Solutions of a Quadratic Inverse Eigenvalue Problem for Damped Gyroscopic Second-Order Systems

Given k pairs of complex numbers and vectors (closed under conjugation), we consider the inverse quadratic eigenvalue problem of constructing n × n real matrices M, D, G, and K, where M > 0, K and D are symmetric, and G is skew-symmetric, so that the quadratic pencil Q(λ) = λ2M + λ(D + G) + K has the given k pairs as eigenpairs. First, we construct a general solution to this problem with k ≤ n. Then, with the special properties D = 0 and K < 0, we construct a particular solution. Numerical results illustrate these solutions.


Introduction
Vibrating structures such as buildings, bridges, highways, and airplanes are distributed parameter systems [1].Very often a distributed parameter system is first discretized to a matrix second-order using techniques of finite element or finite difference, and then an approximate solution is obtained for the discretized model.Associated with the matrix second-order model is the eigenvalue problem of the quadratic pencil,  () ≡  2  +  ( + ) + , where , , , and  are, respectively, mass, damping, gyroscopic and stiffness matrices.
The system represented by ( 1) is called damped gyroscopic system.In general, the gyroscopic matrix  is always skew-symmetric, the damping matrix  and the stiffness matrix  are symmetric, the mass matrix  is symmetric positive definite, and they are all  ×  real matrices.If  = 0, the system is called damped nongyroscopic system, and if  = 0, the system is called undamped gyroscopic system.The damped gyroscopic system has been widely studied in two aspects: the quadratic eigenvalue problem (QEP) and the quadratic inverse eigenvalue problem (QIEP).The QEP involves finding scalars  ∈ C and nonzero vectors x ∈ C  , called the eigenvalues and eigenvectors of the system, to satisfy the algebraic equation ()x = 0, when the coefficient matrices are given.Many authors have been devoted to this kind of problems and a series of good results have been made (see, e.g., [2][3][4][5][6][7][8]).The QIEP determines or estimates the parameters of the system from observed or expected eigeninformation of ().Our main interest in this paper is the corresponding inverse problem: given partially measured information about eigenvalues and eigenvectors, we reconstruct matrices , , , and , satisfied with several conditions, so that () has the given  eigenpairs.The problem we considered is stated as follows.
In [12], Kuo et al. constructed the solutions of the QIEP of the damped nongyroscopic system with  ≤  given eigenpairs.And for the same system, Cai et al. [13] solved the QIEP with 2 ≥  >  given eigenpairs.Meanwhile, for the damped gyroscopic system, Yuan [14] solved the QIEP with  given eigenpairs.In [14], Yuan constructed symmetric positive semidefinite matrix  and skew-symmetric matrix  for (), with  > 0 and  ≥ 0 as given matrices.So, it becomes challenging to construct (, , , ) for damped gyroscopic system (1) with  given eigenpairs, and this is the goal of this paper.
This paper is organized as follows.In Section 2, we establish the solubility theory of the Problem 1.In Section 3, we develop a simple method to compute a particular solution to the Problem 1 with  = 0.Moreover, for  = 0 and  < 0, a simple algorithm is developed to compute a solution in Section 4. Some numerical results are presented in Section 5 to illustrate our main results.In the last section, some conclusions and acknowledgements are given.
Throughout this paper, we use capital letters to denote matrices, and lowercase (bold) letters to denote scalars (vectors).  denotes the transpose of the matrix ,   denotes the  ×  identity matrix, and  † denotes the Moore-Penrose generalized inverse of .We write  > 0 ( ≥ 0) if  is real symmetric positive (semidefinite).The spectrum of  is denoted by ().
For simplicity, we make the following assumptions.
(A1) The eigenvector matrix  in Problem 1 has full column rank, that is, rank () = .
Remark 2. For the case that 0 ∈ (Λ), using the assumption of simple eigenvalues, we can partition Λ = diag{Λ 1 , 0}, where Λ 1 has no zero eigenvalue, and then do discussion with Λ 1 .So, in this paper, we only consider the case that Λ has no zero eigenvalue.

General Solution of the Problem
In this section, we will give a general solution to the Problem 1 for given matrix pair (Λ, ) ∈ R × × R × ( ≤ ) as in (2) and (3).At the beginning, we will introduce some lemmas.
When condition (6) is satisfied, the general solution of (5) is where  ∈ R ×ℓ is arbitrary and  ∈ R × is constrained only by the symmetry requirement that Lemma 3 directly results in the following lemma.
Lemma 4. Let  ∈ R × be a nonsingular matrix and  ∈ R × ; then has a solution  ∈ R × if and only if in which case the general solution is  = (1/2) −  + , where  =   ∈ R × is an arbitrary matrix.
Given matrix pair (Λ, ) ∈ R × × R × ( ≤ ) as in Problem 1: let be the QR-factorization of , where  ∈ R × is orthogonal and  ∈ R × is upper triangular.We may require that  has positive diagonal entries, since  is of full column rank.Let  =  + , so that finding , , , and  which satisfy (4) is equivalent to finding , , and  which satisfy and the relations of , , and  are Let  = Λ −1 ; we can see  −1 exists by using 0 ∉ (Λ).Denoting Furthermore,  and  can be expressed as in (13).
As  and  are required to be symmetric positive definite and symmetric, respectively, so are  11 and  11 in ( 14) and (16).From (17) it follows that Let  11 be an arbitrary symmetric positive definite matrix.We need to find  11 such that  11 is symmetric; that is, it satisfies After rearrangement, (21) becomes and  is nonsingular, we can get from Lemma 4 that where  ∈ R × is arbitrary symmetric.Substituting (23) into (20) yields (v).Furthermore,  and  can be expressed as in (13).
Sufficiency.From the description of (i)-(vi), we can obtain that (12) holds; thus (4) holds with Remark 6.The general solution to the Problem 1 with  = 0 and ( ≤ ) prescribed eigenpairs has been given in [12], here, we generalize its solution to the case of  ̸ = 0.
Remark 7. It is complicated for the more general case rank() < , and we will discuss it in our next work.However, here we provide a simple solution.We can select the linear independent columns and the relevant eigenvalues to construct a new   and Λ  , then do discussion with them.
Remark 8.When  = , by Theorem 5, the general solution of the Problem 1 is given by where with   > 0 and   =    which can be arbitrarily chosen and  ∈ R× is arbitrarily symmetric.
Using Theorem 5, we can construct a solution to the Problem 1 as follows.
Algorithm 9.An algorithm for solving Problem 1 is proposed as follows.

Particular Solutions with 𝐷 = 0
As we all know, the applications of the undamped gyroscopic system (i.e.,  = 0) exist in many fields; for details, see [5].
In this section, we will discuss the particular solutions of Problem 1 with  = 0 and ( ≤ ) prescribed eigenpairs.And in this case, () in (1) becomes It is well known that the eigenvalues of (28) have a Hamiltonian structure; that is, they occur in quadruples (, , −, −), possibly collapsing to real or imaginary pairs or singlezero eigenvalues.In [11], Jia and Wei discussed the eigenvalues of () in (28) and separated them into four categories.From the assumption (A2), we can know  is even.Here we rewrite the given eigeninformation pair (Λ, ) of Problem 1 as with where are eigenpairs, and In this section, the Problem 1 becomes the following problem.
Given an eigeninformation pair (Λ, )( ≤ ) with ( 29)-(31), find  ×  real matrices , , and , with  being symmetric definite,  being symmetric, and  being skewsymmetric, so that, As well as in Section 2, let where   is partitioned conforming with that of  in ( 14), and we can easily calculate that  also satisfies the action of  in Section 2, except that  has an additional property, that is,   = −.In the following theorem, we will discuss the solubility of Problem 1 with  = 0 and  ≤ .
with   and   being arbitrary real numbers.
Proof.Necessity.Same as the proof of Theorem 5, we can get where  21 and  21 are arbitrary.We also have and  11 satisfies After rearrangement, (38) becomes It is easily seen that (39) has a particular solution Next, we consider the homogeneous equation Substituting  = Λ −1 into (41), we get Write Γ ≡    11  = (Γ  ), where Γ is partitioned conforming with that of Λ in (29).Then we observe that When  ̸ = , (43) can be rewritten as in which ⊗ stands for the Kronecker product and vec stands for the column vectorization of a matrix.Because  ̸ = , and assumption (A2), we know that (Λ   ) ⊗  +  ⊗ (Λ   ) is nonsingular; therefore vec(Γ  ) = 0, so Γ  = 0. Now we discuss the structures of matrices Γ  ,  = 1, . . ., ℓ 3 , which are skew-symmetric.For simplicity, we denote Γ  by Γ  .Then we need to solve Since Λ  has the form in (30), we can easily compute that the general solution of (45) has the form Thus, the general solution of the homogeneous equation (41) has the form with Γ defined in (35).This, together with (40), gives rise to the general solution of (39) Substituting (48) into (37) yields (v).
Sufficiency.From the description of (i)-(vi), we can obtain that (33) holds.
Remark 11.When  = , by Theorem 10, the solution of the Problem 1 with  = 0 is given by where with   > 0 and   =    which can be arbitrarily chosen.

Particular Solution with 𝐷 = 0 and 𝐾 < 0
In practice, the matrix  in the Problem 1 with  = 0 is sometimes required to be symmetric negative definite [5].In this section, we will apply Theorem 10 to construct such a solution.We first prove the following lemma.
Lemma 12.For any given matrix defined in (35), we can construct a symmetric positive definite matrix  11 so that  11 defined in Theorem 10 is symmetric negative definite.
Proof.Since  = Λ −1 , it is easy to see that  11 in Theorem 10 is symmetric negative definite if and only if the matrix is symmetric negative definite.By the assumption (A2), we can first construct a symmetric positive definite matrix M so that − M + ΓΛ < 0. Then we use M to construct the desired  11 .
From ( 35) and (30), we denote with Here   and   are arbitrary real numbers.Take Furthermore, Using Lemma 12, we can construct a particular solution to the Problem 1 with  = 0,  < 0 as follows.
Algorithm 13.An algorithm for solving the Problem 1 with  = 0,  < 0 is proposed as follows.
Remark 14.When  = , we only need to choose M by ( 54)-(56), and compute   by (57), that is,  11 is the whole   ; then use the same method described in Remark 8, we can obtain the particular solution of the Problem 1 with  = 0 and  < 0.

Numerical Examples
In this section, we present two numerical examples to illustrate the solutions constructed in Sections 2 and 4, respectively.For presentation, we report all numbers in 5 significant digits only, though all calculations are carried out in full precision.
Example 1.In this example, we use Algorithm 9 to construct the general solution of the Problem 1.The partially prescribed eigeninformation (Λ, ) ∈ R 4×4 × R 6×4 as in ( 2)-( 3) is given by the following eigenvalues and eigenvectors, which are from [12]: It is easy to check that the matrix pair (Λ, ) satisfy the assumptions (A1) and (A2).According to Algorithm 9, by randomly choosing It is easy to check that the matrix pair (Λ, ) satisfy the assumptions (A1) and (A2).According to Algorithm 13, by randomly choosing and the numerical results are shown in Table 2.This shows that Algorithm 13 to construct the particular solution of the Problem 1 with  = 0 and  < 0 is effective.

Conclusions
In this paper, we first use techniques involving matrix decompositions to derive an expression of the general solution to the question, for a set of given ( ≤ ) pairs of complex numbers and vectors (closed under conjugation), under assumptions (A1) and (A2).Then, with the special properties  = 0 and  < 0, we construct a particular solution.Numerical results illustrate these solutions.For another case of  > , it is rather complex, and the proof method in Theorem 5 seems not to be used directly to find a solution of Problem 1. Fortunately, for the damped nongyroscopic system, Cai et al. [13] solved the QIEP with 2 ≥  >  given eigenpairs.However, case  >  has never been discussed in the literature for damped gyroscopic system.It might be an interesting research and needs further investigation.
It is easy to check that  is symmetric positive definite,  is symmetric negative definite, and  is skew-symmetric.We define the residual as res (  , x  ) =      ( 2   +    + ) x      2 ,