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Linear Algebra and its Applications

j o u r n a l h o m e p a g e : w w w . e l s e v i e r . c o m / l o c a t e / l a a

The palindromic generalized eigenvalue problem A x = λ Ax: Numerical solution and applications

Tiexiang Li

a

, Chun-Yueh Chiang

b

, Eric King-wah Chu

c,

, Wen-Wei Lin

d

aDepartment of Mathematics, Southeast University, Nanjing 211189, People’s Republic of China bCenter for General Education, National Formosa University, Huwei 632, Taiwan

cSchool of Mathematical Sciences, Building 28, Monash University, VIC 3800, Australia

dCMMSC and NCTS, Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan

A R T I C L E I N F O A B S T R A C T

Article history:

Received 29 June 2009 Accepted 19 November 2009 Available online 12 January 2010 Submitted by V. Mehrmann

AMS classification:

15A18 15A22 65F15

Keywords:

Palindromic generalized eigenvalue problem

Doubling algorithm Singular descriptor system

In this paper, we propose the palindromic doubling algorithm (PDA) for the palindromic generalized eigenvalue problem (PGEP) Ax=λAx. We establish a complete convergence theory of the PDA for PGEPs without unimodular eigenvalues, or with unimodular eigenvalues of partial multiplicities two (one or two for eigenvalue 1). Some important applications from the vibration analysis and the optimal control for singular descriptor linear systems will be presented to illustrate the feasibility and efficiency of the PDA.

© 2009 Elsevier Inc. All rights reserved.

1. Introduction

In this paper, we develop the palindromic doubling algorithm (PDA) for the numerical solution of the palindromic generalized eigenvalue problem (PGEP)

Corresponding author. Tel.: +61 3 99054480; fax: +61 3 99054403.

E-mail addresses: [email protected] (T. Li), [email protected] (C.-Y. Chiang), [email protected] (E.K.-w. Chu), [email protected] (W.-W. Lin).

0024-3795/$ - see front matter © 2009 Elsevier Inc. All rights reserved.

doi:10.1016/j.laa.2009.12.020

(2)

Ax

= λ

Ax, (1.1) whereAis a real or complexN

×

Nmatrix,

λ

Candx

CN

\{

0

}

are eigenvalue and the correspond- ing eigenvector of (1.1), respectively. Here, the symbol “

=

(Transpose) or H (Hermitian). The pencil A

λ

Aand the pair

(

A, A

)

are usually called a palindromic linear pencil and a palindromic matrix pair, respectively. It is easily seen that the eigenvalues of (1.1) satisfy the reciprocal property, i.e., they appear in pairs as in

, 1

}

.

The PGEPs with complex coefficient matrices were firstly suggested as “good” linearizations [5,6]

of palindromic polynomial/quadratic matrix pencils, arising from the study of vibration analysis [2, 4,12]. A PGEP with real coefficient matrices can also be shown to be equivalent to the generalized continuous/discrete-time algebraic Riccati equations, associated with the continuous/discrete-time, linear-quadratic optimal control problems (see [11] for details).

The standard approach for solving the PGEP is to compute its generalized Schur form (e.g., byqzin MATLAB), ignoring its symmetric or palindromic structure in

(

A, A

)

. However, the reciprocal property of eigenvalues of (1.1) is not preserved by computation generally, producing large numerical errors [7].

Recently, a QR-like algorithm [8] and a hybrid method [7] (which combines Jacobi-type method with the Laub’s trick) were proposed for the PGEP. The QR-like algorithm generally requiresO

(

N4

)

flops and the hybrid method requiresO

(

N3log

(

N

))

flops. Alternatively, for methods of cubic complexity, a URV-decomposition based structured method [9] and a structure-preserving algorithm [3] for PGEPs were proposed, producing eigenvalues which are paired to working precision. Unfortunately for PGEPs, these more efficient (and equivalent) methods require the transformation of the PGEP to the quadratic form

2A

+ μ ·

0

+

A

)

x

=

0, leading to operations in larger 2N

×

2Nmatrices. The PDA is a unique and more direct, thus more efficient, algorithm for the PGEP.

The purpose of this paper is to develop the PDA for solving the PGEP structurally. We establish quadratic convergence and linear convergence with rate 1

/

2 of the PDA, respectively, when

(

A, A

)

has no unimodular eigenvalues and has unimodular eigenvalues with partial multiplicities two. In appli- cation to discrete-time optimal control problems, we especially develop a new algorithm combined with the PDA (as in Algorithm 4.1) for solving the optimal control ofsingulardescriptor linear systems.

To our knowledge, the associated generalized discrete-time algebraic Riccati equation (GDARE) has not been solved successfully in a structure-preserving manner.

This paper is organized as follows. In Section 2 we develop the palindromic doubling algorithm (PDA) for solving PGEPs. In Section 3 we establish the convergence theory for the PDA. In Section 4 we use the PDA to compute numerical solutions structurally in different applications in PGEPs, GCAREs and GDAREs. Concluding remarks are given in Section 5.

Throughout this paper,Cm×n and Rm×n denote the sets ofm

×

n complex and real matrices, respectively. For convenience, we denoteCn

=

Cn×1,C

=

C1,Rn

=

Rn×1andR

=

R1. The open left-half plane and the open unit disk are denoted byCandD1, respectively; 0m×n

(

0m

)

andImare them

×

n

(

m

×

m

)

zero matrix and them

×

midentity matrix, respectively. We use

σ(

A, B

)

to denote the spectrum of the matrix pair

(

A, B

)

, and

·

denotes the 2-norm of a matrix.

2. Palindromic doubling algorithm

For a given palindromic matrix pair

(

A, A

)

, we shall develop a doubling algorithm for solving the associated PGEP which preserves the palindromic structure at each iterative step.

Suppose

1

/ σ(

A, A

)

(the assumption can be removed later in Remark 3.1). We then have A

(

A

+

A

)

1A

= ((

A

+

A

)

A

)(

A

+

A

)

1

(

A

+

A

A

)

= (

I

A

(

A

+

A

)

1

)((

A

+

A

)

A

)

=

A

(

A

+

A

)

1A

.

(2.1)

From (2.1), it is easily seen that

A

(

A

+

A

)

1, A

(

A

+

A

)

1

AA

=

0

.

(2.2)

(3)

We now define the doubling transformA

Aby

A

=

A

(

A

+

A

)

1A

.

(2.3)

Theorem 2.1.The matrix pair

(

A,A

)

has the doubling property

;

i

.

e

.

, if

AU

=

AUS, (2.4)

where U

CN×and S

C×, then

AU

=

AUS2

.

(2.5)

Proof.Multiplying the both sides of (2.4) byA

(

A

+

A

)

1, and (2.1) and (2.4) imply (2.5).

From Theorem2.1, we see that the doubling transform (2.3) preserves the palindromic structure.

So, for a palindromic matrix pair A0, A0

withA0

CN×N, we can develop the PDA to generate the sequence Ak, Ak

if no breakdown occurs in the iterative process.

PDA AlgorithmGivenA0

CN×N,

τ

(a small tolerance), fork

=

0,1,2,

. . .

,compute

Ak+1

=

Ak

Ak

+

Ak

1

Ak, (2.6)

ifdist

(

Null

(

Ak+1

)

,Null

(

Ak

)) < τ

, then stop, end for

Here, “Null

(·)

” denotes the null space of the given matrix and “dist

,

·)

” denotes the distance between two subspaces.

To develop the PDA further, denote

Ak

=

Hk

+

Kk, (2.7a)

where Hk

=

1

2

Ak

+

Ak

=

Hk, Kk

=

1 2

Ak

Ak

= −

Kk (2.7b)

are the

-symmetric and

-anti-symmetric parts ofAk, respectively. Then the iteration (2.6) can be rewritten as

Ak+1

=

Hk+1

+

Kk+1

=

1

2

(

Hk

+

Kk

)

Hk1

(

Hk

+

Kk

)

=

1 2

Hk

+

KkHk1Kk

+

Kk

.

The iteration (2.6) in the PDA can be simplified to Hk+1

=

1

2

Hk

+

K0Hk1K0 , Kk+1

=

Kk

= · · · =

K0

.

3. Convergence of PDA

LetA0

CN×N. Suppose the eigenvalue “1" of A0, A0

(if exists) has partial multiplicity one or two, and the other unimodular eigenvalues of

A0, A0

(if exist) have exactly partial multiplicities two. By the theorem of Kronecker canonical form there are nonsingular matricesQandZsuch that

(4)

QA0Z

=

J0

Ω0 0,˜

Ir

0n,n˜ I˜

Ω0

C0, (3.1a)

QA0Z

=

0In 0n,n˜

˜

n,n J0

Ir

D0, (3.1b)

whereΩ0

=

diageiω1,

. . .

, eiωr,J0

C×consists of stable Jordan blocks (i.e.,

ρ(

J0

) <

1, where

ρ(·)

is the radius of the spectrum) andJ0

=

J0

Imwithn

= +

r,

˜ = +

m,n

˜ =

n

+

m

= +

r

+

mandN

=

2n

+

m. Here “

" denotes the direct sum of matrices.

SinceC0D0

=

D0C0, from (3.1b) we have thatA0ZD0

=

A0ZC0. From Theorem2.1and steps in the PDA, it follows that

AkZD20k

=

AkZC02k

.

(3.2)

Substituting (3.1b) into (3.2), we get AkZ

In 0n,n˜ 0n,n˜ J02k

Ir

=

AkZ

J02k

Ω02k 0,˜

k

0˜n,n I

Ω02k

, (3.3)

whereΓk

=

2kΩ02k1. On the other hand, we can interchange the role of A0, A0

by considering the pair

A0, A0

which has the same Kronecker structure as A0, A0

. Therefore, there are nonsingularP andYsuch that

PA0Y

=

J0

Ω0 0,˜

Ir

0n,n˜ I˜

Ω0

E0, (3.4a)

PA0Y

=

I0n 0n,˜n

˜

n,n J0

Ir

F0, (3.4b)

SinceE0F0

=

F0E0, we deduce thatA0YF0

=

A0YE0. Using the similar arguments as in (3.2) and (3.3), we obtain

AkY

In 0n

0n J02k

Ir

=

AkY

J02k

Ω0

2k

0,˜

Γk

0n,n˜ I˜

Ω0

2k

.

(3.5)

We partitionAk,HkandK0in (2.7a) into four sub-blocks as in Ak

=

AAk1 Ak3

k2 Ak4

, Hk

=

HHk1 Hk2

k2 Hk4

, K0

=

K01

K02

K02 K04

, (3.6a)

whereAk1,Hk1,K01

Cn×n,Ak2, Ak3, Hk2, K02

CnטnandAk4, Hk4, K04

C˜nטn. From (2.7a) and (3.6a), we also have

Ak1

=

Hk1

+

K01, Ak2

=

Hk2

+

K02, (3.6b)

Ak3

=

Hk2

K02, Ak4

=

Hk4

+

K04

.

(3.6c)

Furthermore, we partitionZin (3.3) andYin (3.5) as in Z

=

ZZ1 Z3

2 Z4

, Y

=

YY1 Y3

2 Y4

, (3.7)

whereZ1,Y1

Cn×n;Z2, Z3, Y2, Y3

CnטnandZ4, Y4

Cn˜×˜n. For convenience, we denote

Zi,a

Zi

(:

,1

: )

, Yi,a

Yi

(:

,1

: )

, i

=

3,4, (3.8) Zi,b

Zi

(:

,

+

1

:

n

)

, Yi,b

Yi

(:

,

+

1

:

n

)

, i

=

1,2

.

(3.9)

(5)

Theorem 3.1.Let A0

CN×N

.

Suppose that the eigenvalue “1" of A0, A0

(

if exists

)

has partial multiplicity one or two, the other unimodular eigenvalues of

A0, A0

(

if exist

)

have exactly partial multiplicities two, and

(

3.1b

)

and

(

3.4b

)

hold withn

˜

2

(

i

.

e

.

, r

+

m

).

Suppose that Z1and Y1in

(

3.7

)

are invertible, and W

≡ [

Z3,a

Z4,a

|

Y3,a

Y4,a

] ∈

Cn˜×2is of full row rank, whereΦ

Z2Z11

Y2Y11

.

If

1

/

rj=1

e2kiωj, k0

, then the sequence Ak, Ak

generated by the PDA is well defined and satisfies Ak

Z1 Z2

0, linearly ask

→ ∞

, (3.10a)

Ak Y1

Y2

0,linearly ask

→ ∞

, (3.10b)

with convergence rate1

/

2, wherespanZZ12andspanYY34form the weakly stable and the unstable invariant subspaces of

A0, A0

corresponding to

(

J0

Ω0, In

)

andIn, J0

Ω0

, respectively

.

Proof.Since

1

/

rj=1

e2kiωj, k0

, from (2.6) we see that

1

/ σ

Ak, Ak

, thus,Ak

+

Akis in- vertible for allk.

From (3.6a), (3.3) and (3.7), we have Ak1Z1

+

Ak2Z2

=

Ak1Z1

J02k

Ω02k

+

Ak3Z2

J02k

Ω02k

, (3.11)

Ak3Z1

+

Ak4Z2

=

Ak2Z1

J02k

Ω02k

+

Ak4Z2

J02k

Ω02k

, (3.12)

Ak1Z3J02k

Ir

+

Ak2Z4J02k

Ir

=

Ak1

Z1

(

0,˜

Γk

) +

Z3

I˜

Ω02k

+

Ak3

Z2

(

0,˜

Γk

) +

Z4

I˜

Ω02k

,(3.13) Ak3Z3J02k

Ir

+

Ak4Z4J02k

Ir

=

Ak2

Z1

(

0,˜

Γk

) +

Z3

I˜

Ω02k

+

Ak4

Z2

(

0,˜

Γk

) +

Z4

I˜

Ω02k

.

(3.14)

Post-multiplying (3.13) by 0˜,

Γk1Ω02k, we get Ak1Z3

0˜,

Γk1Ω02k

+

Ak2Z4

0˜,

Γk1Ω02k

= (

Ak1Z1

+

Ak3Z2

)

0

Ω02k

+ (

Ak1Z3

+

Ak3Z4

)

0˜,

Ω02kΓk1Ω02k

.

(3.15)

Substituting (3.15) into (3.11) and usingΩ02kΓk1Ω02k

=

2kΩ02k+1, we have Ak1Z1

+

Ak2Z2

= (

Ak1Z1

+

Ak3Z2

)

J02k

0r

+ (

Ak1Z1

+

Ak3Z2

)

0

Ω02k

= (

Ak1Z1

+

Ak3Z2

)

J02k

0r

+

Ak1Z3

+

Ak2Z4 0˜,

Γk1Ω02k

(

Ak1Z3

+

Ak3Z4

)

0˜,

2kΩ02k+1

.

(3.16)

Using (3.6a) and re-arranging (3.16), we get Hk1Z

Z1

In

J02k

0r

Z3

0˜,

2kΩ0

Ir

Ω02k

(6)

+

Hk2

Z2

In

J02k

0r

Z4

0˜,

2kΩ0

Ir

Ω02k

=

K01

Z1

In

+

J02k

0r

Z3

0˜,

2kΩ0

Ir

+

Ω02k

K02

Z2

In

+

J02k

0r

Z4

0˜,

2kΩ0

Ir

+

Ω02k

.

(3.17)

Denote

k

max

ρ(

J0

)

2k,2k

0, ask

→ ∞.

(3.18)

SinceΩ02kis bounded andZ1is invertible, by lettingΦ

Z2Z11, (3.17) can be simplified to Hk1

= −

Hk2

(

Φ

+

O

(

k

)) +

K01

K02Φ

+

O

(

k

).

(3.19) Post-multiplying (3.13) byI˜

Γk1, we have

Ak1Z3J02k

Γk1

+

Ak2Z4J02k

Γk1

=

Ak1

Z1

(

0,˜

Ir

) +

Z3

I˜

Ω02kΓk1

+

Ak3

Z2

(

0,˜

Ir

) +

Z4

I˜

Ω02kΓk1

.

(3.20)

From (3.6a) and (3.18), (3.20) becomes

Hk1

[

Z3

(

I

0m+r

) +

Z1

(

0,˜

Ir

) +

O

(

k

)] +

Hk2

[

Z4

(

I

0m+r

) +

Z2

(

0,˜

Ir

) +

O

(

k

)]

= −

K01

[

Z3

(

I

2Im

0r

) +

Z1

(

0,˜

Ir

) +

O

(

k

)]

+

K02

[

Z4

(

I

2Im

0r

) +

Z2

(

0,˜

Ir

) +

O

(

k

)].

(3.21) Substituting (3.19) into (3.21) we get

Hk2

(

Φ

+

O

(

k

))[

Z3

(

I

0m+r

) +

Z1

(

0,˜

Ir

) +

O

(

k

)] −

Z4

(

I

0m+r

)

Z2

(

0,˜

Ir

) +

O

(

k

)

=

O

(

1

).

SinceΦZ1,b

=

Z2,b, it holds that

Hk2

([

ΦZ3,a

Z4,a

] +

O

(

k

)) =

O

(

1

)

Cn˜×

.

(3.22) On the other hand, from (3.6a), (3.5) and (3.7), we have

Ak1Y1

+

Ak3Y2

=

Ak1Y1

J02k

Ω0

2k

+

Ak2Y2

J02k

(

Ω0

)

2k, (3.23) Ak2Y1

+

Ak4Y2

=

Ak3Y1

J02k

Ω0

2k

+

Ak4Y2 J02k

Ω0

2k

, (3.24)

Ak1Y3

J20k

Ir

+

Ak3Y4

J02k

Ir

=

Ak1Y3

+

Ak2Y4 I˜

(

Ω0

)

2k

+

Ak1Y1

+

Ak2Y2 0,˜

Γk

, (3.25)

Ak2Y3

J20k

Ir

+

Ak4Y4

J02k

Ir

=

Ak3Y3

+

Ak4Y4 I˜

(

Ω0

)

2k

+

Ak3Y1

+

Ak4Y2 0,˜

Γk

.

(3.26)

(7)

As in (3.15) and (3.17), post-multiplying (3.25) by 0˜,

Γk

1 Ω0

2k

and substituting it into (3.23), we have

Ak1Y1

+

Ak3Y2

=

Ak1Y1

+

Ak2Y2 J02k

0r

+

Ak1Y3

+

Ak3Y4 0˜,

2kΩ0

Ak1Y3

+

Ak2Y4 0˜,

2k

(

Ω0

)

2k+1

.

(3.27) From (3.6a) and (3.18), (3.27) becomes

Hk1

Y1

In

J0

2k

0r

Y3

0˜,

2kΩ0

Ir

Ω0

2k

+

Hk2

Y2

In

J0

2k

0r

Y4

0˜,

2kΩ0

Ir

Ω0

2k

= −

K01

Y1

In

+

J0

2k

0r

Y3

0˜,

2kΩ0

(

Ir

+ (

Ω0

)

2k

)

+

K02

Y2

In

+

J0

2k

0r

Y40˜,

2kΩ0

Ir

+ (

Ω0

)

2k, (3.28) and then

Hk1

= −

Hk2

(

Ψ

+

O

(

k

))

K01

+

K02Ψ

+

O

(

k

)

, whereΨ

=

Y2Y11.

Post-multiplying (3.25) byI˜

Γk

1

and substituting (3.28) into it, we have Hk2

Y4

(

I

0m+r

) +

Y2

(

0,˜

Ir

) +

O

(

k

)(

Ψ

+

O

(

k

))[

Y3

(

I

0m+r

) +

Y1

(

0,˜

Ir

) +

O

(

k

)]

=

O

(

1

).

SinceΨY1,b

=

Y2,b, it holds that

Hk2

([

ΨY3,a

Y4,a

] +

O

(

k

)) =

O

(

1

)

Cn˜×

.

(3.29) Combining (3.22) and (3.29) we get

Hk2

([

ΦZ3,a

Z4,a

|

ΨY3,a

Y4,a

] +

O

(

k

)) =

O

(

1

)

Cn˜×2

.

(3.30) By the assumption thatW

≡ [

ΦZ3,a

Z4,a

|

ΨY3,a

Y4,a

] ∈

Cn˜×2is of full row rank, it follows that Hk2is uniformly bounded onk. Consequently, (3.19) implies that

Hk1

, and in turn

Ak1

andAk2, are uniformly bounded onk. From (3.16), it follows that

Ak1Z1

+

Ak2Z2

=

O

(

k

)

0, ask

→ ∞.

(3.31)

Applying the similar argument as in (3.15) and (3.17) to (3.24) and (3.26), we deduce that Hk4

= −

Hk2

Y2Y11

+

O

(

k

)

+

K04

K02Y2Y11

+

O

(

k

).

Thus, (3.30) implies that

Hk4

, and in turn

Ak4

, are uniformly bounded onk.

To showAk3Z1

+

Ak4Z2

=

O

(

k

)

, we post-multiply (3.14) by 0˜,

Γk1Ω02kand obtain Ak3Z3

0˜,

Γk1Ω02k

+

Ak4Z4

0˜,

Γk1Ω02k

= (

Ak2Z1

+

Ak4Z2

)

0

Ω02k

+ (

Ak2Z3

+

Ak4Z4

)

0˜,

2kΩ02k+1

.

(3.32)

Références

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