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Volume 2009, Article ID 620758,18pages doi:10.1155/2009/620758

Research Article

New Trace Bounds for the Product of Two Matrices and Their Applications in the Algebraic Riccati Equation

Jianzhou Liu and Juan Zhang

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 September 2008; Accepted 19 February 2009

Recommended by Panayiotis Siafarikas

By using singular value decomposition and majorization inequalities, we propose new inequalities for the trace of the product of two arbitrary real square matrices. These bounds improve and extend the recent results. Further, we give their application in the algebraic Riccati equation. Finally, numerical examples have illustrated that our results are effective and superior.

Copyrightq2009 J. Liu and J. Zhang. 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

In the analysis and design of controllers and filters for linear dynamical systems, the Riccati equation is of great importance in both theory and practicesee1–5. Consider the following linear systemsee4:

xt ˙ Axt But, x0 x0, 1.1

with the cost

J

0

xTQxuTu

dt. 1.2

Moreover, the optimal control rateuand the optimal costJof1.1and1.2are uP x, PBTK,

JxT0Kx0,

1.3

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where x0Rn is the initial state of the systems1.1 and1.2,K is the positive definite solution of the following algebraic Riccati equationARE:

ATKKAKRK−Q, 1.4

with R BBT andQ are symmetric positive definite matrices. To guarantee the existence of the positive definite solution to1.4, we shall make the following assumptions: the pair A, Ris stabilizable, and the pairQ, Ais 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 solve1.4, 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 theARE 6–12.

In addition, from2,6, we know that an interpretation of trKis that trK/nis the average value of the optimal costJasx0varies over the surface of a unit sphere. Therefore, consider its applications, it is important to discuss trace bounds for the product of two matrices.

Most available results are based on the assumption that at least one matrix is symmetric 7,8,11,12. However, it is important and difficult to get an estimate of the trace bounds when any matrix in the product is nonsymmetric in theory and practice. There are some results in13–15.

In this paper, we propose new trace bounds for the product of two general matrices.

The new trace bounds improve the recent results. Then, for their application in the algebraic Riccati equation, we get some upper and lower bounds.

In the following, letRn×ndenote the set ofn×nreal matrices. Letx x1, x2, . . . , xnbe a real n-element array which is reordered, and its elements are arranged in nonincreasing order. That is, x1x2 ≥ · · · ≥ xn. Let |x| |x1|,|x2|, . . . ,|xn|. For A aijRn×n, let dA d1A, d2A, . . . , dnA, λA λ1A, λ2A, . . . , λnA, σA σ1A, σ2A, . . . , σnAdenote the diagonal elements, the eigenvalues, the singular values ofA, respectively, Let trA, AT denote the trace, the transpose ofA, respectively. We define AiiaiidiA,A AAT/2. The notationA >0A≥0is used to denote thatAis a symmetric positive definitesemidefinitematrix.

Letα, βbe two realn-element arrays. If they satisfy k

i1

αik

i1

βi, k1,2, . . . , n, 1.5

then it is said thatαis controlled weakly byβ, which is signed byα≺wβ.

Ifα≺wβand

n i1

αin

i1

βi, 1.6

then it is said thatαis controlled byβ, which is signed byαβ.

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Therefore, considering the application of the trace bounds, many scholars pay much attention to estimate the trace bounds for the product of two matrices.

Marshall and Olkin in16have showed that for anyA, BRn×n,then

n

i1

σiiB≤trAB≤n

i1

σiiB. 1.7

Xing et al. in13have observed another result. LetA, BRn×nbe arbitrary matrices with the following singular value decomposition:

BUdiag

σ1B, σ2B, . . . , σnB

VT, 1.8

whereU, VRn×nare orthogonal. Then

λnASn

i1

σiB≤trAB≤λ1ASn

i1

σiB, 1.9

whereSUVT is orthogonal.

Liu and He in14have obtained the following: letA, BRn×nbe arbitrary matrices with the following singular value decomposition:

BUdiag

σ1B, σ2B, . . . , σnB

VT, 1.10

whereU, VRn×nare orthogonal. Then

1≤i≤nmin

VTAU

ii

n i1

σiB≤trAB≤max

1≤i≤n

VTAU

ii

n i1

σiB. 1.11

F. Zhang and Q. Zhang in15have obtained the following: letA, BRn×nbe arbitrary matrices with the following singular value decomposition:

BUdiag

σ1B, σ2B, . . . , σnB

VT, 1.12

whereU, VRn×nare orthogonal. Then n

i1

σin−i1AS≤trAB≤n

i1

σiiAS, 1.13

whereSUVT is orthogonal. They show that1.13has improved1.9.

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2. Main Results

The following lemmas are used to prove the main results.

Lemma 2.1see16, page 92, H.2.c. Ifx1 ≥ · · · ≥ xn, y1 ≥ · · · ≥ yn andxy, then for any real arrayu1≥ · · · ≥un,

n i1

xiuin

i1

yiui. 2.1

Lemma 2.2see16, page 95, H.3.b. Ifx1 ≥ · · · ≥ xn, y1 ≥ · · · ≥ynandx≺wy, then for any real arrayu1≥ · · · ≥un0,

n i1

xiuin

i1

yiui. 2.2

Remark 2.3. Note that ifx≺wy, then fork 1,2, . . . , n,x1, . . . , xkwy1, . . . , yk. Thus fromLemma 2.2, we have

k i1

xiuik

i1

yiui, k1,2, . . . , n. 2.3

Lemma 2.4see16, page 218, B.1. LetAATRn×n, then

dAλA. 2.4

Lemma 2.5see16, page 240, F.4.a. LetARn×n, then

λ

AAT 2

w

λ

AAT 2

wσA. 2.5

Lemma 2.6see17. Let 0< m1akM1,0< m2bkM2, k1,2, . . . , n,1/p1/q1.

Then

n k1

akbkn

k1

apk

1/p n

k1

bqk 1/q

cp,q

n k1

akbk, 2.6

where

cp,q Mp1M2qmp1mq2 p

M1m2Mq2m1M2mq21/p q

m1M2Mp1M1m2mp11/q. 2.7

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Note that if m1 0, m2/0 or m2 0, m1/0, obviously,2.6 holds. If m1 m2 0, choose cp,q ∞, then2.6also holds.

Remark 2.7. Ifpq2, then we obtain Cauchy-Schwartz inequality

n k1

akbkn

k1

a2k

1/2 n

k1

b2k 1/2

c2

n k1

akbk, 2.8

where

c2 M1M2

m1m2

m1m2

M1M2

. 2.9

Remark 2.8. Note that

p→ ∞lim

ap1ap2· · ·apn

1/p

max

1≤k≤n

ak

,

limp→ ∞

q1

cp,qlim

p→ ∞

q1

Mp1M2qmp1mq2 p

M1m2Mq2m1M2mq21/p q

m1M2Mp1M1m2mp11/q

lim

p→ ∞

q1

Mp1

Mq2−m1/M1pmq2 M1/p1

p

m2M2q−m1/M1M2mq21/p

Mq/p1 q

m1M2−M1m2m1/M1p1/q

lim

p→ ∞

q1

M2

M1/pp/q−p1 m1M2

lim

p→ ∞

q1

1 M11/p−1m1

M1

m1.

2.10

Letp → ∞, q → 1 in2.6, then we obtain

m1

n k1

bkn

k1

akbkM1

n k1

bk. 2.11

Lemma 2.9. Ifq1,ai≥0 i1,2, . . . , n, then

1 n

n i1

ai

q

≤ 1 n

n i1

aqi. 2.12

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Proof. 1Note thatq1, orai0i1,2, . . . , n, 1

n n

i1

ai

q 1

n n

i1

aqi. 2.13

2Ifq > 1,ai > 0, forx > 0, choosefx xq, thenfx qxq−1 > 0 and f x qq−1xq−2 > 0. Thus,fxis a convex function. Asai > 0 and1/nn

i1ai > 0, from the property of the convex function, we have

1 n

n i1

ai

q

f 1 n

n i1

ai

≤ 1 n

n i1

fai

1 n

n i1

aqi. 2.14

3Ifq >1, without loss of generality, we may assumeai 0 i1, . . . , r, ai>0i r1, . . . , n. Then from2, we have

1 nr

q n

i1

ai

q

1 nr

n i1

ai

q

≤ 1 nr

n i1

aqi. 2.15

Sincen−r/nq≤n−r/n, thus 1

n n

i1

ai

q

nr n

q 1 nr

q n

i1

ai

q

nr n

1 nr

n i1

aqi 1 n

n i1

aqi. 2.16

This completes the proof.

Theorem 2.10. Let A, BRn×n be arbitrary matrices with the following singular value decomposition:

BUdiagσ1B, σ2B, . . . , σnBVT, 2.17

whereU, VRn×nare orthogonal. Then n

i1

σiBdn−i1

VTAU

≤trAB≤n

i1

σiBdi

VTAU

. 2.18

Proof. By the matrix theory we have trAB tr

AUdiag

σ1B, σ2B, . . . , σnB VT tr

VTAUdiag

σ1B, σ2B, . . . , σnB n

i1

σiB

VTAU

ii.

2.19

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Sinceσ1B≥σ2B≥ · · · ≥σnB≥0, without loss of generality, we may assumeσB σ1B, σ2B, . . . , σnB. Next, we will prove the left-hand side of2.18:

n i1

σiBdn−i1

VTAU

n

i1

σiBdi

VTAU

. 2.20

If

d

VTAU

dn

VTAU , dn−1

VTAU

, . . . , d1

VTAU

, 2.21

we obtain the conclusion. Now assume that there exists j < k such that djVTAU >

dkVTAU,then σjBdk

VTAU

σkBdj

VTAU

σjBdj

VTAU

σkBdk

VTAU

σjB−σkB dk

VTAU

dj

VTAU

≤0.

2.22

We usedV TAUto denote the vector ofdVTAUafter changingdjVTAUanddkVTAU, then

n i1

σiBdi

VTAU

n

i1

σiBdi

VTAU

. 2.23

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

i1

σiBdi

VTAU

n

i1

σiBdi

VTAU

. 2.24

If

d

VTAU

d1VTAU , d2

VTAU

, . . . , dn

VTAU

, 2.25

we obtain the conclusion. Now assume that there exists j > k such that djVTAU <

dkVTAU,then σjBdk

VTAU

σkBdj

VTAU

σjBdj

VTAU

σkBdk

VTAU

σjB−σkB dk

VTAU

dj

VTAU

≥0.

2.26

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We usedV TAUto denote the vector ofdVTAUafter changingdjVTAUanddkVTAU, then

n i1

σiBdi

VTAU

n

i1

σiBdi

VTAU

. 2.27

After limited steps, we obtain the right-hand side of2.18. Therefore, n

i1

σiBdn−i1

VTAU

≤trAB≤n

i1

σiBdi

VTAU

. 2.28

This completes the proof.

Since trAB trBA,applying2.18withBin lieu ofA,we immediately have the following corollary.

Corollary 2.11. Let A, BRn×n be arbitrary matrices with the following singular value decomposition:

APdiagσ1A, σ2A, . . . , σnAQT, 2.29

whereP, QRn×nare orthogonal. Then n

i1

σiAdn−i1 QTBP

≤trAB≤n

i1

σiAdiQTBP. 2.30

Now using2.18and2.30, one finally has the following theorem.

Theorem 2.12. Let A, BRn×n be arbitrary matrices with the following singular value decompositions, respectively:

APdiag

σ1A, σ2A, . . . , σnA QT, BUdiag

σ1B, σ2B, . . . , σnB VT,

2.31

whereP, Q, U, VRn×nare orthogonal. Then

max n

i1

σiAdn−i1 QTBP

, n

i1

σiBdn−i1

VTAU

≤trAB≤min n

i1

σiBdi

VTAU ,

n i1

σiAdi

QTBP .

2.32

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Remark 2.13. We point out that2.18improves1.11. In fact, it is obvious that

1≤i≤nmin

VTAU

ii

n i1

σiB≤n

i1

σiBdn−i1

VTAU

≤trAB≤n

i1

σiBdi

VTAU

≤max

1≤i≤n

VTAU

ii

n i1

σiB.

2.33

This implies that2.18improves1.11.

Remark 2.14. We point out that2.18improves1.13. Since for i 1, . . . , n,σiB ≥ 0 and diVTAU diVTAU VTAUT/2, from Lemmas2.1and2.4, then2.18implies

n i1

σin−i1 VTAU

VTAUT

2

n

i1

σiBdn−i1 VTAU

VTAUT

2

≤trAB

n

i1

σiBdi VTAU

VTAUT

2

n

i1

σii VTAU

VTAUT

2

.

2.34

In fact, fori1,2, . . . , n, we have

λi VTAU VTAUT 2

λi

VTAUVT AUVTT

2 V

λi AUVT AUVTT 2

λiAS.

2.35

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Then2.34can be rewritten as n i1

σin−i1AS≤n

i1

σiBdn−i1

VTAU

≤trAB

n

i1

σiBdi

VTAU

n

i1

σiiAS.

2.36

This implies that2.18improves1.13.

Remark 2.15. We point out that1.13improves1.7. In fact, fromLemma 2.5, we have

λAS≺wσAS. 2.37

SinceSis orthogonal,σAS σA. Then2.37is rewritten as follows:λAS≺wσA.By usingσ1B≥σ2B≥ · · · ≥σnB≥0 andLemma 2.2, we obtain

n i1

σiiAS≤n

i1

σiiA. 2.38

Note thatλi−AS −λn−i1AS, fromLemma 2.2and2.38, we have

n

i1

σin−i1AS n

i1

σii−AS

n

i1

σiBλiAS≤n

i1

σiiA.

2.39

Thus, we obtain

n

i1

σiiA≤n

i1

σin−i1AS. 2.40

Both2.38and2.40show that1.13is tighter than1.7.

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3. Applications of the Results

Wang et al. in6have obtained the following: letKbe the positive semidefinite solution of the ARE1.4. Then the trace of matrixKhas the lower and upper bounds given by

λnA

λnA2

λ1RtrQ

λ1R ≤trK≤ λ1A

λ1A2

λnR/ntrQ

λnR/n .

3.1

In this section, we obtain the application in the algebraic Riccati equation of our results including3.1. Some of our results and3.1cannot contain each other.

Theorem 3.1. If 1/p1/q1 andKis the positive semidefinite solution of the ARE1.4, then

1the trace of matrixKhas the lower and upper bounds given by

λnA

λnA2

n

i1λpiR1/p

trQ n

i1λpiR1/p

≤trK≤ λ1A

λ1A2

1/cp,qn2−1/qn

i1λpiR1/p trQ 1/cp,qn2−1/qn

i1λpiR1/p .

3.2

2IfAAT/2≥0, then the trace of matrixKhas the lower and upper bounds given by

1/cp,qn1−1/qn

i1λpiA1/p

1/cp,qn1−1/qn

i1λpiA1/p2 n

i1λpiR1/p

trQ n

i1λpiR1/p

≤trK

n

i1λpiA1/p

n

i1λpiA2/p

1/cp,qn2−1/qn

i1λpiR1/p trQ 1/cp,qn2−1/qn

i1λpiR1/p .

3.3

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3IfAAT/2≤0, then the trace of matrixKhas the lower and upper bounds given by

n

i1λn−i1Ap1/p

n

i1λn−i1Ap2/p

1/cp,qn2−1/qn

i1λpiR1/p trQ 1/cp,qn2−1/qn

i1λpiR1/p

≤trK

−1/cp,qn1−1/qn

i1λn−i1Ap1/p

n

i1λpiR1/p

1/cp,qn1−1/qn

i1λiAp1/p2n

i1λpiR1/p

trQ n

i1λpiR1/p ,

3.4 where

cp,q MprMqkmprmqk p

MrmkMkqmrMkmqk1/p q

mrMkMprMrmkmpr

1/q,

Mr λ1R, mrλnR, Mkλ1K, mkλnK,

cp,q Mp1Mqkmp1mqk p

M1mkMqkm1Mkmqk1/p q

m1MkMp1M1mkmp11/q, M1λ1A, m1λnA.

3.5

Proof. 1Take the trace in both sides of the matrix ARE1.4to get

tr ATK

trKA−trKRK −trQ. 3.6

SinceK is symmetric positive definite matrix,λK σK, trK n

i1σiK,and from Lemma 2.9, we have

trK n1−1/q

tr Kq1/q

≤trK, 3.7

n i1

σiKK n

i1

σi2 K≤ n

i1

σiK 2

trK2

. 3.8

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By the Cauchy-Schwartz inequality2.8, it can be shown that n

i1

σiKK n

i1

σi2 K≥ n

i1σiK2

n

trK n

2

. 3.9

Note that

K2Udiagλ21K, λ22K, . . . , λ2nKUT, 3.10

K, Q, R > 0, λiUTRU λiR i 1, . . . , n,then by2.34, use2.6, considering3.7 and3.9, we have

trKRK tr K2R

n

i1

λii K2

≥ 1 cp,q

n

i1

λpiR 1/p n

i1

σiq

K21/q

≥ 1

cp,qn2−1/q n

i1

λpiR 1/p

trK2

.

3.11

From2.34, note thatλiUTAU λiAandλiUTATU λiAT i 1, . . . , n,then we obtain

trAK≤n

i1

λiiK≤λ1An

i1

σiK,

trATKn

i1

λiATσiK≤λ1ATn

i1

σiK.

3.12

It is easy to see that

trATK trKA≤ λ1

AT

trK λ1AtrK trK 2λ1

ATA 2

trK 2λ1AtrK.

3.13

Combine3.11and3.13, we obtain 1

cp,qn2−1/q n

i1

λpiR 1/p

trK2−2trKλnA−trQ≤0. 3.14

Solving3.14for trKyields the right-hand side of the inequality3.2. Similarly, we can obtain the left-hand side of the inequality3.2.

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2 Note that when A AT/2 ≥ 0, λiUTAU λiA and λiUTATU λiAT i1, . . . , n,by2.34,2.6and3.7, we have

tr ATK

n

i1

λpi

AT1/p trK,

trKA≤ n

i1

λpiA 1/p

trK.

3.15

Thus,

tr ATK

trKA≤ n

i1

λpi

AT1/p

n

i1

λpiA

trK

≤2 n

i1

λpi

ATA 2

1/p trK

2 n

i1

λpiA 1/p

trK.

3.16

From3.11and3.16, with similar argument to1, we can obtain3.3easily.

3 Note that when A AT/2 ≤ 0, by 3.3, we obtain 3.4 immediately. This completes the proof.

Remark 3.2. FromRemark 2.7andTheorem 3.1, letp2, q2 in3.2, then we obtain

λnA

λnA2

n

i1λ2iR1/2trQ n

i1λ2iR1/2

≤trK≤ λ1A

λ1A2

1/c1n3/2n

i1λ2iR1/2 trQ 1/c1n3/2n

i1λ2iR1/2 ,

3.17

wherec1

MrMk/mrmk

mrmk/MrMk.

Remark 3.3. FromRemark 2.7andTheorem 3.1, letp → ∞, q → 1 in3.2, then we obtain 3.1immediately.

4. Numerical Examples

In this section, firstly, we will give two examples to illustrate that our new trace bounds are better than the recent results. Then, to illustrate the application in the algebraic Riccati

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equation of our results will have different superiority if we choose differentpandq, we will give two examples whenp2, q2,andp → ∞, q → 1.

Example 4.1see13. Now let

A

⎜⎜

⎜⎝

0.9140 0.6989 0.6062 0.2309 0.0169 0.04501 0.3471 0.5585 0.0304

⎟⎟

⎟⎠,

B

⎜⎜

⎜⎝

0.9892 0.1140 0.1233 0.0410 0.3096 0.5125 0.0476 0.7097 0.0962

⎟⎟

⎟⎠.

4.1

NeitherAnorBis symmetric. In this case, the results of6–12are not valid.

Using1.9we obtain

0.78≤trAB≤1.97. 4.2

Using1.11yields

0.8611≤trAB≤1.9090. 4.3

By2.18, we obtain

1.0268≤trAB≤1.7524, 4.4

where both lower and upper bounds are better than those of4.2and4.3.

Example 4.2. Let

A

⎜⎜

⎜⎜

⎜⎜

0.0624 0.8844 0.2782 0.0389 0.7163 0.6565 0.2923 0.5980 0.5502 0.2660 0.5486 0.3376 0.1134 0.5739 0.3999 0.2792

⎟⎟

⎟⎟

⎟⎟

,

B

⎜⎜

⎜⎜

⎜⎜

1.7205 0.6542 1.3030 0.8813 0.6542 0.0631 0.6191 0.2696 1.3030 0.6191 0.4715 0.7551 0.8813 0.2696 0.7551 0.4584

⎟⎟

⎟⎟

⎟⎟

.

4.5

NeitherAnorBis symmetric. In this case, the results of6–12are not valid.

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Using1.7yields

−6.1424≤trAB≤6.1424. 4.6

From1.9we have

−1.5007≤trAB≤5.0110. 4.7

Using1.11yields

−3.1058≤trAB≤6.0736. 4.8

By1.13, we obtain

−0.7267≤trAB≤4.3399. 4.9

The bound in2.18yields

−0.5375≤trAB≤4.2659. 4.10

Obviously,4.10is tighter than4.6,4.7,4.8and4.9.

Example 4.3. Consider the systems1.1,1.2with

A

⎜⎜

⎜⎝

−5 −2 4 2 3 −1 1 0 −3

⎟⎟

⎟⎠, BBT

⎜⎜

⎜⎝ 8 2 3 2 7 4 3 4 9

⎟⎟

⎟⎠, Q

⎜⎜

⎜⎝

538 440 266 440 441 321 266 321 296

⎟⎟

⎟⎠. 4.11

Moreover, the corresponding ARE1.4with R BBT,A, Ris stabilizable and Q, Ais observable.

Using3.17yields

8.5498≤trK≤47.9041. 4.12

Using3.1we obtain

9.0132≤trK≤19.0099, 4.13

where both lower and upper bounds are better than those of4.12.

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Example 4.4. Consider the systems 1.1,1.2with

A

⎜⎜

⎜⎝

−6.0 1.5 2.0 0.0 −2.0 −3.0 2.5 4.0 −1.5

⎟⎟

⎟⎠, BBT

⎜⎜

⎜⎝

4.0 1.0 2.0 1.0 2.0 0.5 2.0 0.5 2.5

⎟⎟

⎟⎠, Q

⎜⎜

⎜⎝

17.5 7.45 3.465 7.45 9.7 7.845 3.465 7.845 9.905

⎟⎟

⎟⎠.

4.14 Moreover, the corresponding ARE1.4with R BBT,A, Ris stabilizable and Q, Ais observable.

Using3.1we obtain

1.6039≤trK≤5.6548. 4.15

Using3.17yields

1.6771≤trK≤5.5757, 4.16

where both lower and upper bounds are better than those of4.15.

5. Conclusion

In this paper, we have proposed lower and upper bounds for the trace of the product of two arbitrary real matrices. We have showed that our bounds for the trace are the tightest among the parallel trace bounds in nonsymmetric case. Then, we have obtained the application in the algebraic Riccati equation of our results. Finally, numerical examples have illustrated that our bounds are better than the recent results.

Acknowledgments

The author thanks the referee for the very helpful comments and suggestions. The work was supported in part by National Natural Science Foundation of China10671164, Science and Research Fund of Hunan Provincial Education Department06A070.

References

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4 M.-L. Ni, “Existence condition on solutions to the algebraic Riccati equation,” Acta Automatica Sinica, vol. 34, no. 1, pp. 85–87, 2008.

5 K. Ogata, Modern Control Engineering, Prentice-Hall, Upper Saddle River, NJ, USA, 3rd edition, 1997.

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6 S.-D. Wang, T.-S. Kuo, and C.-F. Hsu, “Trace bounds on the solution of the algebraic matrix Riccati and Lyapunov equation,” IEEE Transactions on Automatic Control, vol. 31, no. 7, pp. 654–656, 1986.

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