Yunfeng Cai · Yuen-Cheng Kuo · Wen-Wei Lin · Shu-Fang Xu
Quadratic pencil solutions to the
quadratic inverse eigenvalue problem
Received: date / Revised: date
Abstract In this paper, we are interested in the study of solvability of the quadratic inverse eigenvalue problem (QIEP) of dimension n. Let k∗= (1 +
√
1 + 8n)/2 and 0 ≤ k < k∗, and for m := n + k prescribed eigenpairs
{(λj, xj)}mj=1, we prove that, generically, there is a constructible nonsingular
symmetric quadratic pencil solution Q(λ)≡ λ2M +λC +K to the QIEP such
that Q(λj)xj = 0 (j = 1, . . . , m). If k∗ ≤ k ≤ n, we show that, generically,
all symmetric quadratic pencil solutions are singular. We also derive the dimension of the solution subspace of the QIEP for both cases. Furthermore, we develop an algorithm for finding a symmetric positive definite M for the QIEP if it exists.
Keywords Quadratic inverse eigenvalue problem, symmetric quadratic pencil
Yunfeng Cai
LMAM, School of Mathematical Sciences, Peking University, Beijing, 100871, China E-mail: [email protected]
Yuen-Cheng Kuo
Department of Applied Mathematics, National University of Kaohsiung, Kaohsi-ung, 811, Taiwan
E-mail: [email protected] Wen-Wei Lin
Department of Mathematics, National Tsinghua University, Hsinchu, 300, Taiwan E-mail: [email protected]
Shu-Fang Xu
LMAM, School of Mathematical Sciences, Peking University, Beijing, 100871, China E-mail: [email protected]
1 Introduction
We consider the quadratic eigenvalue problem (QEP)
Q(λ)x≡ (λ2M + λC + K)x = 0, (1.1)
where M = M⊤, C = C⊤ and K = K⊤ are real symmetric n× n matrices
with M being nonsingular. The scalar λ∈ C and the nonzero vector x ∈ Cn
are called, respectively, the eigenvalue and the eigenvector corresponding to λ. The quadratic matrix polynomial Q(λ) in (1.1) is generally called a nonsingular quadratic pencil. As M is nonsingular, it is well known that Q(λ) has 2n finite eigenvalues.
QEPs appear in many applications. Some early theoretical results appear in Lancaster [12] and Gohberg, Lancaster and Rodman [7,8]. A good survey of applications, mathematical properties and variety of numerical algorithms for the QEP can be found in Tisseur and Meerbergen [17].
There are two aspects of the QEP (1.1), namely the direct problem and the inverse problem, associated with a second-order vibrating system
M ¨v + C ˙v + Kv = f (t). (1.2) Both are worthy of consideration. It is well known that if v(t) = xeλt
repre-sents a fundamental solution to (1.2), then (λ, x) must solve (1.1). The direct problem analyzes and computes the spectral (eigen-)information, hence de-ducing the dynamical behavior of the system from a priori known physical parameters such as mass, elasticity, inductance and capacitance. The in-verse problem determines or estimates the parameters of the system from its observed or expected eigen-information. Both problems are of significant importance in application. A quadratic inverse eigenvalue problem (QIEP) can be formulated as follows :
(QIEP) Given m (n < m ≤ 2n) prescribed eigenpairs in matrix form (Λ, X) ∈ Rm×m× Rn×m, where Λ is in real Jordan canonical form with
at most 2× 2 blocks along the diagonal and X ∈ Rn×m represents the
“eigen-vector matrix”, find a nontrivial quadratic pencil Q(λ) = λ2M + λC + K with
M = M⊤, C = C⊤ and K = K⊤∈ Rn×n such that
M XΛ2+ CXΛ + KX = 0 n×m. (1.3) Here a 2× 2 block αj βj −βj αj
and the corresponding columns [xjR, xjI] in X
represent or store the complex conjugate pairs of eigenvalues αj ± iβj and
the corresponding eigenvectors xjR± ixjI.
Throughout this paper, we make the following assumptions on (Λ, X): A1. Let m := n + k > n, ¯k := n− k, and
Λ = diag(λ[2]1 , . . . , λ[2]ℓ ; λ2ℓ+1, . . . , λm), (1.4a) where λj∈ R, λ[2]j = αj βj −βj αj ; αj, βj∈ R, βj > 0, (1.4b)
and
X = [x1R, x1I, . . . , xℓR, xℓI; x2ℓ+1, . . . , xm]∈ Rn×m. (1.4c)
A2. All m eigenvalues of Λ are distinct, rank(X) = n, and rank
X XΛ
= m. To solve the QIEP with m ≤ n, a general solution (M, C, K) with the m prescribed eigenpairs and other particular solutions with additional eigen-structures have been derived in [10]. For the model updating problem with a given quadratic pencil (M0, C0, K0), a general solution replacing ¯k (¯k < n)
original eigenpairs while preserving the remaining 2n− ¯k eigenpairs has been solved in [4]. The second condition above is known as the no spill-over phe-nomenon [3]. Recently, a series of papers [11,13,14,16] study the solvability of QIEPs when complete spectral information is given. That is, for m = 2n, the QIEP with M and K being symmetric positive definite is solvable, pro-vided that an admissible set of the complete eigen-information satisfies a very special structure with a one to one correspondence between Jordan pairs and structure preserving similarities. Other types of inverse QEPs, known as model updating problems and partial pole assignments, can be found in [6, 9,10,18] and [2,5,15], respectively.
We now consider the algebraic linear system (1.3) for the triplet (M, C, K). There are nm equations in 3n(n + 1)/2 unknowns in the homogeneous linear system (1.3). If m < 3(n + 1)/2, then the system (1.3) is under-determined and it is intuitively true that the QIEP is solvable; otherwise the QIEP has only the trivial solution. However, this assertion is made without considering any linear dependence in (1.3). Moreover, it neglects whether the nontrivial solution (M, C, K) forms a nonsingular quadratic pencil. We note a singular quadratic pencil Q(λ), with det(Q(λ))≡ 0, is impractical.
The main purpose of this paper is to prove that the generic solvability of the QIEP is characterized by the number k∗≡ (1+√1 + 8n)/2 — generically,
when 0≤ k < k∗, the solutions to the QIEP form a subspace of dimension 1
2(n− k)(n − k + 1) + n + 1
2k(1− k); when k∗≤ k ≤ n, the solutions form a
subspace of dimension 1
2(n− k)(n − k + 1). Moreover, when 0 ≤ k < k∗, there
exists a triplet (M, C, K) which forms a nonsingular quadratic pencil. On the contrary, when k∗ ≤ k ≤ n, all quadratic pencil solutions are singular.
This paper is organized as follows. In Section 2, we characterize the solv-ability by nonsingular/singular quadratic pencil solutions for the QIEP. In Section 3, we propose a simple algorithm to construct a symmetric positive definite M for the QIEP if it exists. Finally in Section 4, we present selected numerical results to illustrate our main results.
2 Solvability of QIEPs
To solve the QIEP, we first show two equivalence conditions for the solu-tion of (1.3). We derive the general solusolu-tion to the QIEP represented in a parameterized form.
Lemma 1 Given a matrix pair (Λ, X)∈ Rm×m× Rn×m satisfying A1–A2
with m≥ n. There exist real symmetric matrices M, C and K satisfying the equation (1.3) if and only if
X⊤CX =−(Λ⊤X⊤M X + X⊤M XΛ) + D, (2.1)
X⊤KX = Λ⊤X⊤M XΛ− Λ⊤D (2.2)
for some D in the form D = diag ξ1 η1 η1−ξ1 , . . . , ξℓ ηℓ ηℓ −ξℓ ; ξ2ℓ+1, . . . , ξm (2.3) with ξi and ηi being arbitrarily real numbers and satisfies
DΛ = Λ⊤D. (2.4)
Proof (N ecessity) Suppose there are real symmetric matrices M , C and K satisfying (1.3). We have [X⊤, Λ⊤X⊤] C M M 0 X XΛ = [X⊤CX + X⊤M XΛ + Λ⊤X⊤M X]≡ D, (2.5) and [X⊤, Λ⊤X⊤] −K 0 0 M X XΛ = [−X⊤KX + Λ⊤X⊤M XΛ]≡ E. (2.6)
Taking the transpose of (1.3), we get
Λ2⊤X⊤M + Λ⊤X⊤C + X⊤K = 0
m×n. (2.7)
With similar techniques as in [5], eliminating the K-term in (1.3) and (2.7) by X⊤× (1.3) − (2.7) × X⊤, and adding Λ⊤X⊤M XΛ to both sides, we get
(X⊤M XΛ + Λ⊤X⊤M X + X⊤CX)Λ
= Λ⊤(X⊤M XΛ + Λ⊤X⊤M X + X⊤CX). (2.8)
Since all eigenvalues of Λ are distinct, it is easily seen from (1.4a), (1.4b) and the Lyapunov equation (2.8) that D in (2.5) has the form in (2.3) and satisfies DΛ = Λ⊤D.
Similarly, we eliminate the C-term in (1.3) and (2.7) by Λ⊤X⊤× (1.3) −
(2.7)× XΛ and obtain
(Λ⊤X⊤M XΛ− X⊤KX)Λ = Λ⊤(Λ⊤X⊤M XΛ− X⊤KX). (2.9)
The Lyapunov equation (2.9) then implies that E satisfies EΛ = Λ⊤E.
Post-multiplying (2.5) by Λ and with the help of (1.3), we obtain
From (2.5), (2.6) and (2.10), (2.1) and (2.2) follow immediately. (Suf f iciency) Consider
X⊤(M XΛ2+ CXΛ + KX)
= X⊤M XΛ2+ X⊤CXΛ + X⊤KX (2.11)
= X⊤M XΛ2
− (Λ⊤X⊤M X + X⊤M XΛ)Λ + DΛ + Λ⊤X⊤M XΛ− Λ⊤D
= 0m×m.
Equation (2.11) and the fact that X is of full row rank, we arrive at M XΛ2+
CXΛ + KX = 0n×m. ⊓⊔
From the QR-factorization of X⊤, we have
XQ = [R, 0], (2.12)
where Q ∈ Rm×m is orthogonal, and R ∈ Rn×n is lower triangular with
positive diagonal entries. Partition Q by Q = [Q1, Q2], where Q1 ∈ Rm×n
and Q2∈ Rm×k. Then from (2.12), it follows that
X = RQ⊤
1 and XQ2= 0n×k. (2.13)
Denoting
Mr:= R⊤M R, Cr:= R⊤CR, Kr:= R⊤KR, (2.14)
we are going to prove the following Lemmas.
Lemma 2 Given (Λ, X)∈ Rm×m× Rn×m satisfying A1–A2 with m ≥ n.
Let Mr, Cr and Kr be defined as in (2.14). Then there are real symmetric
matrices M , C and K satisfying (1.3) if and only if Cr=− (Q⊤ 1Λ⊤Q1)Mr+ Mr(Q1⊤ΛQ1)+ Q⊤1DQ1, (2.15a) Kr= (Q⊤1Λ⊤Q1)Mr(Q1⊤ΛQ1)− Q⊤1Λ⊤DQ1, (2.15b) Mr(Q⊤1ΛQ2) = Q⊤1DQ2 (2.15c)
for some D∈ D(Λ,X), where
D(Λ,X):=D∈ Rm×m D is of the form (2.3) and Q⊤
2DQ2= 0 . (2.16)
Proof (N ecessity) Using (2.1), (2.13) and (2.14), we have Q⊤ 1DQ1Q⊤1DQ2 Q⊤ 2DQ1Q⊤2DQ2 = Q⊤DQ = Q⊤(X⊤CX + Λ⊤X⊤M X + X⊤M XΛ)Q = Cr+ (Q⊤1Λ⊤Q1)Mr+ Mr(Q⊤1ΛQ1) Mr(Q⊤1ΛQ2) (Q⊤ 2Λ⊤Q1)Mr 0n×k . (2.17) Then it follows that
Cr=− (Q⊤ 1Λ⊤Q1)Mr+ Mr(Q⊤1ΛQ1) + Q⊤ 1DQ1, (2.18a) Mr(Q⊤1ΛQ2) = Q⊤1DQ2, (2.18b) Q⊤2DQ2= 0. (2.18c)
Similarly, using (2.2), (2.13) and (2.14), we have
Kr= (Q⊤1Λ⊤Q1)Mr(Q1⊤ΛQ1)− Q⊤1Λ⊤DQ1, (2.19a)
Q⊤1Λ⊤DQ2= (Q⊤1Λ⊤Q1)Mr(Q⊤1ΛQ2), (2.19b)
Q⊤
2Λ⊤DQ2= (Q⊤2Λ⊤Q1)Mr(Q⊤1ΛQ2). (2.19c)
Notice that the last two equalities in (2.18) satisfy (D− Q1MrQ⊤1Λ)Q2= 0,
which implies the last two equalities in (2.19) immediately. Combining (2.18) and (2.19), we get (2.15).
(Suf f iciency) It follows from (2.18a), (2.19a) and Lemma 1. ⊓⊔ Lemma 3 The matrix B≡ Q⊤
1ΛQ2∈ Rn×k is of full column rank.
Proof Since
X XΛ
is of full column rank by A2, the matrix X XΛ Q = XQ XQQ⊤ΛQ = R 0 RQ⊤ 1ΛQ1RQ⊤1ΛQ2
is of full column rank. So, the matrix B ≡ Q⊤
1ΛQ2 is of full column rank
because R is nonsingular. ⊓⊔
Lemma 4 Let B ≡ Q⊤
1ΛQ2. For any D ∈ D(Λ,X), the matrix equation
MrB = Q⊤1DQ2 for Mr is solvable. Moreover, Mris given by
Mr= U B⊤Q⊤ 1DQ2Q⊤2DQ1Z Z⊤Q⊤ 1DQ2 W U⊤, (2.20)
where W⊤= W ∈ R(n−k)×(n−k) is arbitrary, and U = [B(B⊤B)−1, Z] with
Z∈ Rn×(n−k) satisfying that B⊤Z = 0 and Z⊤Z = I n−k.
Proof By Lemma 3, U is well-defined and satisfies [B, Z]⊤U = I.
Conse-quently, MrB = Q⊤1DQ2 can be rewritten as
U−1M rU−⊤U⊤B = U−1Q⊤1DQ2, or U−1M rU−⊤ I 0 = B⊤Q⊤ 1DQ2 Z⊤Q⊤ 1DQ2 . (2.21) Note that B⊤Q⊤ 1DQ2= Q⊤2Λ⊤Q1Q⊤1DQ2 = Q⊤2Λ⊤(I− Q2Q2⊤)DQ2= Q⊤2Λ⊤DQ2 (2.22)
is symmetric, since D ∈ D(Λ,X) implies that Λ⊤D = DΛ. From (2.21), we
can deduce that MrB = Q⊤1DQ2 is solvable and its solution is given by
From (2.20) and the definition of U , it follows that
Mr=B(B⊤B)−1Q⊤2Λ⊤DQ2(B⊤B)−1B⊤+ ZZ⊤Q⊤1DQ2(B⊤B)−1B⊤
+ B(B⊤B)−1Q⊤2DQ1ZZ⊤+ ZW Z⊤. (2.23)
From (2.13), we see that Q1 and R are uniquely determined by X, B ≡
Q⊤
1ΛQ2 and Q2 are uniquely determined up to a right orthogonal
transfor-mation. Therefore, B(B⊤B)−1Q⊤
2 in (2.23) is uniquely determined by X and
Λ. Furthermore, the choice of the orthonormal basis Z for the null space of B⊤ in (2.23) is independent of the representation of M
r. Thus, we conclude
that Mr in (2.23) is only parameterized by W = W⊤ ∈ R(n−k)×(n−k) and
D∈ D(Λ,X).
With Lemmas 2 and 4, we have finally proved our first main result which completely characterizes the general solution to the QIEP.
Theorem 1 Given (Λ, X) ∈ Rm×m× Rn×m satisfying A1–A2 with m =
n+k≥ n. Let V = Q1R−1and U be defined as in Lemma 4. Then the general
solution to the QIEP can be represented in the following parameterized forms in terms of W and D: M = R−⊤U B⊤Q⊤ 1DQ2Q⊤2DQ1Z Z⊤Q⊤ 1DQ2 W U⊤R−1, (2.24a) C = V⊤DV − V⊤Λ⊤X⊤M − MXΛV, (2.24b) K = V⊤Λ⊤X⊤M XΛV − V⊤Λ⊤DV, (2.24c)
where W⊤ = W ∈ R(n−k)×(n−k) is arbitrary, D is of the form (2.3) and
satisfies Q⊤
2DQ2 = 0, (i.e., D ∈ D(Λ,X)). Here Q2 is an orthonormal basis
for the null space of X.
Remark 1 (i) From Theorem 1 we see that if (W1, D1) and (W2, D2)
charac-terize two solutions to the QIEP, where W1= W1⊤, W2= W2⊤∈ R(n−k)×(n−k),
and D1, D2∈ D(Λ,X), then any linear combination of them also forms a
so-lution to the QIEP. Thus, the dimension of the soso-lution space of the QIEP is
1
2(n− k)(n − k + 1) + d, where d = dim D(Λ,X).
(ii) In Theorem 2.1 of [10] and Theorem 1, we construct general solutions (M, C, K) to the QIEP for k ≤ 0 and k ≥ 0. In fact, k = 0 plays a pivot role in both two cases. The technique, employed in [10] and Theorem 1, is to use the QR factorization of X ∈ Rn×(n+k) to reduce the quadratic pencil to
“invertible” systems from which parameters can be introduced. The differ-ence is that in the case when k≤ 0, the QR factorization assumes the form X = Q
R
0
, where R is (n+k)×(n+k) nonsingular, while in the current case when k≥ 1, the QR factorization assumes X = [R, 0]Q⊤, where R is n× n
nonsingular. More precisely, for k≤ 0, we have the partitions C =
C11C12
C⊤ 12C22
and K = K11K12 K⊤ 12K22
, with C11, K11 ∈ R(n+k)×(n+k) (see [10]); for k ≥ 1,
we have M = R−⊤U M11M12 M⊤ 12M22 U⊤R−1 with M 22 ∈ R(n−k)×(n−k) (c.f.
(2.24a)). The constructions of C11, K11 and K12 (for k≤ 0), and C and K
(for k ≥ 1) are the same, determined by D and M using the orthogonality relations in (2.5) and (2.6). Furthermore, for k≤ 0, M, C22 and K22 is
ar-bitrary and symmetric, and C12 is arbitrary. However, for k≥ 1, M cannot
be arbitrary. From (2.24a), we see that only the submatrix M22is arbitrary
and symmetric, and the other submatrices M11 and M12 are determined by
D using Lemmas 2–4. ⊓⊔
As an application of Theorem 1, we next characterize when the solution to the QIEP is (non-)singular.
Theorem 2 If d = dim D(Λ,X)= 0, the QIEP has only solutions of singular
quadratic pencils which form a subspace of dimension 12(n− k)(n − k + 1). Proof If d = dim D(Λ,X)= 0, D(Λ,X)={ 0m×m} and k ≥ 1. Therefore, by
Theorem 1, the general solution of QIEP becomes
M = R−⊤ZW Z⊤R−1, (2.25a)
C =−Y⊤M− MY, (2.25b)
K = Y⊤M Y, (2.25c)
where Y = XΛV . Since W = W⊤ ∈ R(n−k)×(n−k), it is easily seen form
(2.25a) that M is singular. From (2.25b)–(2.25c), we obtain Q(λ) = λ2M + λC + K = (λI− Y )⊤M (λI− Y ).
Therefore, det (Q(λ))≡ 0, i.e., Q(λ) is a singular quadratic pencil. ⊓⊔ Remark 2 Notice that D ∈ D(Λ,X) means that D is in the form (2.3) and
satisfies
Q⊤
2DQ2= 0, (2.26a)
which is a homogeneous linear system for D with 12k(k + 1) equations in (n + k) unknowns. Furthermore, the equation (2.26a) can also be rewritten as a homogeneous linear system
AQ2d = 0, (2.26b)
where AQ2is a suitable
1
2k(k+1)×(n+k) real matrix and d = (ξ1, η1, . . . , ξℓ, ηℓ,
ξ2ℓ+1, . . . , ξm)⊤. So, generically, the equation (2.26) has only the trivial
so-lution D = 0, provided that
n + k≤1 2k(k + 1). (2.27) That is, if k≥ k∗≡ 1 +√1 + 8n 2 , (2.28)
By Theorem 2, we conclude that
Corollary 1 Given (Λ, X) ∈ R(n+k)×(n+k) × Rn×(n+k) satisfying A1–A2
with k∗ ≤ k ≤ n, the QIEP, generically, has only solutions of singular
quadratic pencils which form a subspace of dimension 1
2(n− k)(n − k + 1).
Remark 3 (i) From here on, “generically” means that “the statement” holds for almost any (Λ, X) satisfying A1–A2. That is, the set of those (Λ, X) in which “the statement” holds is open and dense. It is easily seen that (2.26) has only the trivial solution if and only if rank(AQ2) = n + k. In fact, the set
of those AQ2 with rank(AQ2) < n + k is of measure zero.
(ii) If we now choose (Λ, X)∈ R(n+k)×(n+k)× Rn×(n+k) as an eigenmatrix
pair from a prescribed nonsingular quadratic pencil, then (2.24a) implies that D cannot be a zero matrix. Consequently, (2.26) always has a nontrivial solution D for k≥ k∗. Therefore, the set of those (Λ, X) chosen this way is of measure zero because of rank(AQ2) < n + k. In this case, the dimension of
quadratic pencil solutions to the QIEP is at least 1
2(n−k)(n−k+1)+1, which
is the sum of the dimension (= 12(n− k)(n − k + 1)) of singular quadratic pencil solutions and the dimension (≥ 1) of nonsingular quadratic pencil
solutions to the QIEP. ⊓⊔
In the following, we shall study the nonsingular quadratic pencil solutions to the QIEP.
Theorem 3 The QIEP has a solution (M, C, K) with M being nonsingular if and only if there exists a D∈ D(Λ,X) such that the matrix Q⊤1DQ2 is of
full column rank. In this case, the QIEP has a nonsingular quadratic pencil solution.
Proof If the QIEP has a solution (M, C, K) with a nonsingular M , then (2.24a) implies [B, Z]⊤Q⊤1DQ2= B⊤Q⊤ 1DQ2 Z⊤Q⊤ 1DQ2
is of full column rank for some D ∈ D(Λ,X). So there exists a D ∈ D(Λ,X)
such that Q⊤
1DQ2is of full column rank, since [B, Z] is nonsingular.
Conversely, if there exists a D∈ D(Λ,X)such that the matrix Q⊤1DQ2 is
of full column rank, then B⊤Q⊤ 1DQ2 Z⊤Q⊤ 1DQ2 = [B, Z]⊤Q⊤ 1DQ2
is of full column rank. Since B⊤Q⊤
1DQ2is symmetric by (2.24a), there exists a k× k orthogonal
matrix P1 such that
P⊤ 1 (B⊤Q⊤1DQ2)P1= Σs0 0 0 , (2.29)
where Σsis an s× s nonsingular matrix. Denote
E1= P1 0 0 Ik¯ ,
then the matrix E⊤ 1 B⊤Q⊤ 1DQ2 Z⊤Q⊤ 1DQ2 P1= Σ0 0s0 Z⊤Q⊤ 1DQ2P1
has full column rank. Partition
Z⊤Q⊤1DQ2P1= [Z1⊤, Z2⊤],
where Z1∈ Rrׯk, then Z2⊤is of full column rank. Thus there exists a matrix
P⊤ 2 ∈ R
¯
k×k and a nonsingular matrix P⊤ 3 ∈ R ¯ kׯk such that Ik 0 P⊤ 2 Ik¯ Σ0s Z⊤ 1 = Σ0s 0 and P⊤ 3 Z2⊤= Is¯ 0 , where ¯s = k− s. Now let P = E1 Ik P2 0 I¯k Ik 0 0 P3 . We then have P⊤ B⊤Q⊤ 1DQ2Q⊤2DQ1Z Z⊤Q⊤ 1DQ2 W P = Σs0 0 0 0 0 I¯s0 0 Is¯ 0 0 Wc ≡M ,c (2.30)
where cW ∈ R¯kׯk is an arbitrarily symmetric matrix. It is easily seen that
there exists a symmetric matrix cW such that cM is nonsingular. Therefore, from (2.30) and (2.24), the QIEP has a solution (M, C, K) with M being
nonsingular. ⊓⊔
Remark 4 If 0≤ k < k∗ (i.e., n + k > 12k(k + 1)), we have
d = dim D(Λ,X)≥ (n + k) −
1
2k(k + 1) > 0.
In this case, generically, there exists a D ∈ D(Λ,X) such that the matrix
Q⊤
1DQ2 is of full column rank, and so, by Theorem 3, the QIEP has a
nonsingular quadratic pencil solution. So in this case, generically, we have d = dim D(Λ,X)= (n + k)− 1 2k(k + 1) = n +1 2k(1− k), ⊓ ⊔ By Remarks 2 and 4, we conclude that
Corollary 2 Given (Λ, X) ∈ R(n+k)×(n+k) × Rn×(n+k) satisfying A1–A2
with 0 ≤ k < k∗, the QIEP, generically, has a nonsingular quadratic pencil
solution and the solutions of quadratic pencils form a subspace of dimension 1 2(n− k)(n − k + 1) + d = 1 2(n− k)(n − k + 1) + n + 1 2k(1− k). Theorem 4 The QIEP has a solution (M, C, K) with M being symmetric positive definite if and only if there exists a D∈ D(Λ,X)such that the matrix
Q⊤
2Λ⊤DQ2 is symmetric positive definite.
Proof If the QIEP has a solution (M, C, K) with M being symmetric positive definite, (2.22) and (2.24a) imply
Q⊤
2Λ⊤DQ2= B⊤Q⊤1DQ2
is symmetric positive definite for some D∈ D(Λ,X).
Conversely, if there exists a D∈ D(Λ,X)such that the matrix Q⊤2Λ⊤DQ2
is symmetric positive definite, then from (2.22), it follows that B⊤Q⊤
1DQ2= Q⊤2Λ⊤DQ2
is symmetric positive definite, and Ik 0 −M21M11−1 I¯k M11M21⊤ M21 W Ik 0 −M21M11−1I¯k ⊤ = M 11 0 0 fW = fM , where M11 = B⊤Q⊤1DQ2, M21 = Z⊤Q⊤1DQ2, fW ∈ R ¯ kׯk is an arbitrarily
symmetric matrix. It is easily seen that there exists a symmetric matrix fW such that fM is symmetric positive definite. Therefore, by Theorem 1, the QIEP has a solution (M, C, K) with M being symmetric positive definite.
⊓ ⊔ QIEP with complete eigen-information.
Suppose we are given the complete eigen-information (Λ, X)∈ R2n×2n×
Rn×2n with
X XΛ
being nonsingular. For B = Q⊤
1ΛQ2, Theorem 1 (with
m = 2n) implies that the general solution of the QIEP is given by M = R−⊤B−⊤(Q⊤ 2Λ⊤DQ2)B−1R−1, (2.31a) C = V⊤DV − V⊤Λ⊤X⊤M− MXΛV, (2.31b) K = V⊤Λ⊤X⊤M XΛV − V⊤Λ⊤DV =−Ψ−⊤(Q⊤ 2DΛ−1Q2)Ψ−1, (2.31c)
where D∈ D(Λ,X) is arbitrary, V = Q1R−1 and Ψ = XΛ−1Q2. The second
equality in (2.31c) can be derived from (2.6), (2.10) and (2.13).
In [13] and [14], alternative solvability conditions and general solutions for the QIEP were presented. We now compare the general solution in (2.31) and [13,14]. In [14], the complete eigen-information (J, XL) is stored in
J = diagΛ1, U2, U3, Λ1
where Λ1 = diag{α1+ ιβ1, . . . , αn−r+ ιβn−r}, U2 and U3 ∈ Rr×r are real
diagonal, X1= [x1R+ιx1I, . . . , x(n−r)R+ιx(n−r)I]∈ Cn×(n−r)and XR2, XR3 ∈
Rn×r. Let P = 0 0 0 In−r 0 Ir 0 0 0 0 −Ir 0 In−r 0 0 0 . (2.33) If the conditions (P J)H= P J, X LP XLH= 0 (2.34)
hold, the nontrivial symmetric triplet (M, C, K) for the QIEP is solvable [14]. As in (1.4), we define Λ = diagnλ[2]1 , . . . , λ [2] n−r; U2, U3 o ∈ R2n×2n, (2.35a) X = [x1R, x1I, . . . , x(n−r)R, x(n−r)I; XR2, XR3]∈ Rn×2n, (2.35b) D = D−1= diag 1 0 0−1 , . . . , 1 0 0−1 ; Ir,−Ir ∈ R2n×2n, (2.35c) where λ[2]j = αj βj −βj αj
, for j = 1, . . . , n− r. It is easy to find a unitary transformation Ω such that XLΩ = X, ΩHP Ω = D and ΩHJΩ = Λ. From
(2.34) and (2.35c), we have
Λ⊤D = DΛ, XD−1X⊤= 0. (2.36)
From (2.13), it follows that Q⊤
1D−1Q1= 0 and Q⊤ 1 Q⊤ 2 D−1[Q 1, Q2] = 0 (Q⊤ 2DQ1)−1 (Q⊤ 1DQ2)−1 (Q⊤1DQ2)−1(Q⊤1DQ1)(Q⊤2DQ1)−1 . (2.37) Taking the inverse of (2.37), we get
Q⊤ 1 Q⊤ 2 D[Q1, Q2] = Q⊤ 1DQ1Q⊤1DQ2 Q⊤ 2DQ1 0 . (2.38)
From (2.36) and (2.38), the matrix D in (2.35a) satisfying Q⊤
1DQ2is
nonsin-gular, Λ⊤D = DΛ and Q⊤
2DQ2= 0. By Theorem 3, the QIEP has a solution
(M, C, K) with M being nonsingular.
We now study the positivity of M and K. We consider the case of real spectrum. Suppose we are given
where U2, U3 ∈ Rn×n are real diagonal and XR3 = XR2Θ ∈ Rn×n with Θ
being an n× n orthogonal matrix. Then, from (2.39), (2.35c) and (2.13), it is easily seen that
Q1= √1 2 Θ⊤, I n ⊤ , Q2=√1 2 −Θ⊤, I n ⊤ , (2.40a) R =√2XR2Θ, D = diag{In,−In} , (2.40b) and B = Q⊤1ΛQ2=−1 2(Θ ⊤U 2Θ− U3). (2.40c) If max uj∈diag(U3) (uj) < min uk∈diag(U2) (uk) (2.41)
holds (see [14]), then from (2.40a) and (2.40b), we have
Q⊤ 2Λ⊤DQ2= 1 2(Θ ⊤U 2Θ− U3) > 0 (2.42a) (2.42b) and −Q⊤ 2DΛ−1Q2= 1 2(−Θ ⊤U−1 2 Θ + U3−1) > 0. (2.42c)
It follows from (2.31a), (2.31c) and (2.42) that M and K are symmetric positive definite which coincides with the result in [14]. Furthermore, if
U22≤ ΘU32Θ⊤, (2.43)
(see also [14]) then (2.40), (2.31a) and (2.31b) imply C =V⊤DV − V⊤Λ⊤X⊤M − MXΛV =− R−⊤(Θ⊤U 2Θ + U3)(Θ⊤U2Θ− U3)−1R−1 (2.44) − R−⊤(Θ⊤U 2Θ− U3)−1(Θ⊤U2Θ + U3)R−1 =− R−⊤(Θ⊤U 2Θ− U3)−1(Θ⊤U22Θ− U32)(Θ⊤U2Θ− U3)−1R−1≥ 0.
For the case with complex eigenvalues, the positivity of M , C and K derived in [13] can also be established using a similar derivation. We omit the details here.
3 Numerical algorithm
In many applications, it is more practical to seek a symmetric positive definite M for the QIEP than a nonsingular one. Let {D1, . . . , Dr} be a basis for
D(Λ,X), where r ≡ dim D(Λ,X) ≥ 1. Inspired by Theorems 1 and 4, it is natural to ask if there exist scalars a1, . . . , ar such that
M11(a) = r
X
i=1
ai Q⊤2Λ⊤DiQ2 (3.1)
is symmetric positive definite. If so, one can choose a suitable W in (2.24a) so that M is symmetric positive definite.
To this end, we denote
Sn={A ∈ Rn×n| A⊤ = A},
Sn+={A ∈ Sn| A is positive definite}, Sn
0,+={A ∈ Sn| A is positive semidefinite}.
We define a linear transformation vec : Sn → Rn(n+1)2 by
vec(A) = 1 √ 2a11, a21, . . . , an1; 1 √ 2a22, a22, . . . , an2; . . . , 1 √ 2ann ⊤ , (3.2) where A = [aij]ni,j=1. For convenience, we multiply aii by a factor √12, i =
1, . . . , n. We then have kAkF = q tr(A⊤A) =√2kvec(A)k2. (3.3) Let Hi= Q⊤2Λ⊤DiQ2, i = 1, . . . , r, (3.4) and H = [vec(H1), . . . , vec(Hr)]∈ R n(n+1) 2 ×r, H = span(H). (3.5)
The following lemma is useful for our algorithm for finding {a1, . . . , ar} so
that M11(a) in (3.1) is symmetric positive definite.
Lemma 5 If 06= B ∈ Sn
0,+ and vec(B)∈ H⊥, then vec(Sn+)∩ H = ∅.
Proof Suppose u∈ vec(Sn
+)∩ H . Then A = vec−1(u) is symmetric positive
definite. Let A = LL⊤ be the Cholesky decomposition of A. Since vec(B)∈
H⊥ and vec(A)∈ H , we obtain
0 = vec(A)⊤vec(B) = 1 2tr(AB) = 1 2tr(LL ⊤B) =1 2tr(L ⊤BL⊤), which contradicts B6= 0. ⊓⊔
To find the orthonormal bases for H and H⊥, we compute the QR-factorization H = U R 0
, where U is an n(n+1)2 ×n(n+1)2 orthogonal matrix. Partition U = [U1, U2] with U1being 12n(n + 1)× r, we have H =span(U1),
H⊥=span(U
2). We now choose Z∈ Sn+withkZkF= 1 and set z = vec(Z).
We then project z orthogonally onto span(U1) and span(U2) to obtain
u = U1U1⊤z∈ span(U1),
v = U2U2⊤z∈ span(U2). (3.6)
Let A = vec−1(u) and B = vec−1(v). We see that if λmin(A) > 0, then a
solution is given by
[a1, . . . , ar]⊤= R−1(U1⊤z)
with the matrix in (3.1) satisfying
M11(a) = r X i=1 aiHi= vec−1 r X i=1 vec(Hi)ai !
= vec−1(Ha) = vec−1(u) = A≥ 0.
If λmin(B) ≥ 0, then by Lemma 5 there is no symmetric positive definite
solution for (3.1).
If λmin(A) ≤ 0 and λmin(B) < 0, we lift A and B back to Sn0,+ (as bA
and bB), which bA and bB solve minY ∈Sn
0,+kY − AkF and minY ∈Sn0,+kY − BkF,
respectively. Note that by the well-known Wielandt-Hoffman Theorem, if A = U⊤diag(λ
1, . . . , λn)U with U being orthogonal and λ1≥ . . . ≥ λp> 0≥
λp+1≥ . . . ≥ λn, then bA is given by
b
A = U⊤diag(λ
1, . . . , λp, 0, . . . , 0)U. (3.7)
Because Sn
0,+ is convex, we choose the arithmetic mean of bA and bB with
Frobenius norm one, or Znew:= ( bA + bB)/k bA + bBkF, and continue the above
process by setting znew = vec(Znew). A geometrical interpretation of this
procedure is illustrated in Fig. 3.1.
Based on these steps, we develop an algorithm to find a symmetric positive definite solution for (3.1).
Algorithm 3.1.
Input: H1, . . . , Hr∈ Sn as in (3.4), a maximal number imaxand a tolerance
T ol.
Output: Either{a1, . . . , ar} when the matrix in (3.1) is symmetric positive
definite, otherwise no solution. Set H = [vec(H1), . . . , vec(Hr)];
Compute a QR-factorization of H with H = U R 0 ; Set U1= U ( : , 1 : r) and U2= U ( : , r + 1 : n(n+1)2 );
vec(Sn 0,+) z u znew unew v ˆ u= vec(b A) ˆ v= vec(B)b H⊥ H
Fig. 3.1 Geometrical presentation for two steps of Algorithm 3.1.
Repeat: until stop, Compute
w1= U1⊤z, w2= U2⊤z;
u = U1w1, v = U2w2.
Set A = vec−1(u), B = vec−1(v);
If λmin(A) > 0, the solution is [a1, . . . , ar]⊤= R−1w1, stop;
else if λmin(B)≥ 0, no solution (by Lemma 5), stop;
else compute bA and bB by (3.7), respectively, and set bZ := 1
2( bA+ bB), bZ :=
b Z/k bZkF,
Ifk bZ− ZkF < T ol or i > imax, fails, stop;
else Z = bZ, z = vec(Z), i = i + 1. Go to Repeat.
Remark 5 The orthogonal projection step (as in (3.6)) of Algorithm 3.1 is borrowed from the idea for maximizing the minimal eigenvalue of a linear combination of symmetric matrices proposed by [1]. Lemma 5 guarantees, if λmin(B) ≥ 0 in some step of Algorithm 3.1, that there is no symmetric
positive definite solution M for the QIEP. In our numerical experience, Al-gorithm 3.1 converges well to a symmetric positive definite M , if it exists. However, because it is difficult to control the distance between the projec-tion vector and the lifting vector during the process, a complete theory of
4 Numerical results
For an arbitrarily given (Λ, X)∈ R(n+k)×(n+k)×Rn×(n+k)satisfying A1–A2
with 0≤ k ≤ n, we solve the QIEP, seeking a symmetric triplet (M, C, K) satisfying (1.3), via the homogeneous linear system
Λ2⊤X⊤⊗ I n, Λ⊤X⊤⊗ In, X⊤⊗ In I3⊗ Π⊤ vec(M )vec(C) vec(K) = 0, (4.1)
where “⊗ ” denotes the Kronecker product between two matrices, vec(X) = [x⊤ 1| . . . |x⊤n]⊤ with X = [x1| . . . |xn]∈ Rn×n, and Π = [e11, . . . , en1| . . . |e1n, . . . , enn] with e⊤ij = [0⊤n(i−1)| e⊤j| 0n⊤(j−i−1)| e⊤i | 0⊤n(n−j)]∈ R1×n 2 , (4.2)
for i, j = 1, . . . , n. Here ej∈ Rn denotes the jth column of In.
By Corollaries 1 and 2, we define d(k) := 1 2(n− k)(n − k + 1) + n + 1 2k(1− k), for 0 ≤ k ≤ k∗ 1 2(n− k)(n − k + 1), for k∗≤ k ≤ n . (4.3)
The homogeneous system (4.1) is under-determined and has a nontrivial solution if m≡ n + k < 3(n+1)2 . Let
m∗=
3ℓ + 1, for n = 2ℓ
3ℓ + 2, for n = 2ℓ + 1 (4.4) be the critical number for the under- or over-determined linear system of (4.1). For a given matrix pair (Λ, X)∈ R(n+k)×(n+k)×Rn×(n+k)(m≡ n+k ≤
m∗) representing m prescribed eigenpairs closed under complex conjugation,
we define
e(k) := 3
2n(n + 1)− n(n + k), for n ≤ n + k ≤ m∗
0, for m∗≤ n + k ≤ 2n . (4.5)
In the following example, we study the relation among d(k), e(k) and the dimension of the solution space for the QIEPs.
Example 1. Let n = 11. Then k∗ ≈ 5.217 by (2.28) and m∗ = 17 by (4.4).
Given (Λ, X)∈ R(n+k)×(n+k)×Rn×(n+k)where 0≤ k ≤ 11, we list e(k), d(k)
and dims (the dimensions of the solution subspaces for the QIEPs computed by (4.1)) in Table 4.1. Furthermore, we also compute rank(Q⊤
1DQ2) if there
is a nontrivial solution D∈ D(Λ,X) and set rank(Q⊤1DQ2) = 0 for a trivial
D. In Table 4.1 we see that the dimensions of the solution subspaces for the QIEPs coincide with d(k) in (4.3), which is consistent with Corollary 2 (for 0 ≤ k < k∗) and Corollary 1 (for k∗ ≤ k ≤ n). The dimensions coincide
with e(k) only when 0 ≤ k ≤ k∗ ≈ 5.217. For k > k∗ ≈ 5.217, we have
Table 4.1 e(k), d(k), dimensions and rank(Q⊤ 1DQ2), for 0 ≤ k ≤ 11. k 0 1 2 3 4 5 6 7 8 9 10 11 e(k) 77 66 55 44 33 22 11 0 0 0 0 0 d(k) 77 66 55 44 33 22 15 10 6 3 1 0 dims 77 66 55 44 33 22 15 10 6 3 1 0 rank(Q⊤ 1DQ2) ∅ 1 2 3 4 5 0 0 0 0 0 0
solution in general. However, the dimensions of the solution subspaces for (4.1) coincide with d(k), but not e(k), for 11 > k > k∗.
Example 2. Given an 11× 11 triplet (M0, C0, K0) with M0= I11 and C0,
K0 being randomly generated symmetric matrices. We first compute all 22
eigenpairs of Q0(λ) = λ2I + λC0+ K0 and (Λ, X) ∈ R14×14× R10×14 are
chosen from those 22 computed eigenpairs of Q0(λ). We compute a basis
{D1, . . . , D5} for nontrivial solutions D ∈ D(Λ,X). Using Algorithm 3.1 we
find {a1, . . . , a5} so that the matrix M11(a) in (3.1) is symmetric positive
definite. That is, there is a nontrivial solution D =P5i=1aiDi∈ D(Λ,X)such
that Q⊤
2Λ⊤DQ2 is symmetric positive definite. We then choose a symmetric
W = I7+ (Z⊤Q⊤1DQ2)(Q⊤2Λ⊤DQ2)−1(Z⊤Q⊤1DQ2)⊤ ∈ R7×7 such that the
matrix M in (2.24a) is symmetric positive definite. We compute the sym-metric C and K using (2.24b). (2.24c), respectively. The relative residual is estimated by kMXΛ2+ CXΛ + KX k2 kMXΛ2k 2+kCXΛk2+kKXk2 = 1.171× 10 −14. 5 Conclusions
Solving quadratic inverse eigenvalue problems for some partially prescribed eigen-information is a challenging task in many applications. Many attempts have been made, both theoretical and computational. Thus far, the results are somewhat limited. One of the most fundamental challenges is to characterize when the QIEP has a nonsingular quadratic pencil solution, and when the QIEP has only singular quadratic pencil solutions.
This paper provides a complete theory on when a nonsingular quadratic pencil solution to the QIEP is possible. In particular, two contributions made in this paper are significant. First, we describe a parameterized matrix rep-resentation for the general solution to the QIEP. An important characteristic in our construction for the general solution is that the parameters involve an (n− k) × (n − k) symmetric matrix W and an m × m quasi-diagonal matrix D so that any basis for the null space of the given eigenvector matrix is D-orthogonal (see (2.26a)). Secondly, we prove that the QIEP, generically, has a nonsingular quadratic pencil solution and, generically, has only singular quadratic pencil solutions when the number of the given eigenpairs, respec-tively, smaller and larger than a critical number k∗. Furthermore, the generic
Because the QIEP has important applications in many disciplines, the results in this paper, especially those fully addressing the issue of solvability, should be of interest.
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