A generalized Hermite–Biehler theorem and non-Hermitian perturbations of Jacobi matrices

The classical Hermite–Biehler theorem describes the zero conﬁguration of a complex linear combination of two real polynomials whose zeros are real, simple, and strictly interlace. We provide the full characterization of the zero conﬁguration for the case when this interlacing is broken at exactly one location. We apply this result to solve the direct and inverse spectral problem for non-Hermitian rank-one multiplicative perturbations and rank-two additive perturbations of ﬁnite Hermitian and Jacobi


Introduction
The classical Hermite-Biehler [5,14] theorem states that all the zeros of the complex polynomial of the form p(z) − ilq(z) lie in the upper (lower) half-plane of the complex plane if and only if l > 0 (l < 0) and the zeros of the real polynomials p and q strictly interlace, that is, all the zeros are real, simple, and between two consecutive zeros of one polynomial there lies a unique zero of the other polynomial.
This theorem is of fundamental importance in many areas such as stability theory, dynamical systems, and control theory.Its rich connections with other areas of analysis and algebra have been exposed in many subsequent works.Monographs by Chebotarev and Meiman [9] and by Postnikov [27] give a detailed account of such related topics, including the amplitude-phase interpretation of stability, Sturm chains, Cauchy indices, continued fractions, Routh-Hurwitz criterion, and rational lossless functions.Krein and Naimark's survey [19] explore connections of the Hermite-Biehler scheme with Bezoutians.The analogue of the Hermite-Biehler theorem for entire functions is presented in [9] and [23].
There are numerous further generalizations and applications of this theorem in the literature, see [6,7,10,15,16,25,26,28,29] and references therein.This note is not a survey of the field.Here we generalize the Hermite-Biehler theorem to the situation when strict interlacing of the zeros of the real and imaginary parts of a given complex polynomial is broken at exactly one point, see Theorems 3.1 and 4.1 and additional Remarks 3.3 and 4.2.
The motivation for our generalization is twofold.Firstly, one often wants to understand the locations of roots of f + ig in terms of the arrangement of the roots of two real polynomials f and g.This is natural, e.g., in the study of critical points of rational functions and zeros of Wronskians, see, e.g., [20].
Secondly, as we show in Section 5, the characteristic polynomials for rank-one multiplicative non-Hermitian perturbations of Hermitian and/or Jacobi matrices are precisely of the type described in Theorems 3.1 and 4.1.This realization allows us to solve the direct and inverse spectral problem for such perturbations and establish the uniqueness of the solutions.Such perturbations were recently considered in random matrix theory [2,24].
The paper is organized as follows.In Section 2 we set the notation and remind the readers the amplitudephase technique for proving the Hermite-Biehler theorem.Section 3 is devoted to the generalization of the Hermite-Biehler theorem to the setting when the real and imaginary parts of a given complex polynomial have real, simple zeros that strictly interlace except at a certain known point.In this case, the zeros of this polynomial may be located either in the upper or lower complex half-planes and are subject to an explicit restriction on their arguments, Theorem 3.1.The case when the interlacing of zeros of the real and imaginary part is broken on an interval around a given point is considered in Section 4 whose main result is Theorem 4.1.In the final Section 5, we apply the results of Sections 2-4 to the direct and inverse problems for non-Hermitian perturbations of Hermitian and Jacobi matrices.

Hermite-Biehler theorem
Let (2.1) be a monic complex polynomial.In the complex plane C, consider the curve Γ h ≡ { h(t) : t ∈ R } with the direction of increasing t.This oriented curve is called the (Mikhailov-Nyquist) hodograph of the polynomial h, see.e.g., [4,27].Let us assume that the polynomial h has no zeros on the real axis.In that case, Γ h does not pass through the origin and the function where Arg z ∈ [−π, π) is the principal value of the argument of z ∈ C \ {0}, is continuous at each point of the real axis.It is clear that ϕ h (t) is a branch of arg h(t), t ∈ R. Let us introduce the increment 3) The existence of the limits (2.3) will be shown below.
Proof.Let s j (z) := z − z j .Then ϕ s j (t) = Arg(t − z j ).It is easy to see that the hodograph Γ s j of s j is a horizontal line traversed from left to right that intersects the imaginary axis at the point −i Im z j .Thus, as t runs from −∞ to ∞, the radius-vector of a point on the hodograph makes a counterclockwise turn of magnitude π if Im z j > 0 (clockwise if Im z j < 0), that is, Δ s j = π sgn Im z j .Now from (2.2)-(2.3)we obtain From the proof of the theorem we obtain that the number Δ h is well defined, and Let us consider the extremal case when all zeros of the polynomial h lie in C + .In this case, the polynomial h has remarkable properties.In particular, from Theorem 2.1 we immediately have the following consequences.
Corollary 2.2.Let h be as in (2.1) with z j / ∈ R for all j.Then h has all zeros in C + if and only if Δ h = πn.
Lemma 2.3 (Hermite).If all zeros of h lie in C + , then ϕ h (t) is a monotone (strictly) increasing function on R.
Proof.Indeed, for s j (z) = z − z j , with Im z j > 0, one has Thus, ϕ s j (t) is strictly increasing function on R. Then so is the function As was independently established by C. Hermite [14] and M. Biehler [5] in 1879, such a property of the polynomial h can be restated in terms of the root locations of the real and imaginary parts of h.Namely, after normalization we can express the polynomial h (2.1) as where p and q are monic real polynomials with deg p = n, deg q n − 1, and l ∈ R. Then the following theorem holds whose brief proof we provide for completeness.
Theorem 2.4 (Hermite-Biehler [14,5]).Let a polynomial h be given by (2.5) with p and q monic and real, deg p = n, deg q n − 1, l ∈ R. Then all zeros of h belong to C + if and only if l > 0 and zeros of p and q strictly interlace, that is, between any two zeros of one polynomial there lies exactly on zero of the other one.
Proof.Let all the zeros of h lie in C + .Then by (2.4)  with ϕ h (λ j ) = π 2 + kπ and n − 1 real values {μ j } n−1 j=1 with ϕ h (μ j ) = kπ, and these two sets strictly interlace, that is, all the elements are distinct and ordered by increasing values, and between any two consecutive elements of one set there is exactly one element of another set.From (2.5) it follows that p(λ j ) = 0 and q(μ j ) = 0 for each j.Finally, since l = n j=1 Im z j we get l > 0.
Conversely, suppose p(λ j ) = 0 and q(μ j ) = 0, where the distinct real numbers {λ j } n j=1 and {μ j } n−1 j=1 strictly interlace, so that λ j < μ j < λ j+1 .Notice that z j / ∈ R for any j, otherwise, p(z j ) = q(z j ) = 0 contrary to strict interlacing, so that ϕ h (t) is well defined for all t ∈ R. It is easy to see that p (t)q(t) − p(t)q (t) > 0 (see, e.g., [18,Thm 3.4]).Since ϕ h (t) is a branch of arg h(t) we get for all t ∈ R. Thus, ϕ h (t) is strictly increasing on R, and its graph intersects the imaginary axis exactly n times at the points λ j , j = 1, . . ., n, and the real axis n − 1 times at the points μ j , j = 1, . . ., n − 1.This means that Δ h = πn, so that all zeros of h(z) lie in C + according to Corollary 2.2.
Remark 2.5.Theorem 2.4 is just one step away from the full classification of complex spectrum of rankone non-Hermitian perturbations of Jacobi matrices (or any Hermitian matrices with a cyclic vector).We elaborate it in Section 5.1 below.
Remark 2.6.The same proof shows that Theorem 2.4 holds if we replace (2.5) with h(z) = p(z) + βq(z), where β with Im β < 0 is fixed.One can also prove an analogue of Theorem 2.4 in the case when deg q = n.Namely, zeros of the polynomial h(z) = αp(z) + (1 − α)q(z) belong to C + if and only if the zeros of p and q strictly interlace and Im α • A < 0, where A is the leading z n−1 -coefficient of the polynomial p(z) − q(z).
If one assumes that p has s negative and n −s non-negative zeros then we can say more about the possible locations of zeros of h in (2.5).This was already discussed in [21,Sect 3], but we provide a more general statement and a more transparent proof here.It is worth noting that (2.6)-(2.7)are related to the condition that appears in the case of broken interlacing (see (3.3) if and only if p has s negative zeros and n − s positive zeros.On the other hand, The result in (i) follows by combining these two statements.For (ii) observe that λ s+1 = 0, which holds if and only if ) follows from (2.8), as required.

Generalized Hermite-Biehler theorem: broken interlacing at the origin
As mentioned above (see also Section 5.1 for details), the Hermite-Biehler theorem appears naturally in the solution of the inverse spectral problem for a non-Hermitian rank-one additive perturbation of Jacobi and Hermitian matrices.However, higher rank perturbations require analogues of the Hermite-Biehler theorem where we no longer have interlacing of zeros of the real and imaginary parts of the given polynomial.
In particular, multiplicative rank-one perturbations and additive rank-two perturbations (see Sections 5.2 and 5.3) lead to situations where interlacing of zeros is broken at exactly one location.We therefore need to consider a complex linear combination of two such polynomials and study possible configurations of its zeros.
In the sequel, we use the following function Geometrically Arg [0,π) z ∈ [0, π) measures the angle between the radius-vector of z and the positive half-axis R + if z ∈ C + and the negative half-axis R − if z ∈ C − .In other words, The following theorem is the one of the main results of the paper.It characterizes possible configurations of zeros of linear combinations of two monic real polynomials whose positive and negative zeros interlace except for broken interlacing at the origin.Theorem 3.1.Let p and q be monic real polynomials, deg p = n, deg q = n − 1, p(0) = 0, and α ∈ C + .Given the monic polynomial if the zeros of p and q are real, simple, and strictly interlace, then h has all of its zeros Conversely, if α ∈ C + and all the zeros of a monic complex polynomial h then there exists a unique pair of monic real polynomials p and q with strictly interlacing zeros such that The number of z j 's in C + (respectively, in C − ) coincides with the number of positive (respectively, negative) zeros of p. Remark 3.2.If all zeros of p are positive or all negative, then we actually have strict interlacing of zeros of p(z) and zq(z).After dividing h(z) by α we end up in the setting of Remark 2.6.The extra restriction (3.3) on possible configurations of zeros comes from the fact that zq(z) vanishes at the origin.Compare with Proposition 2.7(ii), s = 0. Proof.
[⇒] Suppose that p(λ j ) = 0 and q(μ j ) = 0, where the distinct real numbers {λ j } n j=1 and {μ j } n−1 j=1 strictly interlace, so that λ j < μ j < λ j+1 .Assume also that s is the number of negative zeros of p, i.e., λ s < 0 and λ s+1 > 0. Since p(0) = 0, Im α = 0, and zeros of p and q do not coincide, the polynomial h defined in (3.2) has no real zeros.Therefore z j ∈ C \ R for each j and the function ϕ h (t) in (2.2) is well defined for all t ∈ R.
If s = n, then by the same arguments as above we obtain (3.9) for all j and k.Thus, ϕ h (0) < 0 by (3.8), and In this case ϕ h (λ j ) > 0, ϕ h (μ k ) > 0, and [⇐] Conversely, let the zeros of the polynomial h satisfy (3.3) for a given complex number α with Im α > 0. Then we have r j z j , where p j 's and r j 's are to be determined from the condition By equating the coefficients of z j , we get h j = αp j + (1 − α)r j which gives us the system of linear equations Thus, r(z) = zq(z) for a polynomial q of degree n − 1, so the formula (3.2) holds with the polynomials p and q being uniquely determined by the polynomial h and the number α.
Let us show that the zeros {λ j } n j=1 of p and {μ j } n−1 j=1 of q are strictly interlacing.To this end, define A 1 and A 2 with π > A 1 > A 2 > 0 as in (3.4) and observe that (3.5)-(3.6)still hold by the identical arguments as above (notice that p(t) and tq(t) cannot vanish simultaneously for t ∈ R since h has no real zeros).
Moreover, the identity (3.12) holds with n + being the number of zeros of h in C + .Combining this with (3.3) and (3.4), we get ϕ h (0) = A 2 −πn + .Furthermore, lim by (2.4).Now the intermediate value theorem for continuous functions implies that ϕ h has to attain each value of the form A 1 + πm (m ∈ Z) in the range ϕ h (0) < y < ϕ h (−∞) at least once on t < 0. There are exactly n − of such numbers, namely, A 1 − (n + − j)π, 0 j n − − 1 (this includes the case n − = 0 with ϕ h (0) > ϕ h (−∞)).From (3.5) it follows that there are at least n − distinct negative zeros of p.Similarly, there are n + values of the form A 1 + kπ in the range ϕ h (0) < y < ϕ h (+∞), namely, A 1 − (n + − j)π, 0 j n + − 1.From the intermediate value theorem and (3.5) we obtain that there are at least n + distinct positive zeros of p. Since n − + n + = n, and deg p = n, this implies that p has only real and simple zeros {λ j } n j=1 .In the same way, one can show that the zeros of zq(z) are real and simple.Moreover, there is at least one zero of q(z) on each interval (λ j , λ j+1 ) for j = n − .
If n − > 0 then we need to inspect the interval (λ n − , λ n − +1 ) where we have . This means that apart from t = 0 there is at least one more zero of tq(t) (possibly at t = 0, as well, a double zero).Thus, the zeros of q are strictly interlacing with the zeros of p.
If n − = 0 then there are at least n − 1 positive zeros of tq(t), one in each interval (λ j , λ j+1 ) ⊂ R + as obtained above.These are, therefore, n − 1 zeros of q which are all real, simple, and strictly interlacing with {λ j } n j=1 , the zeros of p.
Remark 3.3.One can change the origin to an arbitrary point by considering the polynomial h in the following form where ξ is an arbitrary given number.In this case, the result of Theorem 3.1 remains the same with condition (3.3) naturally replaced by

Generalized Hermite-Biehler theorem: broken interlacing at a point around the origin
In the present section, we establish a generalization of Theorem 3.1 for the case when interlacing of zeros of two polynomials is broken at one location around the origin.Precisely, we consider the following polynomial where the zeros of zp(z) and r(z) strictly interlace.This type of broken interlacing naturally appears for multiplicative perturbations of singular Jacobi matrices, see Section 5.2.
Note that if in 3.1 we allow p(z) to vanish at the origin, then we have p(z) = z p(z) and therefore h(z) = z α p(z) + (1 − α)q(z) , so that h(z)/z satisfies the conditions of the following theorem.if the zeros {λ j } n j=1 of p and {μ j } n j=1 of r satisfy the inequalities (for some 0 s n − 1), then the zeros of h(z) belong to Conversely, if α ∈ C + and the zeros of a monic complex polynomial h(z) = n j=1 (z − z j ) satisfy the condition (4.3), then there exists a unique pair of monic real polynomials p and r with h(z) = αp(z) + (1 − α)r(z) whose zeros {λ j } n j=1 and {μ j } n j=1 satisfy (4.2).The number of z j 's in C + (respectively, in C − ) coincides with the number of positive (respectively, negative) zeros of p.Note that if s = 0, then we obtain the conditions of Theorem 2.4 and Proposition 2.7, and (4.2) becomes (partially) the condition (2.6).Moreover, as it follows from the proof below, if μ s → 0 or μ s+1 → 0, then the condition (4.3) approaches (3.3), since (4.4) becomes (3.10).

Proof of Theorem 4.1.
[⇒] Given α ∈ C + , suppose that the zeros of the polynomials p and r satisfy the inequalities (4.2), and consider the polynomial h defined in (4.1).Repeating the arguments in the [⇒] proof of Theorem 3.1 with r(z) instead of zq(z) and {μ j } n j=1 instead of {0} ∪ {μ j } n−1 j=1 , we get that the function ϕ h defined in (2.2) is well defined on R (since p and r do not vanish on R simultaneously) and satisfies the identities (3.7), (3.9) and (3.11).The analogue of (3.10)  Arg [0,π) z j < Arg α.
[⇐] Conversely, let the zeros of the polynomial h belong to (4.3) for a given complex number α ∈ C + .Then from (3.12) and (4.3) we have where A 1 and A 2 are defined in (3.4), and n + is the number of zeros of h in the C + .In the same way as in the proof of Theorem 3.1, one can show that there exists a unique pair of real monic polynomials p and r of degree n such that (4.1) holds.Following the proof of Theorem 3.1 and using (4.5), we establish that the polynomial p has real and simple zeros λ j 's satisfying where n − is the number of zeros of h in C − , n − + n + = n while the polynomial r(z) = n j=1 (z − μ j ) has an odd number of zeros, counting multiplicities, on every interval (λ j , λ j+1 ), j = n − , and an even number of zeros (at least two), counting multiplicities, on (λ n − , λ n − +1 ) since ϕ h (λ n − ) = ϕ h (λ n − +1 ) > ϕ h (0).Moreover, due to ϕ h (0) < A 2 − πn + , the polynomial r must have exactly two distinct zeros on (λ n − , λ n − +1 ) such that μ n − < 0 < μ n − +1 , as required.
Remark 4.2.One can change the origin to an arbitrary point by considering the polynomial h in the form (4.1) with (4.2) replaced by where ξ is an arbitrary given number.In this case, the result of Theorem 4.1 remains the same with condition (4.3) naturally replaced by From (3.12) it follows that Arg z j < Arg α − πs.

Applications
In this section, we show how the Hermite-Biehler theorem and its extensions from the previous sections can be applied to solve the direct and inverse spectral problems for additive and multiplicative rank-one (and specific rank-two) perturbations of certain classes of Hermitian and Jacobi matrices.
Given a Jacobi matrix let us define J (j) to be the (n − j) × (n − j) (Jacobi) submatrix obtained from (5.1) by removing its first j rows and first j columns.Let p j , j = 1, . . ., n − 1, be the characteristic polynomials of J (j) , respectively, and p 0 is the characteristic polynomial of J .It is easy to check that where p n (z) ≡ 1 and p n+1 (z) ≡ 1.It is well-known that the zeros of p 0 (z) and p 1 (z) are real, simple, and strictly interlacing (see, e.g., [12]), and given two strictly interlacing sets of real numbers {λ j } n j=1 and {μ k } n−1 k=1 there exists a unique matrix J of the form (5.1) such that these two sets are the eigenvalues of J and J (1) , respectively (see [32], as well as [8,11,13,17]).

Additive non-Hermitian rank one perturbations
Let e 1 ∈ R n be the first coordinate vector.Consider the non-Hermitian (additive) perturbation J l,+ = J + ile 1 e * 1 (5.3) of J .In other words, J l,+ is (5.1) with the (1, 1)-entry replaced by b 1 + il.See [3] for an in-depth study of such perturbations, and [30] and [31, Chapter X] for a related topic in theory of stable polynomials.
Let h(z) be the characteristic polynomial h(z) of J l,+ .Using the Laplace expansion for the determinants we obtain the following representation for h(z) As mentioned above, the zeros of p 0 and p 1 strictly interlace.Therefore, by the Hermite-Biehler theorem, Theorem 2.4, the eigenvalues of J l,+ lie in the open upper (lower) half-plane whenever l > 0 (l < 0).The inverse problem for matrix J l,+ requires to find the matrix from its spectrum.Namely, let the numbers (z − z j ) can be uniquely represented as h = p − ilq for some l ∈ R and monic real polynomials p and q with deg p = n, deg q = n − 1.From Theorem 2.4 it follows that l > 0, and the zeros of p and q strictly interlace.These p and q allow to recover J uniquely, and the number l determines the perturbation magnitude, see (5.3).Thus, J l,+ (5.3) can be reconstructed from its spectrum, and the solution of the inverse problem is unique.This also allows us to study the spectrum of rank-one perturbations of generic Hermitian matrices: recall that any n × n Hermitian matrix H = H * with a cyclic vector v can be reduced to a Jacobi form J (5.1) via H = S * J S for some unitary S with Sv = e 1 (the Lanczos algorithm [22]).Using this, one can reduce any as required.Moreover, assuming that the signature of H is given, one ends up in the setting of Proposition 2.7 where we know the number of positive/negative zeros of p.In particular, if v is cyclic and l > 0 then eigenvalues of (5.4) satisfy (2.7) or (2.6), depending on whether λ = 0 is an eigenvalue of H or not, respectively.
If instead v is not cyclic, i.e., dim span{H j v : j 0} = k < n then the Gram-Schmidt procedure in the Lanczos algorithm terminates early.We still get (5.5) but now with a k = 0 in (5.1), and so we get k eigenvalues in C + and n − k real eigenvalues.For applications of the results discussed above, see [1,21].

Multiplicative non-Hermitian rank one perturbations
Let us now consider multiplicative rank-one perturbation with k > 0.Here I is the n × n identity matrix.In other words, J k,× is (5.1) with the (1, 1)-entry being replaced by b 1 (1 + ik) and the (2, 1)-entry replaced by a 1 (1 + ik).
Let us find the location of the eigenvalues of J k,× .To this end, we find its characteristic polynomial h: (5.7) In the next two statements we solve the direct and inverse spectral problems for J k,× .The cases det J = 0 and det J = 0 are considered separately.Theorem 5.1.Given a matrix J as in (5.1) with det J = 0.The spectrum of the matrix J k,× defined in (5.6) with k > 0, satisfy the condition Conversely, each configuration of points from (5.8) occurs as a spectrum of a unique J k,× with some k > 0.
The number of positive and negative eigenvalues of J coincides with the number of eigenvalues of J k,× in C + and C − , respectively.
Proof.The characteristic polynomial h of J k,× has the form (5.7), which coincides with (3.2) for α = 1 + ik.Clearly, as k varies in (0, ∞) we get Arg α ∈ (0, π 2 ), and each such value of Arg α is achieved exactly once.The result now follows from Theorem 3.1.
Conversely, given a configuration of points {z j } n j=1 in (5.8), we set α = 1 + ik where k = tan Arg [0,π) z j > 0, so that α ∈ C + with Arg α = Arg [0,π) z j .By Theorem 3.1, the set {z j } n j=1 is the zero set of a unique polynomial h of the form (5.7) with two polynomials p 0 and p 1 of degrees n and n − 1, respectively, with strictly interlacing zeros.As discussed above, p 0 and p 1 uniquely determine J such that p 0 and p 1 are the characteristic polynomials of J and J (1) .The uniqueness and existence of J and k > 0 implies the uniqueness and existence of J k,× whose spectrum is {z j } n j=1 .
If the matrix J under perturbation is degenerate, the following fact holds true.
Theorem 5.2.Given a matrix J as in (5.1) with det J = 0, the spectrum of the matrix J k,× defined in (5.6) with k > 0, contains a simple eigenvalue 0, while the remaining eigenvalues satisfy the condition (5.9) Conversely, for any k > 0, each configuration of points of the form (5.9) occurs as a spectrum of a unique J k,× with det J = 0.
The number of positive and negative eigenvalues of J coincides with the number of eigenvalues of J k,× in C + and C − , respectively.
Proof.Since det J = 0 we have p 0 (z) = z p 0 (z), so by (5.7) the characteristic polynomial of J k,× has the form then the zeros {λ j } n−1 j=1 of p 0 and {μ j } n−1 j=1 of p 1 satisfy the inequalities for some integer s, 0 s n − 1.As Arg(1 + ik) = arctan k, by Theorem 4.1 we obtain that the n − 1 nonzero eigenvalues of J k,× belong to (5.9).
Conversely, given k > 0 and a configuration of points {z j } n−1 j=1 satisfying (5.9), from Theorem 4.1 with α := 1 + ik one obtains two polynomials p and r of degree n − 1 whose zeros {λ j } n−1 j=1 and {μ j } n−1 j=1 satisfy (5.10), so zp(z) and r(z) have strictly interlacing zeros.Now we can reconstruct a unique Jacobi matrix J of the form (5.1) such that zp(z) and r(z) are the characteristic polynomials of J and J (1) .
Remark 5.3.Note that unlike the situation in Theorem 5.1, for a fixed k > 0 given a set {z j } n−1 j=0 , z 0 = 0, satisfying (5.9), there are infinitely many matrices J κ,× of the form (5.6) with various κ > 0 whose spectrum is {z j } n−1 j=0 , z 0 = 0.However, if κ > 0 is fixed, then such a matrix J κ,× is unique by Theorem 4.1.In particular, for κ = k we have the converse statement of Theorem 5.2.Theorems 5.1-5.2 can be used to study eigenvalues of multiplicative rank-one perturbations of generic Hermitian matrices of the form H(I + iΓ), (5.11)where H = H * , Γ = Γ * , rank Γ = 1.Provided that a non-zero vector from Ran Γ is cyclic for H, the matrix (5.11) can be reduced to (5.6) via a unitary conjugation.See [2] for an application of such results to multiplicative perturbations of random matrices.

Additive non-Hermitian rank-two perturbations
Consider now an additive perturbation of J of the following type with m > 0 and l ∈ R. A similar calculation as in the previous section shows that the characteristic polynomial of J l,m is (5.13)So it can be represented in the form with α = 1 + im a 1 and ξ = b 1 − la 1 m .Thus, as above we can solve the direct and inverse spectral problems for J l,m .The cases det(J − ξI) = 0 and det(J − ξI) = 0 are considered separately.Theorem 5.4.Given the numbers m > 0 and l ∈ R, the spectrum of the matrix J l,m defined in (5.12) belongs to with ξ = b 1 − la 1 m provided det(J − ξI) = 0 where J = J 0,0 .Conversely, given a number ξ ∈ R, each configuration of points from (5.14) occurs as the spectrum of a unique matrix J l,m with some m > 0 and l ∈ R.
The number of eigenvalues of J greater than ξ and less than ξ coincides with the number of eigenvalues of J l,m in C + and C − , respectively.
Proof.The characteristic polynomial h of J l,m has the form (5.13) where the zeros of the polynomials p and q strictly interlace.Now we are in the situation of Remark 3. Arg [0,π) (z j − ξ).By Theorem 3.1 and Remark 3.3, the set {z j } n j=1 is the zero set of a unique polynomial h of the form (5.13) with two polynomials p and q of degrees n and n − 1, respectively, with strictly interlacing zeros.As discussed above, p and q uniquely determine J such that p and q are the characteristic polynomials of J and J (1) .In particular, they uniquely determine the numbers a 1 > 0 and b 1 ∈ R as follows: Uniqueness and existence of J , m > 0, and l ∈ R implies uniqueness and existence of J l,m whose spectrum is {z j } n j=1 .
If the matrix J − ξI, where ξ = b 1 − la 1 m with given m > 0 and l ∈ R, is degenerate, then one obtains the following result.where A = m a 1 > 0.
Conversely, given ξ ∈ R, for any A > 0, each such a configuration of points occurs as the spectrum of a unique J l,m with det(J − ξ) = 0, and m > 0, l ∈ R.
The number of eigenvalues of J greater than ξ and less than ξ coincides with the number of eigenvalues of J l,m in C + and C − , respectively.
C. de Boor and G. Golub [8] found a numerically stable algorithm to reconstruct J from the sets {λ j } n j=1 and {μ k } n−1 k=1 of distinct strictly interlacing real numbers such that the first set is the spectrum of J while the second one is the spectrum of its submatrix J (1) .We believe that on the base of that algorithm one can develop a numerically stable algorithm to reconstruct the matrices J l,+ , J l,× , and J l,m from the respective data.

. 14 )
Determinant of the coefficient matrix is equal to − Im α = 0, so that p j and r j are uniquely determined.Since h n = 1 we have p n = r n = 1, p and r are monic of degree n.By (3.13) and the fact that h 0 = (−1) n n j=1 z j , the solution of the system (3.14) for j = 0 has the form p 0 = (−1)

. 4 )
Then using s = n − , n − s = n + (the same arguments as in the proof of Theorem 3.1) and (3.12), we get n j=1

7 . 4 . 3 .Proposition 4 . 4 .
z j ) + π(n + − n − ) = A 2 − πn − This formula and Theorems 3.1 and 4.1 imply the following analogues of Proposition 2.Proposition In the setting of Theorem 3.1, p has s negative and n − s positive zeros if and only if the polynomial h(z) = n j=1 (z − z j ) has s zeros in C − , n − s zeros in C + , and n j=1 Arg z j = Arg α − πs.In the setting of Theorem 4.1, p has s negative and n − s positive zeros if and only if the polynomial h(z) = n j=1 (z − z j ) has s zeros in C − , n − s zeros in C + , and −πs < n j=1

) for some integer s, 0 s n − 1 . As Arg 1 + i m a 1 = arctan m a 1 ,
by Theorem 4.1 and Remark 4.2, we get that the n − 1 nonzero eigenvalues of J l,m belong to (5.16) with A = m a 1 .

Let the polynomial h defined in (2.5) have all zeros {z j } n j=1 in C + . The following statements hold. (i) p has s negative and n − s positive zeros if and only if the zeros of h satisfy the inequalities
and Remark 3.2).