . . . α1,N−1 α1,N 0 . . ........... . .......... . . .α0,N α1,N−1 α2,N−2 . . . . . . . . .0 α1,N α2,N−1 . . . . . . . . ....................... . .,Notice that the case of equation (3) with Hermitian (2N + 1)-diagonal matrixJ is closely related to the orthogonal polynomials on radial rays in the complexplane. The measure of orthogonality plays a role of a spectral function. For thehistory of this type equations and recent results see [2] and references therein (forthe case of five-diagonal matrices see [3, 4]). Symmetric, non-Hermitian matriceshave a much more complicated structure (see [5]) and we can hardly expect forintegral representations of their spectral functions.Direct and inverse spectral problems. . . 473In Section 3 we shall give necessary and sufficient conditions for a linearfunctional σ to be a spectral function of (2). If we know a sequence of signs, wecan reconstruct difference equation (2) from its spectral function. The procedureof reconstruction is easy and transparent.2. Spectral problems for three-diagonal matrices. Let{pn(λ)}∞n=0 be a solution of (1) with the initial condition p0(λ) = 1. It is nothard to see that pn(λ) is a polynomial of degree n. Let us define a linear withrespect to the both arguments functional σ(u, v), u, v ∈ P, where P is the spaceof all polynomials. We define it in the following way:(5) σ(pn(λ), pm(λ)) = δn,m, n,m = 0, 1, 2, . . . .Using linearity we extend the definition of σ for arbitrary u, v ∈ P. We noticethat from linearity it follows that(6) σ(u, v) = σ(v, u), u, v ∈ P.Using recurrence relation (1) we write:σ(λpn(λ), pm(λ)) = σ(an−1pn−1 + bnpn + anpn+1, pm) =(7) = an−1δn−1,m + bnδn,m + anδn+1,m;σ(pn(λ), λpm(λ)) = σ(pn, am−1pm−1 + bmpm + ampm+1) =(8) = am−1δn,m−1 + bmδn,m + amδn,m+1, n,m = 0, 1, 2, . . . .Comparing (7) with (8) for n = m − 1;m;m + 1, we see that the expressionscoincide. But for n < m − 1 or n > m + 1 we obtain zero in equations (7) and(8). So, we get(9) σ(λpn(λ), pm(λ)) = σ(pn(λ), λpm(λ)), n,m = 0, 1, 2, . . . .Using linearity, from (9) we derive(10) σ(λu(λ), v(λ)) = σ(u(λ), λv(λ)), u, v ∈ P.Conversely, let a sequence of polynomials {pn(λ)}∞n=0, deg pn = n, p0 = 1, begiven. Also let a linear functional σ(u, v) which satisfies (5), (10) be given. Thenwe can derive (1). Really, since deg pn = n, one can write:(11) λpn(λ) =n+1∑k=0an,kpk(λ), an,k ∈ C : an,n+1 6= 0, n = 0, 1, 2, . . . .474 S. M. ZagorodnyukUsing relations (5), (6), (10) we get:σ(λpn, pj) = an,j, j = 0, 1, 2, . . . , n+ 1;σ(λpn, pj) = σ(pn, λpj) = σ(λpj , pn) ={0, j + 1 < naj,n, j + 1 ≥ n,and(12) an,j ={0, j + 1 < naj,n, j + 1 ≥ n, j = 0, 1, 2, . . . , n+ 1.Relation (11) can be written in the following form:(13) λpn(λ) = an,n−1pn−1(λ) + an,npn(λ) + an,n+1pn+1(λ), n = 0, 1, 2, . . . .We put by definition: an = an,n+1, bn = an,n. Using relation (12), from relation(13) we get (1).Definition. A linear with respect to the both arguments functional σ(u, v),u, v ∈ P, which satisfies (5), we shall call the spectral function of differenceequation (1).The following theorem is a reformulation of [1, Theorem 2.1, p. 242], butthe method of proof is quite another:Theorem 1. A linear with respect to the both arguments functionalσ(u, v), u, v ∈ P, is the spectral function of a difference equation of type (1) iff:1) σ(λu(λ), v(λ)) = σ(u(λ), λv(λ)), u, v ∈ P;2) σ(1, 1) = 1;3) For arbitrary polynomial uk(λ) of degree k, there exists a polynomialuˆk(λ) of degree k such that:σ(uk(λ), uˆk(λ)) 6= 0.P r o o f. Necessity. Let σ(u, v) be a spectral function of a differenceequation (1). Relation 1) can be derived in the same manner as relation (10)was derived. Relation 2) is obvious from (5) since p0 = 1. For a polynomialuk(λ),deg uk = k, we can write: uk(λ) =k∑j=0ξjpj(λ), ξj ∈ C : ξk 6= 0. Putuˆk(λ) =k∑j=0ξjpj(λ). Using orthonormality relation (5) we getσ(uk(λ), uˆk(λ)) =k∑j=0|ξj |2 > 0,Direct and inverse spectral problems. . . 475and relation 3) is verified.Sufficiency. Let σ(u, v) be a linear functional satisfying relations 1),2),3).Let us construct a system of polynomials {pn(λ)}∞n=0. Put p0 = 1. Let R1(λ)be arbitrary polynomial of degree 1. Then we can write: R1(λ) = µ1λ +α1,0p0(λ), µ1, α1,0 ∈ C, µ1 6= 0. Thenσ(R1(λ), p0(λ)) = σ(µ1λ+ α1,0p0(λ), p0(λ)) = µ1σ(λ, p0) + α1,0.If we take α1,0 = −µ1σ(λ, p0) thenR1(λ) = µ1(λ− σ(λ, p0)p0(λ)),(14) σ(R1(λ), p0(λ)) = 0,From relation 3) it follows that there exists a polynomial Rˆ1(λ),deg Rˆ1 = 1, suchthatM1 := σ(R1(λ), Rˆ1(λ)) 6= 0.For the polynomial Rˆ1(λ) we can write: Rˆ1(λ) = ν1R1(λ) + β1,0p0(λ), ν1, β1,0 ∈C, ν1 6= 0. ThenM1 = σ(R1(λ), Rˆ1(λ)) = σ(R1(λ), ν1R1(λ) + β1,0p0(λ)) = ν1σ(R1(λ), R1(λ));ν1M1σ(R1(λ), R1(λ)) = 1.Put by definitionµ˜1,1 =√ν1M1,where we take arbitrary branch of the square root. Putp1(λ) = µ˜1,1R1(λ).Then deg p1 = 1 andσ(pi, pj) = δi,j, i, j = 0, 1.Suppose that we have constructed such polynomials p0(λ), p1(λ), . . . , pn−1(λ); deg pi =i, n ≥ 2, that(15) σ(pi, pj) = δi,j , i, j = 0, 1, . . . , n− 1.476 S. M. ZagorodnyukConsider arbitrary polynomial Rn(λ) of degree n. We can write: Rn(λ) = µnλn+n−1∑i=0αn,ipi(λ), µn, αn,i ∈ C, µn 6= 0. Thenσ(Rn(λ), pj(λ)) = σ(µnλn +n−1∑i=0αn,ipi(λ), pj(λ)) = µnσ(λn, pj(λ)) + αn,j,j = 0, 1, . . . , n− 1.Let us take αn,j = −µnσ(λn, pj(λ)), j = 0, 1, 2, . . . , n− 1. Then(16) Rn(λ) = µn(λn −n−1∑i=0σ(tn, pi(t))pi(λ)),σ(Rn(λ), pj(λ)) = 0, j = 0, 1, . . . , n − 1.As it follows from relation 3), there exists a polynomial Rˆn(λ), deg Rˆn = n, suchthatMn := σ(Rn(λ), Rˆn(λ)) 6= 0.For the polynomial Rˆn(λ) we can write the following representation: Rˆn(λ) =νnRn(λ) +n−1∑i=0βn,ipi(λ), νn, βn,i ∈ C, νn 6= 0. ThenMn = σ(Rn(λ), Rˆn(λ)) = σ(Rn(λ), νnRn(λ)+n−1∑i=0βn,ipi(λ)) = νnσ(Rn(λ), Rn(λ));νnMnσ(Rn(λ), Rn(λ)) = 1.Put by definition(17) µ˜n,n =√νnMn,where we take arbitrary branch of the square root. We put(18) pn(λ) = µ˜n,nRn(λ).Then deg pn = n andσ(pi, pj) = δi,j, i, j = 0, 1, . . . , n.Direct and inverse spectral problems. . . 477We continue this process and obtain a sequence of polynomials {pn(λ)}∞n=0,deg pn = n, such that: p0(λ) = 1 and σ(pi, pj) = δi,j, i, j = 0, 1, 2, . . . . As ithas been shown, if we have a linear functional σ(u, v), u, v ∈ P, and a sequenceof polynomials {pn(λ)}∞n=0, deg pn = n, p0 = 1, which satisfy (5), (10), we canderive (1). The functional σ will be a spectral function of this equation as itfollows from the definition of the spectral function.The proof is complete. Let an equation of type (1) be given and using (5) we have constructedits spectral function σ(u, v). How we can reconstruct our difference equation (1)?We must use the procedure of construction of {pn(λ)}∞n=0 in the proof of the lasttheorem. We put p0 = 1 and then begin from relation (15) for n = 1, 2, 3, . . . .During this procedure we must take µn = 1 for Rn(λ) in (16) and take µ˜n,n in (17)such that Arg µ˜n,n ∈ [0, π) if Arg µn,n = Arg1anan−1 . . . a0∈ [0, π), or Arg µ˜n,n ∈[π, 2π) otherwise, where µn,n is the leading coefficient of pn which is connectedwith (1). Really, let {Pn(λ)}∞n=0 be a system of polynomials which is constructedfollowing this procedure. Suppose that Pi(λ) = pi(λ), i = 0, 1, . . . , n − 1;n ≥ 1,i.e. the first n − 1 constructed polynomials coincide with polynomials definedfrom (1). Then for pn(λ) we can write: pn(λ) = µn,nλn +n−1∑i=0α˜n,iPi(λ). Then0 = σ(pn(λ), pi(λ)) = σ(pn(λ), Pi(λ)) = µn,nσ(λn, Pi(λ))+α˜n,i, i = 0, 1, . . . , n−1.We get α˜n,i = −µn,nσ(λn, Pi(λ)) andpn(λ) = µn,n(λn −n−1∑i=0σ(λn, Pi(λ))Pi(λ)) = µn,nRn(λ).Comparing this relation with (18) (keeping in mind that pn in (18) we now denoteby Pn and µn = 1) we getPn(λ) =µ˜n,nµn,npn(λ).Then1 = σ(Pn(λ), Pn(λ)) =(µ˜n,nµn,n)2σ(pn(λ), pn(λ)) =(µ˜n,nµn,n)2,An := µ˜2n,n = µ2n,n 6= 0.So, µ˜n,n and µn,n are square roots of An and according to our choice of branches ofthese roots, they coincide. Hence Pn = pn. Consequently, Pi = pi, i = 0, 1, 2, . . ..But the coefficients of equation (1) are uniquely defined by σ, {pn}∞0 .478 S. M. ZagorodnyukIf we define a sequence of signs {±,±, . . .}, where “+” in the n-th placemeans that Arg1anan−1 . . . a0∈ [0, π), and ”-” means that Arg1anan−1 . . . a0∈[π, 2π), then the spectral function σ and this sequence will uniquely define differ-ence equation (1).3. Spectral problems for difference equations of the (2N)-th order. Let us consider a linear difference equation of the (2N)-th order(2). Let {pn(λ)}∞n=0 be a solution of (2) with initial conditions pi(λ) = λi, i =0, 1, 2, . . . , N − 1. It is clear from recurrence relation (2) that deg pn(λ) = n.Define a linear with respect to the both arguments functional σ(u, v), u, v ∈ P,from relation (5). Using linearity we extend it for arbitrary u, v ∈ P. We noticethat from linearity it follows that relation (6) is fulfilled. From recurrence relation(2) it follows thatσ(λNpn(λ), pm(λ)) = σ(N∑i=1(αn−i,ipn−i + αn,ipn+i) + αn,0pn, pm) =(19) =N∑i=1(αn−i,iδn−i,m + αn,iδn+i,m) + αn,0δn,m;σ(pn(λ), λNpm(λ)) = σ(pn,N∑j=1(αm−j,jpm−j + αm,jpm+j) + αm,0pm) =(20) =N∑j=1(αm−j,jδn,m−j + αm,jδn,m+j) + αm,0δn,m, n,m = 0, 1, 2, . . . .Comparing the right-hand sides of (19) and (20) for n = m−j, j = 1, 2, . . . , N ;n =m;n = m + j, j = 1, 2, . . . , N, we see that they coincide. But if n < m − N orn > m+N the right-hand sides are equal to zero. So, we obtainσ(λNpn(λ), pm(λ)) = σ(pn(λ), λNpm(λ)), n,m = 0, 1, 2, . . . .Using linearity, from this relation we derive that(21) σ(λNu(λ), v(λ)) = σ(u(λ), λNv(λ)), u, v ∈ P.Direct and inverse spectral problems. . . 479Conversely, let {pn(λ)}∞n=0, deg pn = n, be a sequence of polynomials such thatpi(λ) = λi, i = 0, 1, . . . , N − 1, and σ(u, v) be a linear functional which satisfies(5), (21). Then we can write:(22) λNpk(λ) =N+k∑i=0ξk,ipi(λ), ξk,i ∈ C : ξk,N+k 6= 0; k = 0, 1, 2, . . . .Using (5),(21),(6) we get:σ(λNpk(λ), pj(λ)) = ξk,j, j = 0, 1, 2, . . . , k +N ;σ(λNpk(λ), pj(λ)) = σ(pk(λ), λNpj(λ)) = σ(λNpj(λ), pk(λ)) =={ξj,k, j +N ≥ k0, j +N < k.We getξk,j = ξj,k, j = k −N, k −N + 1, . . . , k +N ;ξk,j = 0, j = 0, 1, 2, . . . , k −N − 1; k = 0, 1, 2, . . . .Using these relations, from relation (22) we derive:λNpk(λ) =k+N∑i=k−Nξk,ipi(λ) =N∑j=1ξk,k+jpk+j(λ) +N∑j=1ξk,k−jpk−j(λ)++ξk,kpk(λ) =N∑j=1(ξk−j,kpk−j(λ) + ξk,k+jpk+j(λ)) + ξk,kpk(λ).If we put αk,j = ξk,k+j, j = 1, 2, . . . , N ; k = 0, 1, 2, . . . ; and αk,0 = ξk,k, k =0, 1, 2, . . . ; then we obtain (2).Definition. A linear with respect to the both arguments functional σ(u, v),u, v ∈ P, which satisfies (5), we shall call the spectral function of differenceequation (2).Theorem 2. A linear with respect to the both arguments functionalσ(u, v), u, v ∈ P, is the spectral function of a difference equation of type (2) iff:1) σ(λNu(λ), v(λ)) = σ(u(λ), λNv(λ)), u, v ∈ P;2) σ(λi, λj) = δi,j, i, j = 0, 1, . . . , N − 1;3) For arbitrary polynomial uk(λ) of degree k there exists a polynomialuˆk(λ) of degree k such that:σ(uk(λ), uˆk(λ)) 6= 0.480 S. M. ZagorodnyukP r o o f. Necessity. Let σ(u, v) be a spectral function of a differenceequation (2). Relation 1) can be proved in the same way as we have provedrelation (21). Relation 2) is obvious from (5) since pi(λ) = λi, i = 0, 1, 2, . . . , N−1. For a polynomial uk(λ),deg uk = k, we can write: uk(λ) =k∑j=0ξjpj(λ), ξj ∈C : ξk 6= 0. Put uˆk(λ) =k∑j=0ξjpj(λ). Then, using (5) we get:σ(uk(λ), uˆk(λ)) =k∑j=0|ξj |2 > 0,and relation 3) holds true.Sufficiency. Let σ(u, v) be a linear functional which satisfies conditions1),2),3). Let us construct a system of polynomials {pn(λ)}∞n=0. Put pi(λ) =λi, i = 0, 1, 2, . . . , N − 1. From relation 2) we get:σ(pi, pj) = δi,j , i, j = 0, 1, . . . , N − 1.Suppose that we have built such polynomials p0(λ), p1(λ), . . . , pn−1(λ); deg pi =i, n ≥ N , that(23) σ(pi, pj) = δi,j , i, j = 0, 1, . . . , n− 1.Consider arbitrary polynomial Rn(λ) of degree n. We can write: Rn(λ) = µnλn+n−1∑i=0αn,ipi(λ), µn, αn,i ∈ C, µn 6= 0. Thenσ(Rn(λ), pi(λ)) = σ(µnλn +n−1∑j=0αn,jpj(λ), pi(λ)) = µnσ(λn, pi(λ)) + αn,i,i = 0, 1, . . . , n− 1.If we take αn,i = −µnσ(λn, pj(λ)), j = 0, 1, 2, . . . , n− 1, then(24) Rn(λ) = µn(λn −n−1∑i=0σ(tn, pi(t))pi(λ)),andσ(Rn(λ), pi(λ)) = 0, i = 0, 1, . . . , n− 1.Direct and inverse spectral problems. . . 481From relation 3) it follows that there exists a polynomial Rˆn(λ),deg Rˆn = n, suchthatMn := σ(Rn(λ), Rˆn(λ)) 6= 0.For the polynomial Rˆn(λ) we can write: Rˆn(λ) = νnRn(λ)+n−1∑i=0βn,ipi(λ), νn, βn,i ∈C, νn 6= 0. ThenMn = σ(Rn(λ), Rˆn(λ)) = σ(Rn(λ), νnRn(λ)+n−1∑i=0βn,ipi(λ)) = νnσ(Rn(λ), Rn(λ));νnMnσ(Rn(λ), Rn(λ)) = 1.Put(25) µ˜n,n =√νnMn,where we take arbitrary branch of the square root. We put(26) pn(λ) = µ˜n,nRn(λ).Then deg pn = n andσ(pn, pn) = 1;σ(pi, pj) = δi,j, i, j = 0, 1, . . . , n.Applying this procedure for n = N,N + 1, N + 2, . . . , we build such a sequenceof polynomials {pn(λ)}∞n=0, deg pn = n, that pi(λ) = λi, i = 0, 1, . . . , N − 1, andσ(pi, pj) = δi,j , i, j = 0, 1, 2, . . . .As it has been already shown, if we have a linear functional σ(u, v), u, v ∈ P and asequence of polynomials {pn(λ)}∞n=0, deg pn = n, pi(λ) = λi, i = 0, 1, . . . , N − 1,which satisfy (5),(21), we can derive (2). From the definition of the spectralfunction it follows that σ will be a spectral function of this equation.The proof is complete. Let an equation of type (2) be given and using (5) we have built its spectralfunction σ(u, v). How we can reconstruct our difference equation (2)? We mustuse the procedure of construction of {pn(λ)}∞n=0 in the proof of the previous theo-rem. In this procedure we must take µn = 1 for Rn(λ) in (24) and take µ˜n,n in (25)such that Arg µ˜n,n ∈ [0, π) if Arg µn,n = Arg1αk,Nαk−N,Nαk−2N,N . . . αk−[ kN]N,N∈482 S. M. Zagorodnyuk[0, π), or Arg µ˜n,n ∈ [π, 2π) otherwise, where µn,n is the leading coefficient of pnwhich is defined from (2).We can define a sequence of signs {±,±, . . .} where ”+” appears ifArg1αk,Nαk−N,Nαk−2N,N . . . αk−[ kN]N,N∈ [0, π), and “−” appears ifArg1αk,Nαk−N,Nαk−2N,N . . . αk−[ kN]N,N∈ [π, 2π). This sequence and the spectralfunction allow us uniquely reconstruct difference equation (2).REFERENCES[1] G. S. Guseynov. Definition of an infinite nonself-adjoint Jacobi matrixby its generalized spectral function. Matem. zametki 23, 2 (1978), 237–246(in Russian).[2] S. M. Zagorodnyuk. On generalized Jacobi matrices and orthogonalpolynomials. New York J. Math. 9, (2003), 117–136.[3] S. M. Zagorodnyuk. On five-diagonal Jacobi matrices and orthogonalpolynomials on rays in the complex plane. Serdica Math. J. 24, 3–4 (1998),257–282.[4] S. M. Zagorodnyuk. On a characteristic property of polynomials satisfy-ing difference equation of the 4-th order. Visn. Khark. Univ., Ser. “Matem.,prykl. matem. i mekh.”, 582 (2003), 15–25 (in Russian).[5] F. R. Gantmaher. Theory of matrices. Moscow, Nauka, 1967 (inRussian).School of Mathematics and MechanicsKarazin Kharkiv National University61077 Kharkiv 77Ukrainee-mail: Sergey.M.Zagorodnyuk@univer.kharkov.uaReceived May 20, 2003Revised August 31, 2004

Direct and Inverse Spectral Problems for (2N + 1)-Diagonal, Complex, Symmetric, Non-Hermitian Matrices

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Direct and Inverse Spectral Problems for (2N + 1)-Diagonal, Complex, Symmetric, Non-Hermitian Matrices

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