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Volume 2009, Article ID 101085, 17 pagesdoi:10.1155/2009/101085 Research Article Trace Inequalities for Matrix Products and Trace Bounds for the Solution of the Algebraic Riccati Equatio

Volume 2009, Article ID 101085, 17 pages

doi:10.1155/2009/101085

Research Article

Trace Inequalities for Matrix Products and

Trace Bounds for the Solution of the Algebraic

Riccati Equations

Jianzhou Liu,1, 2 Juan Zhang,2 and Yu Liu1

1 Department of Mathematic Science and Information Technology, Hanshan Normal University,

Chaozhou, Guangdong 521041, China

2 Department of Mathematics and Computational Science, Xiangtan University, Xiangtan,

Hunan 411105, China

Correspondence should be addressed to Jianzhou Liu,liujz@xtu.edu.cn

Received 25 February 2009; Revised 20 August 2009; Accepted 6 November 2009

Recommended by Jozef Banas

By using diagonalizable matrix decomposition and majorization inequalities, we propose new trace bounds for the product of two real square matrices in which one is diagonalizable These bounds improve and extend the previous results Furthermore, we give some trace bounds for the solution of the algebraic Riccati equations, which improve some of the previous results under certain conditions Finally, numerical examples have illustrated that our results are effective and superior

Copyrightq 2009 Jianzhou Liu et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

As we all know, the Riccati equations are of great importance in both theory and practice in the analysis and design of controllers and filters for linear dynamical systemssee 1 5 For example, consider the following linear systemsee 5:

˙xt  Axt  But, x 0  x0, 1.1 with the cost

J 

∞

0



x T Qx  u T u

The optimal control rate u∗the optimal cost J∗of1.1 and 1.2 are

u∗ Px, P  B T K,

J∗ x T

where x0 ∈ R nis the initial state of system1.1 and 1.2 and K is the positive semidefinite solution of the following algebraic Riccati equationARE:

with R  BB T and Q being positive definite and positive semidefinite matrices, respectively.

To guarantee the existence of the positive definite solution to1.4, we will make the following assumptions: the pairA, R is stabilizable, and the pair Q, A is observable.

In practice, it is hard to solve the ARE, and there is no general method unless the system matrices are special and there are some methods and algorithms to solve 1.4; however, the solution can be time-consuming and computationally difficult, particularly as the dimensions of the system matrices increase Thus, a number of works have been presented

by researchers to evaluate the bounds and trace bounds for the solution of the AREsee 6 16 Moreover, in terms of 2,6, we know that an interpretation of trK is that trK/n is the

average value of the optimal cost J∗as x0varies over the surface of a unit sphere Therefore, considering its applications, it is important to discuss trace bounds for the product of two matrices In symmetric case, a number of works have been proposed for the trace of matrix products2,6 8,17–20, and 18 is the tightest among the parallel results

In 1995, Lasserre showed18 the following given any matrix A ∈ Rn×n , B ∈ S n , then

the following

n



i1

λ i



A

λ n−i1 B ≤ trAB ≤n

i1

λ i



A

λ i B, 1.5

whereA  A  A T /2.

This paper is organized as follows In Section 2, we propose new trace bounds for the product of two general matrices The new trace bounds improve the previous results Then, we present some trace bounds for the solution of the algebraic Riccati equations, which improve some of the previous results under certain conditions in Section 3 In Section 4,

we give numerical examples to demonstrate the effectiveness of our results Finally, we get conclusions inSection 5

2 Trace Inequalities for Matrix Products

In the following, let R n×n denote the set of n × n real matrices and let S n denote the

subset of R n×n consisting of symmetric matrices For A  a ij  ∈ R n×n, we assume that trA, A−1, A T , dA  d1A, , d n A, σA  σ1A, , σ n A denote the trace, the inverse, the transpose, the diagonal elements, the singular values of A, respectively,

and define A ii  a ii  d i A If A ∈ R n×n is an arbitrary symmetric matrix, then

λA  λ1A, , λ n A and Re λA  Re λ1A, , Re λ n A denote the eigenvalues

and the real part of eigenvalues of A Suppose x  x1, x2, , x n  is a real n-element array such as dA, σA, λA, Re λA which is reordered, and its elements are arranged in nonincreasing order; that is, x1 ≥ x2 ≥ · · · ≥ x n The notation A > 0 A ≥ 0 is used

to denote that A is a symmetric positive definite semidefinite matrix.

Let α, β be two real n-element arrays If they satisfy

k



i1

α i≤k

i1

then it is said that α is controlled weakly by β, which is signed by α≺ w β.

If α≺ w β and

n



i1

α in

i1

then it is said that α is controlled by β, which is signed by α ≺ β.

The following lemmas are used to prove the main results

Lemma 2.1 see 21, Page 92, H.2.c If x1 ≥ · · · ≥ x n , y1 ≥ · · · ≥ y n and x ≺ y, then for any real array u1≥ · · · ≥ u n ,

n



i1

x i u i≤n

i1

y i u i 2.3

Lemma 2.2 see 21, Page 218, B.1 Let A  AT ∈ R n×n , then

d A ≺ λA. 2.4

Lemma 2.3 see 21, Page 240, F.4.a Let A ∈ Rn×n , then

λ



A  A T

2



≺w





λ



A  A T

2





≺w σ A. 2.5

Lemma 2.4 see 22 Let 0 < m1≤ a k ≤ M1, 0 < m2≤ b k ≤ M2, k  1, 2, , n, 1/p  1/q  1, then

n



k1

a k b k≤

n



k1

a p k

1/p n



k1

b q k

1/q

≤ c p,q n



k1

a k b k , 2.6

p

1M2q − m p

1m q2

p

M1m2M2q − m1M2m q2 1/p

q

m1M2M p1− M1m2m p1 1/q

. 2.7

Note that if m1  0, m2/  0 or m2  0, m1/ 0, obviously, 2.6 holds If m1  m2  0,

choose c p,q ∞, then 2.6 also holds

Remark 2.5 If p  q  2, then we obtain Cauchy-Schwarz inequality:

n



k1

a k b k≤

n

k1

a2k

1/2n

k1

b2k

1/2

≤ c2

n



k1

a k b k , 2.8

where

c2

⎛

⎝ M1M2

m1m2 



m1m2

M1M2

⎞

Remark 2.6 Note that

lim

p → ∞



a p1 a p2 · · ·  a p n

1/p

 max

1≤k≤n{a k },

lim

p → ∞

q → 1

c p,q limp → ∞

q → 1

M p1M2q − m p

1m q2

p

M1m2M q2− m1M2m q2 1/p

q

m1M2M p1− M1m2m p1 1/q

 limp → ∞

q → 1

M p1

M q2− m1/M1p m q2

M 1/p1

p

m2M2q −m1/M1M2m q2 1/p M q/p1 

q

m1M2−M1m2m1/M1p1/q

 limp → ∞

q → 1

M2

M 1/pp/q−p1 m1M2

 limp → ∞

q → 1

1

M 1/p−11 m1

 M1

m1.

2.10

Let p → ∞, q → 1 in 2.6, then we obtain

m1

n



k1

b k≤n

k1

a k b k ≤ M1

n



k1

b k 2.11

Lemma 2.7 If q ≥ 1, a i ≥ 0 i  1, 2, , n, then

 1

n

n



i1

a i

q

≤ 1

n

n



i1

a q i 2.12

Proof 1 Note that for q  1, or a i  0 i  1, 2, , n,

 1

n

n



i1

a i

q

 1

n

n



i1

a q i 2.13

2 If q > 1, a i > 0, for x > 0, choose fx  x q , then f x  qx q−1 > 0 and f x 

qq − 1x q−2 > 0 Thus, fx is a convex function As a i > 0 and 1/nn

i1 a i > 0, from the

property of the convex function, we have

 1

n

n



i1

a i

q

 f

 1

n

n



i1

a i



≤ 1

n

n



i1

f a i  1

n

n



i1

a q i 2.14

3 If q > 1, without loss of generality, we may assume a i  0 i  1, , r, a i > 0 i 

r  1, , n Then from 2, we have

 1

n − r

qn

i1

a i

q



 1

n − r

n



i1

a i

q

≤ 1

n − r

n



i1

a q i 2.15

Sincen − r/n q ≤ n − r/n, thus



1

n

n



i1

a i

q





n − r n

q 1

n − r

qn

i1

a i

q

≤ n − r

n

1

n − r

n



i1

a q i  1

n

n



i1

a q i 2.16

This completes the proof

Theorem 2.8 Let A, B ∈ R n×n , and let B be diagonalizable with the following decomposition:

B  U diag λ1B, λ2B, , λ n BU−1, 2.17

where U ∈ R n×n is nonsingular Then

n



i1

Re λ i Bd n−i1



U−1AU

≤ trAB ≤n

i1

Re λ i Bd i



U−1AU

. 2.18

Proof Note that U−1AU iiis real; by the matrix theory we have

trAB  Re trAB  Re tr AU diag λ1B, λ2B, , λ n BU−1

 Re tr U−1AU diag λ1B, λ2B, , λ n B

 Ren

i1

λ i BU−1AU

ii

n

i1

Re

λ i BU−1AU

ii

n

i1



U−1AU

ii Re λ i B

n

i1

⎡

⎣U−1AU 



U−1AUT

2

⎤

⎦

ii

Re λ i B

n

i1



U−1AU

ii Re λ i B.

2.19

Since Re λ1B ≥ Re λ2B ≥ · · · ≥ Re λ n B ≥ 0, without loss of generality, we may assume

Re λB  Re λ1B, Re λ2B, , Re λ n B Next, we will prove the left-hand side of

2.18:

n



i1

Re λ i Bd n−i1U−1AU

≤n

i1

Re λ i Bd i



U−1AU

. 2.20

If

d

U−1AU

d n



U−1AU

, d n−1



U−1AU

, , d1



U−1AU

, 2.21

then we obtain the conclusion Now assume that there exists j < k such that d j U−1AU >

d k U−1AU, then

Re λ j Bd k



U−1AU

 Re λ k Bd j



U−1AU

− Re λ j Bd j



U−1AU

− Re λ k Bd k



U−1AU

 Re λ j B − Re λ k B d k



U−1AU

− d j



U−1AU ≤ 0.

2.22

We use dU−1AU to denote the vector of dU−1AU after changing d j U−1AU and

d k U−1AU, then

n



i1

σ i B  d i



U−1AU

≤n

i1

σ i Bd i



U−1AU

After a limited number of steps, we obtain the left-hand side of2.18 For the right-hand side

of2.18

n



i1

Re λ i Bd i



U−1AU

≤n

i1

Re λ i Bd iU−1AU

. 2.24

If

d

V T AU

d1

U−1AU

, d2

U−1AU

, , d n

U−1AU

, 2.25

then we obtain the conclusion Now assume that there exists j > k such that d j U−1AU <

d k U−1AU, then

σ j Bd k



U−1AU

 σ k Bd j



U−1AU

− σ j Bd j



U−1AU

− σ k Bd k



U−1AU

 σ j B − σ k B d k



U−1AU

− d j



U−1AU ≥ 0.

2.26

We use dU−1AU to denote the vector of dU−1AU after changing d j U−1AU and

d k U−1AU, then

n



i1

σ i Bd i



U−1AU

≤n

i1

σ i B  d i



U−1AU

After a limited number of steps, we obtain the right-hand side of2.18 Therefore, we have

n



i1

Re λ i Bd n−i1U−1AU

≤ trAB ≤n

i1

Re λ i Bd iU−1AU

. 2.28

Since trAB  trBA, applying 2.18 with B in lieu of A, we immediately have the following corollary

Corollary 2.9 Let A, B ∈ R n×n , and let A be diagonalizable with the following decomposition:

A  V diag λ1A, λ2A, , λ n AV−1, 2.29

where V ∈ R n×n is nonsingular Then

n



i1

Re λ i Ad n−i1V−1BV

≤ trAB ≤n

i1

Re λ i Ad iV−1BV

. 2.30

Theorem 2.10 Let A ∈ R n×n , B ∈ R n×n be normal Then

n



i1

Re λ i Bλ n−i1A

≤ trAB ≤n

i1

Re λ i Bλ iA

. 2.31

Proof Since B is normal, from 23, page 101, Theorem 2.5.4, we have

B  U diag λ1B, λ2B, , λ n BU−1, 2.32

where U ∈ R n×n is orthogonal Since U T  U−1and UU T  I, then for i  1, 2, , n, we have

λ i



U−1AU

 λ i



U T AU

 λ i

⎛

⎝U T AU U T AUT

2

⎞

⎠

 λ i

⎛

⎝U T

⎛

⎝AUU TAUU TT

2

⎞

⎠U

⎞

⎠

 λ i

⎛

⎝AUU T



AUU TT 2

⎞

⎠  λ i



A

.

2.33

In terms of Lemmas2.1and2.2,2.18 implies

n



i1

Re λ i Bλ n−i1



A

n

i1

Re λ i Bλ n−i1



U−1AU

≤n

i1

Re λ i Bd n−i1



U−1AU

≤ trAB ≤n

i1

Re λ i Bd i



U−1AU

≤n

i1

Re λ i Bλ iU−1AU

n

i1

Re λ i Bλ iA

.

2.34

This completes the proof

Note that if B ∈ S n , Re λ i B  λ i B, then from 2.34 we obtain 1.5 immediately.

This implies that2.18 improves 1.5

Since trAB  trBA, applying 2.31 with B in lieu of A, we immediately have the following corollary

Corollary 2.11 Let B ∈ R n×n , A ∈ R n×n be normal, then

n



i1

Re λ i Aλ n−i1B

≤ trAB ≤n

i1

Re λ i Aλ iB

. 2.35

3 Trace Bounds for the Solution of the Algebraic Riccati Equations

Komaroff 1994 in 16 obtained the following Let K be the positive semidefinite solution

of the ARE1.4 Then the trace of K has the upper bound given by

trK ≤ n

2λ1S  n

2 λ2

1S 4tr



QR−1

where S  R−1A T  AR−1.

In this section, by appling our new trace bounds inSection 2, we obtain some lower trace bounds for the solution of the algebraic Riccati equations Furthermore, we obtain some upper trace bounds which improve3.1 under certain conditions

Theorem 3.1 If 1/p  1/q  1, and K is the positive semidefinite solution of the ARE 1.4.

1 There are both, upper and lower, bounds:

λ n Rλ n S  λ n R





λ n S24/λ n R n

i1 λ p i R 1/p tr

QR−1

2 n

i1 λ p i R 1/p

≤ trK ≤ λ1S 



λ2

1S 4/c p,q n2−1/qλ1R n

i1 λ p i R 1/p tr

QR−1

2 n

i1 λ p i R 1/p /c p,q n2−1/qλ1R

.

3.2

2 If S ≥ 0, then the trace of K has the lower and upper bounds given by



1/c p,q n1−1/q

H 

1/c p,q n1−1/q

H24/λ n Rtr

QR−1

2/λ n R

≤ trK ≤ H 

 n

i1 λ p i S 2/p4/c p,q n2−1/qλ1Rtr

QR−1

3.3

where H denotes n

i1 λ p i S 1/p and  denotes n

i1 λ p i R 1/p

3 If S ≤ 0, then the trace of K has the lower and upper bounds given by

−n

i1λ i Sp1/p





n

i1λ i Sp2/p

 4/λ n R n

i1 λ p i R 1/p tr

QR−1

2 n

i1 λ p i R 1/p /λ n R

≤ trK ≤ c p,q n2−1/qλ1R

2 n

i1 λ p i R 1/p

×

⎧

⎨

⎩

1

c p,q n1−1/q

!

−n

i1

λ i Sp

"1/p



#

$! 1

c p,q n1−1/qN

"2

c p,q n2−1/qλ1R Str



QR−1

⎫

⎪

⎪,

3.4

where N denotes n

i1 |λ i S| p1/p

and S denotes n

i1 λ p i R 1/p ,

We have

p

r M q k − m p

r m q k

p

M r m k M k q − m r M k m q k 1/p

q

m r M k M p r − M r m k m p r

1/q ,

M r  λ1R, m r  λ n R, M k  λ1K, m k  λ n K,

p

s M q k − m p

s m q k

p

M s m k M q k − m s M k m q k 1/p

q

m s M k M p s − M1m k m p s

1/q ,

M s  λ1S, m s  λ n S, S  R−1A T  AR−1.

3.5

Proof. 1 Multiply 1.4 on the right and on the left by R−1/2to get

R −1/2 QR −1/2  K T

1K1− R −1/2

A T K  KA

R −1/2 , 3.6

where K1 R 1/2 KR −1/2 Take the trace of all terms in3.6 to get

tr

K1T K1



− trR−1A T K  KAR−1

− trQR−1

 0. 3.7

Since K is positive semidefiniteness, λK  Re λK, trK n

i1 λ i K n

i1 Re λ i K,

and fromLemma 2.7, we have

trK

n



i1

λ i KK n

i1

λ2i K ≤

! n



i1

λ i K

"2

 trK2

By Cauchy-Schwarz inequality2.8, it can be shown that

n



i1

λ i KK n

i1

λ2i K ≥

n

i1 λ i K2

n  trK2

n . 3.10

Since K, Q are positive semidefiniteness, R is positive definiteness, then by 1.5, note that for

i  1, 2, , n, λ i R−1  λ i R−1  1/λ n−i1 R, and considering 2.6, 3.8, and 3.9, we

have

tr

K T1K1



 trR−1KRK

≤n

i1

λ i



R−1

λ i KRK

n

i1

λ i KRK

λ n−i1 R ≤

1

λ n RtrKRK

≤ 1

λ n R

! n



i1

λ p i R

"1/p

trK2

.

3.11

Note that S  R−1A T  AR−1, λ i S  λ i S, then from 1.5 we have

λ n StrK ≤n

i1

λ n−i1



R−1A T  AR−1

λ i K

≤ trR−1A T K  AR−1K

 trR−1A T K  KAR−1

≤n

i1

λ i

R−1A T  AR−1

λ i K ≤ λ1StrK.

3.12

Combining3.11 with 3.12, we obtain

1

λ n R

!n

i1

λ p i R

"1/p

trK2− trKλ n S − trQR−1

≥ 0. 3.13

Solving3.13 for trK yields the left-hand side of 3.2

Since K, Q are positive semidefiniteness, R is positive definiteness, then by 1.5, note that for i  1, 2, , n, λ n−i1 R−1  λ n−i1 R−1  1/λ i R, and considering 2.6, 3.8,

and3.10, we have

tr

K T1K1



 trR−1KRK

≥n

i1

λ n−i1

R−1

λ i KRK

n

i1

λ i KRK

λ i R ≥

1

λ1RtrKRK

≥ 1

c p,q λ1R

!n

i1

λ p i R

"1/p!n

i1

λ q i K2

"1/q

c p,q n2−1/qλ1R

!n

i1

λ p i R

"1/p

trK2

.

3.14