Let H∈ℂn×n be an n×n unitary upper Hessenberg matrix whose subdiagonal elements are all positive, let Hk be the kth leading principal submatrix of H, and let H˜k be a modified submatrix of Hk. It is shown that when the minimal and maximal eigenvalues of H˜k (k=1,2,…,n) are known, H can be constructed uniquely and efficiently. Theoretic analysis, numerical algorithm, and a small example are
given.

1. Introduction

Direct matrix eigenvalue problems are concerned with deriving and analyzing the spectral information and, hence, predicting the dynamical behavior of a system from a priori known physical parameters such as mass, length, elasticity, inductance, and capacitance. Inverse eigenvalue problems (IEPs), in contrast, are concerned with the determination, identification, or construction of the parameters of a system according to its observed or expected behavior.

The inverse eigenvalue problems arise in a remarkable variety of applications, such as mathematics physics, control theory, vibration project, structure design, system parameter identification, and the revise of mathematics models [1–8]. Recent years, inverse eigenvalue problem of matrices has become an active topic of computational mathematics for needs of project and technology, and it has resolved a great deal of concrete problem. Especially, the inverse eigenvalue problems have many applications in engineering design, for example, they arise in aviation, civil structure, nucleus engineering, bridge design, shipping construction, and so on. Pole assignment problem have been of major interest in system identification and control theory, we can use optimization techniques to get a solution which is least sensitive to perturbation of problem data. Byrnes [9], Kautsky et al. [10], and Chu and Li [11] gave an excellent recount of activities in this area. Joseph [7] presented a method for the design of a structure with specified low-order natural frequencies, and the method can further be used to generate initial feasible designs for optimum design problems with frequency constraints. By measuring the changes in the natural frequencies, the IEP idea can be employed to detect the size and location of a blockage in a duct or a crack in a beam, see [12–15] for additional references. Starek and Inman [16] discussed the applications of IEPs to model updating problems and fault detection problems for machine and structure diagnostics. Applications to other types engineering problems can be found in the books [4, 17] and articles [18–23].

Throughout this paper we use Ij to denote the j×j identity matrix, ej to denote the jth column of the identity matrix, Λ(H) to denote the spectrums of a square matrix H, x̅ to denote the complex conjugate of x, and ℋn to denote the set of unitary upper Hessenberg matrices of order n with positive subdiagonal elements.

It is known [24] that any H∈ℋn can be written uniquely as the products
H=G1(γ1)⋯Gn-1(γn-1)G̃n(γn),
where
Gk(γk)=(Ik-1-γkσkσkγ̅kIn-k-1),k=1,2,…,n-1,G̃n(γn)=diag(In-1,-γn).
In (1.1) and (1.2), the parameters γk∈ℂ (k=1,2,…,n) are called reflection coefficients or Schur parameters in signal processing, σk∈ℝ (k=1,2,…,n-1) are said to be complementary parameters and satisfy |γk|2+σk2=1, σk>0, k=1,…,n-1, and |γn|=1. We refer to (1.1) as Schur parametric form of H [25], it plays a fundamental role in the development of efficient algorithms for solving eigenproblems for unitary Hessenberg matrices. However, (1.2) is called the complex Givens matrices. H in (1.1) is of the explicit form
H=(-γ1-σ1γ2-σ1σ2γ3⋯-σ1⋯σn-1γnσ1-γ̅1γ2-γ̅1σ2γ3⋯-γ̅1σ2⋯σn-1γnσ2-γ̅2γ3⋯-γ̅2σ3⋯σn-1γn⋱⋱⋮σn-1-γ̅n-1γn),
and is uniquely determined by γ1,γ2,…,γn. We denote this n×n unitary Hessenberg matrix by H(γ1,γ2,…,γn), each H∈ℋn is therefore determined by the 2n-1 real parameters. Let Hk be the kth leading principal submatrix of H. The matrix Hk is not unitary for k<n and its eigenvalues are inside the unit circle. However, Hk will become unitary if γk is replaced by ρk which is any number on the unit circle [24]. We introduce the following sequence of modified unitary submatrices:
H̃k=(-γ1-σ1γ2⋯-σ1⋯σk-1ρkσ1-γ̅1γ2⋯-γ̅1σ2⋯σk-1ρk⋱⋱⋮σk-1-γ̅k-1ρk),k=1,2,…,n.
Because all ρk are of modulus one, the modified submatrices H̃k are unitary and its eigenvalues lie on the unit circle, H̃k=H(γ1,…,γk-1,ρk). Assume that -1 is not an eigenvalue of H, then λj(k)∈Λ(H̃k) can be described as
λj(k)=exp(iθj(k)),j=1,2,…,k.
If we number the roots of H̃k starting from -π moving counterclockwise along the unit circle, that is,
-π<θ1(k)≤θ2(k)≤⋯≤θk(k)≤π,
then we also call λ1(k)=exp(iθ1(k)), λk(k)=exp(iθk(k)) are, respectively, the minimal and maximal eigenvalues of H̃k.

Hessenberg matrices arise naturally in several signal processing applications including the frequency estimation procedure and harmonic retrieval problem for radar or sonar navigation [26, 27]. Two kinds of inverse eigenvalue problems for unitary Hessenberg matrices have been considered up to now. Ammar et al. [28] discussed H=H(γ1,γ2,…,γn)∈ℋn is uniquely determined by its eigenvalues and the eigenvalues of Ĥ, where Ĥ=H(αγ1,αγ2,…,αγn)=(I-(1-α)e1e1T)H(γ1,γ2,…,γn), that is, Ĥ a multiplicative rank-one perturbation of H, and the methods are described in [28, 29]. Ammar and He in [24] considered that H∈ℋn can also be determined by its eigenvalues and the eigenvalues of a modified (n-1)×(n-1) leading principal submatrix of H.

In this paper, we consider the following inverse eigenvalue problem.

Problem 1.

For 2n-1 given real numbers θ1(k),θk(k)∈(-π,π] (k=1,2,…,n), find unitary Hessenberg matrices H∈ℋn, such that λ1(k)=exp(iθ1(k)), λk(k)=exp(iθk(k)) are, respectively, the minimal and the maximal eigenvalues of H̃k for all k=1,2,…,n.

This paper is organized as follows. In Section 2, we discussed the properties of unitary Hessenberg matrix. Then the necessary and sufficient conditions for solvability of Problem 1 are derived in Section 3. Section 4 gives the algorithm and numerical example for the problem.

2. The Properties of Unitary Hessenberg Matrix

We denote the characteristic polynomials of H̃k by φk, that is, φk(λ)=det(λIk-H̃k). We can appropriately choose ρk such that φk(λ) satisfy the three-term recurrence relations [30, 31], the following lemma give a special method to define ρk.

Lemma 2.1 (see [<xref ref-type="bibr" rid="B32">32</xref>]).

Let H=H(γ1,…,γn)∈ℋn, assume -1 is not an eigenvalue of H, define
ρn=γn,ρk=γk-ρk+11-γ̅kρk+1,k=n-1,n-2,…,1.
Let H̃k∈ℋk (k=1,…,n) be the modified unitary submatrices defined by (1.5). If one number the eigenvalues of H̃k starting from -1 moving counterclockwise along the unit circle, then the eigenvalues of H̃k interlace those of H̃k+1 in the following sense: the jth eigenvalue of H̃k lies on the arc between the jth and the j+1st eigenvalue of H̃k+1.

If ρk are defined by (2.1), we get the following lemma.

Lemma 2.2 (see [<xref ref-type="bibr" rid="B32">32</xref>]).

The characteristic polynomials φk(λ)=det(λIk-H̃k) of H̃k defined by (1.5) satisfy the following three-term recurrence relations:
φ0(λ)=1,φ1(λ)=λ+ρ1,φk(λ)=(λ+ρkρ̅k-1)φk-1(λ)-αk-1λφk-2(λ),k=2,3,…,n,
where
αk=γ̅k-1(ρk-γk)+ρk+1(ρ̅k-γ̅k),γ0=1.

Lemma 2.3.

If ρk defined by (2.1), αk defined by (2.3), then
γ0=1,γk=αk-γ̅k-1ρk+1ρ̅k-γ̅k-1,fork=1,2,…,n-1,γn=ρn.

Proof.

By (2.1), we get
ρk(1-γ̅kρk+1)=γk-ρk+1,
then
ρk+1(ρ̅k-γ̅k)=γkρ̅k-1.
Substituting the above formula into (2.3), we obtain
αk=γ̅k-1(ρk-γk)+γkρ̅k-1.
Because ρk≠γk-1, we have
γk=αk-γ̅k-1ρk+1ρ̅k-γ̅k-1,k=1,2,…,n-1.

Lemma 2.4.

Let x∈ℂ with |x|=1 and φk(λ) be the characteristic polynomials of H̃k, then
φk(x)x̅k=ρkφ¯k(x),k=1,2,…,n.

Proof.

It is easy to verify that
φk(x)x̅k=(x-λ1(k))(x-λ2(k))⋯(x-λk(k))x̅k=(1-x̅λ1(k))(1-x̅λ1(k))⋯(1-x̅λk(k))=λ1(k)λ2(k)⋯λk(k)(λ̅1(k)-x̅)(λ̅2(k)-x̅)⋯(λ̅k(k)-x̅)=(-1)kdet(H̃k)φ¯k(x)=(-1)k(-1)kρkφ¯k(x)=ρkφ¯k(x).

3. The Solution of Problem <xref ref-type="statement" rid="problem1">1</xref>

We now consider the solvability conditions of Problem 1 and give the following theorem.

Theorem 3.1.

For 2n-1 given real number θ1(k),θk(k)∈(-π,π](k=1,2,…,n), there is a unique H(γ1,γ2,…,γn)∈ℋn such that λ1(k)=exp(θ1(k)), λk(k)=exp(λk(k)) are, respectively, the minimal and the maximal eigenvalues of H̃k(k=1,2,…,n), if and only if
-π<θ1(n)<θ1(n-1)<⋯<θ1(2)<θ1(1)<θ2(2)<⋯<θn-1(n-1)<θn(n)≤π.

Proof.

Sufficiency. Notice that
-π<θ1(n)<θ1(n-1)<⋯<θ1(2)<θ1(1)<θ2(2)<⋯<θn-1(n-1)<θn(n)≤π.
By Lemma 2.1 we have that, if λ1(k), λk(k) are the eigenvalues of H̃k, they must be the minimal and the maximal eigenvalues of H̃k, respectively. So Problem 1 having a solution is equivalent to that the following equations:
φk(λ1(k))=0,φk(λk(k))=0,
having solutions αk-1, ρk satisfying |ρk|=1 for all k=1,2,…,n.

For j=1, we get ρ1=λ1(1)=exp(iθ1(1)), so |ρ1|=1.

For 2≤j≤n, by Lemma 2.1, from (2.2) and (3.3), we have
(λ1(k)+ρkρ̅k-1)φk-1(λ1(k))-αk-1λ1(k)φk-2(λ1(k))=0,(λk(k)+ρkρ̅k-1)φk-1(λk(k))-αk-1λk(k)φk-2(λk(k))=0.
Then
αk-1λ1(k)φk-2(λ1(k))-ρkρ̅k-1φk-1(λ1(k))=λ1(k)φk-1(λ1(k)),αk-1λk(k)φk-2(λk(k))-ρkρ̅k-1φk-1(λk(k))=λk(k)φk-1(λk(k)).
Let mk≡λ1(k)φk-2(λ1(k))φk-1(λk(k))-λk(k)φk-2(λk(k))φk-1(λ1(k)), we now show that mk≠0 by contradiction.

Assume that mk=0. Multiplying the first and second equation of (3.5) by φk-1(λk(k)), φk-1(λ1(k)), respectively, we get
(λ1(k)-λk(k))φk-1(λ1(k))φk-1(λk(k))=0,
so we obtain λ1(k)=λk(k) by φk-1(λ1(k))≠0 and φk-1(λk(k))≠0. This is a contradiction with λ1(k)≠λk(k), therefore, mk≠0. By |ρk-1|=1, we get -ρ̅k-1mj≠0. Then (3.5) have the unique solution
αk-1=ρ̅k-1(λk(k)-λ1(k))φk-1(λ1(k))φk-1(λk(k))-ρ̅k-1mk,ρk=λ1(k)λk(k)(φk-2(λ1(k))φk-1(λk(k))-φk-2(λk(k))φk-1(λ1(k)))-ρ̅k-1mk.
We show |ρk|=1 by induction. By ρ1=exp(iθ1(1)), so |ρ1|=1. Assume that |ρj|=1, for j=1,2,…,k-1.

By (3.8), λ1(k)≠0, and λk(k)≠0, we have
ρk=λ1(k)λk(k)(φk-2(λ1(k))φk-1(λk(k))-φk-2(λk(k))φk-1(λ1(k)))(λ̅1(k))k-1(λ̅k(k))k-1-ρ̅k-1(λ1(k)φk-2(λ1(k))φk-1(λk(k))-λk(k)φk-2(λk(k))φk-1(λ1(k)))(λ̅1(k))k-1(λ̅k(k))k-1=(λ̅1(k))k-2(λ̅k(k))k-2(φk-2(λ1(k))φk-1(λk(k))-φk-2(λk(k))φk-1(λ1(k)))-ρ̅k-1ρk-2ρk-1(φ̅k-2(λ1(k))φ̅k-1(λk(k))-φ̅k-2(λk(k))φ̅k-1(λ1(k)))=(λ̅1(k))k-2(λ̅k(k))k-2(φk-2(λ1(k))φk-1(λk(k))-φk-2(λk(k))φk-1(λ1(k)))|ρk-1|2ρk-2(φ̅k-2(λ1(k))φ̅k-1(λk(k))-φ̅k-2(λk(k))φ̅k-1(λ1(k))),
so |ρk|=1.

Now we have αk(k=1,2,…,n-1) and ρk(k=1,2,…,n), by Lemma 2.3, we can get γk, for k=1,2,…,n. Then we obtain the n×n unitary Hessenberg matrix H=H(γ1,γ2,…,γn).

Necessity. Suppose that Problem 1 has a unique solution, that is, λ1(k)=exp(θ1(k)), λk(k)=exp(λk(k)) are, respectively, the minimal and the maximal eigenvalues of H̃k(k=1,2,…,n), using Lemma 2.3, we get
-π<θ1(n)<θ1(n-1)<⋯<θ1(2)<θ1(1)<θ2(2)<⋯<θn-1(n-1)<θn(n)≤π.

Remark 3.2.

Assume that η0 is not the eigenvalue of H, we define
ρn=γn,ρk=γk+η̅0ρk+11+η̅0γ̅kρk+1,k=n-1,n-2,…,1.
Then Lemmas 2.1 and 2.2 still hold true.

4. Algorithm and Example

Based on the above analysis, it is natural that we should propose the following algorithm for solving Problem 1.

Compute αk-1 and ρk by (3.7) and (3.8) for k=2,3,…,n;

Set γ0=1;

Compute γk by (2.4) for k=1,2,…,n-1;

Set γn=ρn.

We present an example to illustrate this algorithm.

Example 4.2.

Let n=5, given θ1(1)=π/6; θ1(2)=-π/8, θ2(2)=π/4; θ1(3)=-π/4, θ3(3)=π/3; θ1(4)=-π/3, θ4(4)=π/2; θ1(5)=-π/2, θ5(5)=2π/3. By λj(k)=exp(iθj(k)), we get
λ1(1)=0.8660+0.5000i;λ1(2)=0.9239-0.3827i,λ2(2)=0.7071+0.7071i;λ1(3)=0.7071-0.7071i,λ3(3)=0.5000+0.8660i;λ1(4)=0.5000-0.8660i,λ4(4)=0.0000+1.0000i;λ1(5)=0.0000-1.0000i,λ5(5)=-0.5000+0.8660i.
Using Algorithm 4.1, we obtain {ρi}i=15, {αi}i=14, {γi}i=15 listed in Table 1. The unitary Hessenberg matrix is given as follows:
H=(0.7588+0.4471i-0.3354-0.1185i0.1490+0.0810i0.0045-0.0151i0.1169+0.2346i0.47360.6493-0.1269i-0.3152+0.0109i0.0071+0.0283i-0.4088-0.2656i00.66010.4024+0.0643i-0.0020-0.0377i0.4526+0.4382i000.84010.0068+0.0317i-0.4436-0.3106i0000.99820.0438-0.0407i).
We recompute the spectrum of H̃k(k=1,2,…,n), and get
Λ(H̃1)=(0.8660+0.5000i̲),Λ(H̃2)=(0.9239-0.3827i̲,0.7071+0.7071i̲),Λ(H̃3)=(0.7071-0.7071i̲,0.9013+0.4332i,0.5000+0.8660i̲),Λ(H̃4)=(0.5000-0.8660i̲,0.9964-0.0850i,0.7329+0.6803i,0.0000+1.0000i̲),Λ(H̃5)=(0.0000-1.0000i̲,0.7937-0.6083i,0.9411+0.3381i,0.6262+0.7796i,-0.5000+0.8660i̲).
These obtained data show that Algorithm 4.1 is quite efficient, Figure 1 illustrates the eigenvalues of H̃k(k=1,2,…,5).

{ρi}i=15,{αi}i=14,{γi}i=15.

i

ρi

αi

γi

1

-0.8660-0.5000i

-0.2265-0.0451i

-0.7588-0.4471i

2

0.9239+0.3827i

-0.4727-0.0442i

0.7083+0.2501i

3

-0.7584-0.6517i

-0.7675-0.3218i

-0.4766-0.2591i

4

0.3747+0.9271i

-1.3652-0.2759i

-0.0169+0.0574i

5

-0.4458-0.8951i

—

-0.4458-0.8951i

The eigenvalues of H̃k.

Acknowledgment

This work is supported by National Natural Science Foundation of China no. 10531080.

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