# An Inverse Eigenvalue Problem of Hermite-Hamilton Matrices in Structural Dynamic Model Updating

Authors: Linlin Zhao; Guoliang Chen
Journal: Mathematical Problems in Engineering
 (2010)

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## Abstract

We first consider the following inverse eigenvalue problem: givenX∈Cn×m and a diagonal matrix Λ∈Cm×m, find n×n Hermite-Hamilton matrices K and M such that KX=MXΛ. We then consider an optimal approximation problem: given n×n Hermitian matrices Ka and Ma, find a solution (K,M) of the above inverse problem such that ∥K-Ka∥2+∥M-Ma∥2=min. By using the Moore-Penrose generalized inverse and the singular value decompositions, the solvability conditions and the representations of the general solution for the first problem are derived. The expression of the solution to the second problem is presented.


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## Body

## 1. Introduction

Throughout this paper, we will adopt the following notations. LetCm×n,HCn×n, and UCn×n stand for the set of all m×n matrices, n×n Hermitian matrices, and unitary matrices over the complex field C, respectively. By ∥·∥ we denote the Frobenius norm of a matrix. The symbols AT,A*,A-1, and A† denote the transpose, conjugate transpose, inverse, and Moore-Penrose generalized inverse of A, respectively.Definition 1.1.
LetJn=(0Ik-Ik0), n=2k, and A∈Cn×n. If A=A* and JnAJn=A*, then the matrix A is called Hermite-Hamilton matrix.We denote byHHCn×n the set of all n×n Hermite-Hamilton matrices.Vibrating structures such as bridges, highways, buildings, and automobiles are modeled using finite element techniques. These techniques generate structured matrix second-order differential equations:(1.1)Maz̈(t)=Kaz(t),
where Ma,Ka are analytical mass and stiffness matrices. It is well known that all solutions of the above differential equation can be obtained via the algebraic equation Kax=λMax. But such finite element model is rarely available in practice, because its natural frequencies and mode shapes often do not match very well with experimentally measured ones obtained from a real-life vibration test [1]. It becomes necessary to update the original model to attain consistency with empirical results. The most common approach is to modify Ka and Ma to satisfy the dynamic equation with the measured model data. Let X∈Cn×m be the measured model matrix and Λ=diag(δ1,δ2,…,δm)∈Cm×m the measured natural frequencies matrix, where n≥m. The measured mode shapes and frequencies are assumed correct and have to satisfy (1.2)KX=MXΛ,
where M,K∈Cn×n are the mass and stiffness matrices to be corrected. To date, many techniques for model updating have been proposed. For undamped systems, various techniques have been discussed by Berman [2] and Wei [3]. Theory and computation of damped systems were proposed by authors of [4, 5]. Another line of thought is to update damping and stiffness matrices with symmetric low-rank correction [6]. The system matrices are adjusted globally in these methods. As model errors can be localized by using sensitivity analysis [7], residual force approach [8], least squares approach [9], and assigned eigenstructure [10], it is usual practice to adjust partial elements of the system matrices using measured response data.The model updating problem can be regarded as a special case of the inverse eigenvalue problem which occurs in the design and modification of mass-spring systems and dynamic structures. The symmetric inverse eigenvalue problem and generalized inverse eigenvalue problem with submatrix constraint in structural dynamic model updating have been studied in [11] and [12], respectively. Hamiltonian matrices usually arise in the analysis of dynamic structures [13]. However, the inverse eigenvalue problem for Hermite-Hamilton matrices has not been discussed. In this paper, we will consider the following inverse eigenvalue problem and an associated optimal approximation problem.Problem 1.
Given thatX∈Cn×m and a diagonal matrix Λ∈Cm×m, find n×n Hermite-Hamilton matrices K and M such that
(1.3)KX=MXΛ.Problem 2.
Given thatKa,Ma∈HCn×n, let SE be the solution set of Problem 1. Find (K̂,M̂)∈SE such that
(1.4)∥K̂-Ka∥2+∥M̂-Ma∥2=min(K,M)∈SE(∥K-Ka∥2+∥M-Ma∥2).We observe that, whenM=I, Problem 1 can be reduced to the following inverse eigenproblem: (1.5)KX=XΛ,
which has been solved for different classes of structured matrices. For example, Xie et al. considered the problem for the case of symmetric, antipersymmetric, antisymmetric, and persymmetric matrices in [14, 15]. Bai and Chan studied the problem for the case of centrosymmetric and centroskew matrices in [16]. Trench investigated the case of generalized symmetry or skew symmetry matrices for the problem in [17] and Yuan studied R-symmetric matrices for the problem in [18].The paper is organized as follows. In Section2, using the Moore-Penrose generalized inverse and the singular value decompositions of matrices, we give explicit expressions of the solution for Problem 1. In Section 3, the expressions of the unique solution for Problem 2 are given and a numerical example is provided.


## 2. Solution of Problem1

Let(2.1)U=12(IkIk-iIkiIk).Lemma 2.1.
LetA∈Cn×n. Then A∈HHCn×n if and only if there exists a matrix N∈Ck×k such that
(2.2)A=U(0NN*0)U*,
where U is the same as in (2.1).Proof.
LetA=(A11A12A12*A22), and let each block of A be square. From Definition 1.1 and (2.1), it can be easily proved.Lemma 2.2 (see [19]).
LetA∈Cm×n, B∈Cp×q, and E∈Cm×q. Then the matrix equation AXB=E has a solution X∈Cn×p if and only if AA†EB†B=E; in this case the general solution of the equation can be expressed as X=A†EB†+Y-A†AYBB†, where Y∈Cn×p is arbitrary.Let the partition of the matrixU*X be(2.3)U*X=(X1X2),X1,X2∈Ck×m,
where U is defined as in (2.1).We assume that the singular value decompositions of the matricesX1 and X2 are(2.4)X1=R(D000)S*,X2=W(Σ000)V*,
where R=(R1,R2)∈UCk×k, S=(S1,S2)∈UCm×m, D=diag(d1,…,dl)>0,l=rank(X1), R1∈Ck×l,S1∈Cm×l, and W=(W1,W2)∈UCk×k, V=(V1,V2)∈UCm×m, Σ=diag(σ1,…,σs)>0,s=rank(X2), W1∈Ck×s,V1∈Cm×s.Let the singular value decompositions of the matricesX2ΛV2 and X1ΛS2 be(2.5)X2ΛV2=P(Ω000)Q*,X1ΛS2=T(Δ000)H*,
where P=(P1,P2)∈UCk×k, Q=(Q1,Q2)∈UC(m-s)×(m-s), Ω=diag(ω1,…,ωt)>0,t=rank(X2ΛV2), P1∈Ck×t,Q1∈C(m-s)×t, and T=(T1,T2)∈UCk×k, H∈UC(m-l)×(m-l), Δ=diag(a1,…,ag)>0,g=rank(X1ΛS2),  T1∈Ck×g.Theorem 2.3.
Suppose thatX∈Cn×m and Λ∈Cm×m is a diagonal matrix. Let the partition of U*X be (2.3), and let the singular value decompositions of X1,X2,X2ΛV2, and X1ΛS2 be given in (2.4) and (2.5), respectively. Then (1.3) is solvable and its general solution can be expressed as
(2.6)M=U(0FF*0)U*,K=U(0FX2ΛX2†+GW2*(FX2ΛX2†+GW2*)*0)U*,
where
(2.7)F=T2JP2*,G=(X1ΛX1†)*FW2+R2Y,
with J∈C(k-g)×(k-t), Y∈C(k-l)×(k-s) being arbitrary matrices, and U is the same as in (2.1).Proof.
By Lemma2.1, we know that (K,M) is a solution to Problem 1 if and only if there exist matrices N,F∈Ck×k such that
(2.8)K=U(0NN*0)U*,M=U(0FF*0)U*,U(0NN*0)U*X=U(0FF*0)U*XΛ.
Using (2.3), the above equation is equivalent to the following two equations:
(2.9)NX2=FX2Λ,(2.10)N*X1=F*X1Λ,i.e.,X1*N=(X1Λ)*F.
By the singular value decomposition ofX2, then the relation (2.9) becomes
(2.11)0=FX2ΛV2,(2.12)NW1Σ=FX2ΛV1.
Clearly, (2.11) with respect to unknown matrix F is always solvable. By Lemma 2.2 and (2.5), we get
(2.13)F=LP2*,
where L∈Ck×(k-t) is an arbitrary matrix. Substituting F=LP2* into (2.12), we get
(2.14)NW1=(LP2*)X2ΛV1Σ-1.
Since W1 is of full column rank, then the above equation with respect to unknown matrix N is always solvable, and the general solution can be expressed as
(2.15)N=(LP2*X2ΛV1Σ-1)W1*+GW2*=LP2*X2ΛX2†+GW2*,
where G∈Ck×(k-s) is an arbitrary matrix.
SubstitutingF=LP2* and (2.15) into (2.10), we get
(2.16)X1*(LP2*X2ΛX2†+GW2*)=(X1Λ)*LP2*.
By the singular value decomposition ofX1, then the relation (2.16) becomes
(2.17)0=S2*(X1Λ)*LP2*,(2.18)DR1*(LP2*X2ΛX2†+GW2*)=S1*(X1Λ)*LP2*.
Clearly, (2.17) with respect to unknown matrix L is always solvable. From Lemma 2.2 and (2.5), we have
(2.19)L=J1-(X1ΛS2)(X1ΛS2)†J1P2*P2=J1-(X1ΛS2)(X1ΛS2)†J1=T2J,
where J∈C(k-g)×(k-t) is arbitrary. Substituting L=T2J into (2.18), we get
(2.20)DR1*GW2*=(X1ΛS1)*T2JP2*-DR1*T2JP2*X2ΛX2†.
Then, we have
(2.21)R1*GW2*=D-1(X1ΛS1)*T2JP2*-R1*T2JP2*X2ΛX2†.
Since R1* is of full row rank, then the above equation with respect to GW2* is always solvable. By Lemma 2.2, we have
(2.22)GW2*=(X1ΛX1†)*T2JP2*-R1R1*T2JP2*X2ΛX2†+(I-R1R1*)Y1,
where Y1∈Ck×k is arbitrary. Then, we get
(2.23)G=(X1ΛX1†)*T2JP2*W2-R1R1*T2JP2*X2ΛX2†W2+(I-R1R1*)Y1W2,=(X1ΛX1†)*T2JP2*W2+R2Y,
where Y∈C(k-l)×(k-s) is arbitrary.
Finally, we have(2.24)F=T2JP2*,N=FX2ΛX2†+GW2*,
where G=(X1ΛX1†)*FW2+R2Y. The proof is completed.From Lemma2.1, we have that if the mass matrix M∈HHCn×n, then M is not positive definite. If M is symmetric positive definite and K is a symmetric matrix, then (1.3) can be reformulated as the following form:(2.25)AX=XΛ,
where A=M-1K. From [20, Theorem 7.6.3], we know that A is a diagonalizable matrix, all of whose eigenvalues are real. Thus, Λ∈Rm×m and X is of full column rank. Assume that X is a real n×m matrix. Let the singular value decomposition of X be(2.26)X=Ũ(Γ0)ṼT,Ũ∈ORn×n,Ṽ∈ORm×m,Γ=diag(γ1,…,γm)>0,
where ORn×n denotes the set of all orthogonal matrices. The solution of (2.25) can be expressed as (2.27)A=Ũ(ΓṼTΛṼΓ-1Z120Z22)ŨT,
where Z12∈Rm×(n-m) is an arbitrary matrix and Z22∈R(n-m)×(n-m) is an arbitrary diagonalizable matrix (see [21, Theorem 3.1]).LetΛ=diag(λ1Ik1,…,λqIkq) with λ1<λ2<⋯<λq. Choose Z22=G¯Λ2G¯-1, where G¯∈R(n-m)×(n-m) is an arbitrary nonsingular matrix and Λ2=diag(λq+1Ikq+1,…,λpIkp) with λp>⋯>λq+1>λq. The solutions to (1.3) with respect to unknown matrices M>0 and K=KT are presented in the following theorem.Theorem 2.4 (see [21]).
Given thatX∈Rn×m, rank(X)=m, and Λ=diag(λ1Ik1,…,λqIkq)∈Rm×m, let the singular value decomposition of X be (2.26). Then the symmetric positive-definite solution M and symmetric solution K to (1.3) can be expressed as
(2.28)M=ŨF̃TF̃ŨT,K=ŨF̃TΔF̃ŨT,
where Δ=diag(Λ,Λ2), F̃=(F11F120F22), F11=diag(L1,…,Lq)VΓ-1∈Rm×m, and F22=diag(Lq+1,…,Lp)G¯-1∈R(n-m)×(n-m), where Li∈Rki×ki is an arbitrary nonsingular matrix (i=1,2,…,p). The matrix F12 satisfies the equation ΛF12G¯-F12G¯Λ2=F11Z12G¯.


## 3. Solution of Problem2

Lemma 3.1 (see [22]).
Given thatA∈Cm×n,B∈Cp×q,C∈Cl×n,D∈Cp×t,E∈Cm×q, and H∈Cl×t, let
(3.1)Sa={Z∣Z∈Cn×p,∥[AZB-E,CZD-H]∥2=min},Sb={Z∣Z∈Cn×p,A*AZBB*+C*CZDD*=A*EB*+C*HD*}.
Then Z∈Sa if and only if Z∈Sb.For the given matricesKa,Ma∈HCn×n, let (3.2)U*MaU=(C1C2C2*C3),U*KaU=(K1K2K2*K3).From Theorem2.3, we know that SE≠∅. The following theorem is for the best approximation solution of Problem 2.Theorem 3.2.
Given thatX∈Cn×m, Λ∈Cm×m, and Ka,Ma∈HCn×n, then Problem 2 has a unique solution and the solution can be expressed as
(3.3)M̂=U(0F̂F̂*0)U*,K̂=U(0F̂X2ΛX2†+K2W2W2*(F̂X2ΛX2†+K2W2W2*)*0)U*,
where
(3.4)F̂=(C2+K2(X2ΛX2†)*)(I+(X2ΛX2†)(X2ΛX2†)*)-1.Proof.
It is easy to verify thatSE is a closed convex subset of HHCn×n×HHCn×n. From the best approximation theorem, we know that there exists a unique solution (K̂,M̂) in SE such that (1.4) holds. From Theorem 2.3 and the unitary invariant of the Frobenius norm, we have
(3.5)∥Ma-M∥2+∥Ka-K∥2=∥(C1C2C2*C3)-(0FF*0)∥2+∥(K1K2K2*K3)-(0FX2ΛX2†+GW2*(FX2ΛX2†+GW2*)*0)∥2,
where G=(X1ΛX1†)*FW2+R2Y. Hence, ∥Ma-M∥2+∥Ka-K∥2=min is equivalent to
(3.6)∥F-C2∥2+∥FX2ΛX2†+(X1ΛX1†)*FW2W2*+R2YW2*-K2∥2=min.
Let(3.7)f=∥F-C2∥2+∥FX2ΛX2†+(X1ΛX1†)*FW2W2*+R2YW2*-K2∥2.
Then from the unitary invariant of the Frobenius norm, we have
(3.8)f=∥F-C2∥2+∥FX2ΛX2†(W1,W2)+(X1ΛX1†)*FW2W2*(W1,W2)+R2YW2*(W1,W2)-K2(W1,W2)∥2=∥F-C2∥2+∥(FX2ΛX2†W1,0)+(0,(X1ΛX1†)*FW2)+(0,R2Y)-(K2W1,K2W2)∥2=∥F-C2∥2+∥FX2ΛX2†W1-K2W1∥2+∥(X1ΛX1†)*FW2+R2Y-K2W2∥2.
Leth=∥(X1ΛX1†)*FW2+R2Y-K2W2∥2. It is not difficult to see that, when
(3.9)R2Y=K2W2-(X1ΛX1†)*FW2,
that is, Y=R2*K2W2-R2*(X1ΛX1†)*FW2, we have h=0. In other words, we can always find Y such that h=0. Let
(3.10)g=∥F-C2∥2+∥FX2ΛX2†W1-K2W1∥2=∥[F-C2,FX2ΛX2†W1-K2W1]∥2.
Then, we have that f=min is equivalent to g=min. According to Lemma 3.1 and (3.10), we get the following matrix equation:
(3.11)F+F(X2ΛX2†W1)(X2ΛX2†W1)*=C2+K2W1(X2ΛX2†W1)*,
and its solution is F̂=(C2+K2(X2ΛX2†)*)(I+(X2ΛX2†)(X2ΛX2†)*)-1. Again from Lemma 3.1, we have that, when F=F̂, g attains its minimum, which gives Ŷ=R2*K2W2-R2*(X1ΛX1†)*F̂W2, and Ĝ=(X1ΛX1†)*F̂W2+R2Ŷ=K2W2. Then, the unique solution of Problem 2 given by (3.3) is obtained.Now, we give an algorithm to compute the optimal approximate solution of Problem2.Algorithm 3.
(1) Input Ka, Ma, X,Λ, and U.
(2) Compute X2 according to (2.3).
(3) Find the singular value decomposition of X2 according to (2.4).
(4) Calculate F̂ by (3.4).
(5) Compute (M̂,K̂) by (3.3).Example 1.
Letn=6,m=3, and the matrices Ma,Ka,X, and Λ be given by
(3.12)Ma=(1.560.660.54-0.39000.660.360.39-0.27000.540.393.1200.54-0.39-0.39-0.2700.720.39-0.27000.540.393.12000-0.39-0.2700.72),Ka=(23-230036-3300-2-340-2333012-3300-2-3400033012),X=(0.0347+0.1507i-0.6975i0.0003+0.0858i0.6715i0.0277+0.0760i-0.0846-0.0101i-0.0009+0.1587i-0.0814+0.0196i0.6967-0.1507+0.0347i0.6975-0.0858+0.0003i-0.6715-0.0760+0.0277i0.0101-0.0846i-0.1587-0.0009i-0.0196-0.0814i0.6967i),Λ=diag(0.3848+0.0126i,2.5545+0.4802i,2.5607).
From the Algorithm, we obtain the unique solution of Problem2 as follows:
(3.13)F̂=(-1.4080+1.1828i1.0322+0.4732i-0.8111-0.0874i0.9537+0.2935i-0.7529-0.0137i-0.6596-0.3106i-0.6624+0.1982i-0.3566-0.0051i-1.0958+1.0040i),N̂=(-4.3706+2.1344i1.6264-0.3128i-2.2882-0.3290i2.4251+1.2137i-0.5229+0.0005i-1.4620-0.7688i-1.6669+0.1663i0.6991-0.6057i-2.6437+2.5190i),M̂=U(0F̂F̂*0)U*,K̂=U(0N̂N̂0)U*,
where U=(1/2)(I3I3-iI3iI3). It is easy to calculate ∥K̂X-M̂XΛ∥=2.1121e-015, and ∥[M̂-Ma,K̂-Ka]∥=19.7467.



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