Trace Inequalities for Matrix Products and Trace Bounds for the Solution of the Algebraic Riccati Equations

Volume 2009, Article ID 101085,17pages 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,

Juan Zhang,

and Yu Liu

1Department of Mathematic Science and Information Technology, Hanshan Normal University, Chaozhou, Guangdong 521041, China

2Department of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan 411105, China

Correspondence should be addressed to Jianzhou Liu,[email protected]

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.

Copyrightq2009 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 systemssee1–5. For example, consider the following linear systemsee5:

xt ˙ Axt But, x0 x0, 1.1

with the cost

J _∞

0

x^TQxu^Tu

dt. 1.2

The optimal control rateu^∗the optimal costJ^∗of1.1and1.2are u^∗P x, P B^TK,

J^∗x^T₀Kx0, 1.3

wherex0 ∈Rⁿis the initial state of system1.1and1.2andK is the positive semidefinite solution of the following algebraic Riccati equationARE:

A^TKKA−KRK−Q, 1.4

withR BB^TandQbeing 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, 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 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 AREsee6–

16. Moreover, in terms of2,6, we know that an interpretation of trKis that trK/nis the average value of the optimal costJ^∗asx0varies 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, and18is the tightest among the parallel results.

In 1995, Lasserre showed18the following given any matrixA∈R^n×n, B∈Sⁿ,then the following.

n i1

λi

A

λn−i1B≤trAB≤ⁿ

i1

λi

A

λiB, 1.5

whereA AA^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 Sⁿ denote the subset of R^n×n consisting of symmetric matrices. For A aij ∈ R^n×n, we assume that trA, A⁻¹, A^T, dA d1A, . . . , dnA, σA σ1A, . . . , σnA denote the trace, the inverse, the transpose, the diagonal elements, the singular values of A, respectively, and define A_ii aii diA. If A ∈ R^n×n is an arbitrary symmetric matrix, then λA λ1A, . . . , λnAand ReλA Reλ1A, . . . ,ReλnAdenote the eigenvalues

and the real part of eigenvalues ofA. Supposex x1, x2, . . . , xnis a realn-element array such as dA, σA, λA, ReλA which is reordered, and its elements are arranged in nonincreasing order; that is,x1 ≥ x2 ≥ · · · ≥ xn. The notation A > 0 A ≥ 0is used to denote thatAis a symmetric positive definitesemidefinitematrix.

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

k i1

α_i≤^k

i1

β_i, k1,2, . . . , n, 2.1

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

Ifα≺wβand

n i1

αiⁿ

i1

βi, 2.2

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

The following lemmas are used to prove the main results.

Lemma 2.1see21, Page 92, H.2.c. Ifx1 ≥ · · · ≥xn, y1 ≥ · · · ≥ ynandx≺ y, then for any real arrayu1≥ · · · ≥un,

n i1

xiui≤ⁿ

i1

yiui. 2.3

Lemma 2.2see21, Page 218, B.1. LetAA^T ∈R^n×n, then

dA≺λA. 2.4

Lemma 2.3see21, Page 240, F.4.a. LetA∈R^n×n, then

λ

AA^T 2

≺w

λ

AA^T 2

≺wσA. 2.5

Lemma 2.4see22. Let 0< m1≤ak≤M1, 0< m2≤bk≤M2, k1,2, . . . , n,1/p1/q1, then

n k1

akbk≤ _n

k1

a^p_k

_{1/p n}

k1

b^q_{k 1/q}

≤cp,q

n k1

akbk, 2.6

where

cp,q M^p₁M₂^q−m^p₁m^q₂ p

M1m2M₂^q−m1M2m^q_21/p q

m1M2M^p₁−M1m2m^p_11/q. 2.7

Note that ifm1 0, m2/0 orm2 0, m1/0, obviously,2.6holds. Ifm1 m2 0, choosecp,q ∞, then2.6also holds.

Remark 2.5. Ifpq2, then we obtain Cauchy-Schwarz inequality:

n k1

akbk≤ _n

k1

a²_k

_1/2n

k1

b²_{k 1/2}

≤c2

n k1

akbk, 2.8

where

c2

⎛

⎝

M1M2

m1m2

m1m2

M1M2

⎞

⎠. 2.9

Remark 2.6. Note that

plim→ ∞

a^p₁a^p₂· · ·a^pn

1/p

max

1≤k≤n{ak},

p→ ∞lim

q→1

cp,q lim

p→ ∞ q→1

M^p₁M₂^q−m^p₁m^q₂ p

M1m2M^q₂−m1M2m^q₂1/p

q

m1M2M^p₁−M1m2m^p₁1/q

lim

p→ ∞ q→1

M^p₁ M^q₂−m1/M1^pm^q₂ M^1/p₁ p

m2M₂^q−m1/M1M2m^q_21/p M^q/p₁

q

m1M2−M1m2m1/M1^p1/q

lim

p→ ∞ q→1

M2

M^1/pp/q−p₁ m1M2

_{p→ ∞}lim

q→1

1 M^1/p−1₁ m1

M1

m1.

2.10

Letp → ∞, q → 1 in2.6, then we obtain

m1

n k1

bk≤ⁿ

k1

akbk≤M1

n k1

bk. 2.11

Lemma 2.7. Ifq≥1,ai≥0 i1,2, . . . , n, then 1

n n

i1

ai

q

≤ 1 n

n i1

a^q_i. 2.12

Proof. 1Note that forq1, orai0 i1,2, . . . , n, 1

n n

i1

ai

_q 1

n n

i1

a^q_i. 2.13

2Ifq > 1, ai >0, forx >0, choosefx x^q, thenfx qx^q−1 > 0 andf x qq−1x^q−2 > 0. Thus,fxis a convex function. Asai >0 and1/n_n

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

a^q_i. 2.14

3Ifq >1, without loss of generality, we may assumeai0i1, . . . , r, ai >0 i r1, . . . , n. Then from2, we have

1 n−r

_qn

i1

ai

_q

1 n−r

n i1

ai

_q

≤ 1 n−r

n i1

a^q_i. 2.15

Sincen−r/n^q≤n−r/n, thus 1

n n

i1

ai

_q

n−r n

_q 1 n−r

_qn

i1

ai

_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. LetA, B∈R^n×n, and letBbe diagonalizable with the following decomposition:

BUdiagλ1B, λ2B, . . . , λnBU⁻¹, 2.17

whereU∈R^n×nis nonsingular. Then

n i1

ReλiBdn−i1

U⁻¹AU

≤trAB≤ⁿ

i1

ReλiBdi

U⁻¹AU

. 2.18

Proof. Note thatU⁻¹AU_iiis real; by the matrix theory we have

trAB Re trAB Re tr AUdiagλ1B, λ2B, . . . , λnBU⁻¹ Re tr U⁻¹AUdiagλ1B, λ2B, . . . , λnB

Re n

i1

λiB

U⁻¹AU

ii

ⁿ

i1

Re λiB

U⁻¹AU

ii

ⁿ

i1

U⁻¹AU

iiReλiB ⁿ

i1

⎡

⎣U⁻¹AU

U⁻¹AUT

2

⎤

⎦

ii

ReλiB

ⁿ

i1

U⁻¹AU

iiReλiB.

2.19

Since Reλ1B≥Reλ2B≥ · · · ≥ReλnB≥0, without loss of generality, we may assume ReλB Reλ1B,Reλ2B, . . . ,ReλnB. Next, we will prove the left-hand side of 2.18:

n i1

Reλ_iBdn−i1

U⁻¹AU

≤ⁿ

i1

Reλ_iBdi

U⁻¹AU

. 2.20

If

d

U⁻¹AU

dn

U⁻¹AU , dn−1

U⁻¹AU

, . . . , d1

U⁻¹AU

, 2.21

then we obtain the conclusion. Now assume that there existsj < k such thatdjU⁻¹AU >

dkU⁻¹AU,then

Reλ^jBdk

U⁻¹AU

Reλ_kBdj

U⁻¹AU

−Reλ^jBdj

U⁻¹AU

−Reλ_kBdk

U⁻¹AU

Reλ^jB−Reλ_kB dk

U⁻¹AU

−dj

U⁻¹AU

≤0.

2.22

We use dU ⁻¹AU to denote the vector of dU⁻¹AU after changing djU⁻¹AU and dkU⁻¹AU, then

n i1

σiBdi

U⁻¹AU

≤ⁿ

i1

σiBdi

U⁻¹AU

. 2.23

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

n i1

Reλ_iBdi

U⁻¹AU

≤ⁿ

i1

Reλ_iBd_i

U⁻¹AU

. 2.24

If

d

V^TAU

d₁

U⁻¹AU , d₂

U⁻¹AU

, . . . , d_n

U⁻¹AU

, 2.25

then we obtain the conclusion. Now assume that there existsj > k such thatdjU⁻¹AU <

dkU⁻¹AU,then

σ^jBdk

U⁻¹AU

σ_kBdj

U⁻¹AU

−σ^jBdj

U⁻¹AU

−σ_kBdk

U⁻¹AU

σ^jB−σkB dk

U⁻¹AU

−dj

U⁻¹AU

≥0.

2.26

We use dU ⁻¹AU to denote the vector of dU⁻¹AU after changing djU⁻¹AU and dkU⁻¹AU, then

n i1

σiBdi

U⁻¹AU

≤ⁿ

i1

σiBdi

U⁻¹AU

. 2.27

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

n i1

Reλ_iBd_n−i1

U⁻¹AU

≤trAB≤ⁿ

i1

Reλ_iBd_i

U⁻¹AU

. 2.28

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

Corollary 2.9. LetA, B∈R^n×n, and letAbe diagonalizable with the following decomposition:

AVdiagλ1A, λ2A, . . . , λnAV⁻¹, 2.29

whereV ∈R^n×nis nonsingular. Then

n i1

Reλ_iAd_n−i1

V⁻¹BV

≤trAB≤ⁿ

i1

Reλ_iAd_i

V⁻¹BV

. 2.30

Theorem 2.10. LetA∈R^n×n,B∈R^n×nbe normal. Then

n i1

ReλiBλ_n−i1 A

≤trAB≤ⁿ

i1

ReλiBλ_i A

. 2.31

Proof. SinceBis normal, from23, page 101, Theorem 2.5.4, we have

BUdiagλ1B, λ2B, . . . , λnBU⁻¹, 2.32

whereU∈R^n×nis orthogonal. SinceU^T U⁻¹andUU^T I, then fori1,2, . . . , n, we have

λi

U⁻¹AU λi

U^TAU

λi

⎛

⎝U^TAU

U^TAUT

2

⎞

⎠

λi

⎛

⎝U^T

⎛

⎝AUU^T

AUU^TT

2

⎞

⎠U

⎞

⎠

λi

⎛

⎝AUU^T

AUU^T_T 2

⎞

⎠λi

A

.

2.33

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

n i1

ReλiBλn−i1

A

ⁿ

i1

ReλiBλn−i1

U⁻¹AU

≤ⁿ

i1

ReλiBdn−i1

U⁻¹AU

≤trAB≤ⁿ

i1

ReλiBdi

U⁻¹AU

≤ⁿ

i1

Reλ_iBλ_i

U⁻¹AU

ⁿ

i1

Reλ_iBλ_i A

.

2.34

This completes the proof.

Note that ifB∈Sⁿ, ReλiB λiB, then from2.34we obtain1.5immediately.

This implies that2.18improves1.5.

Since trAB trBA,applying2.31withBin lieu ofA,we immediately have the following corollary.

Corollary 2.11. LetB∈R^n×n,A∈R^n×nbe normal, then

n i1

Reλ_iAλ_n−i1 B

≤trAB≤ⁿ

i1

Reλ_iAλ_i B

. 2.35

3. Trace Bounds for the Solution of the Algebraic Riccati Equations

Komaroff 1994in16obtained the following. Let Kbe the positive semidefinite solution of the ARE1.4. Then the trace ofKhas the upper bound given by

trK≤ n

2λ1S n 2

λ²₁S 4tr QR⁻¹

n , 3.1

whereSR⁻¹A^TAR⁻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 improve3.1under certain conditions.

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

1There are both, upper and lower, bounds:

λnRλnS λnR

λnS₂

4/λnR ⁿ_i1λ^p_iR_1/p tr

QR⁻¹ 2 ⁿ_i1λ^p_iR1/p

≤trK≤ λ1S

λ²₁S

4/cp,qn^2−1/qλ1R ⁿ_i1λ^p_iR_1/p tr

QR⁻¹ 2 ⁿ_i1λ^p_iR1/p

/cp,qn^2−1/qλ1R

.

3.2

2IfS≥0, then the trace ofKhas the lower and upper bounds given by

1/c_p,qn^1−1/q

H

1/cp,qn^1−1/q H₂

4/λnR tr

QR⁻¹ 2/λnR

≤trK≤ H ⁿ_i1λ^p_iS2/p

4/cp,qn^2−1/qλ1R tr

QR⁻¹

2/cp,qn^2−1/qλ₁R ,

3.3

whereHdenotes_n

i1λ^p_iS^1/panddenotes_n

i1λ^p_iR^1/p.

3IfS≤0, then the trace ofKhas the lower and upper bounds given by

−_n

i1λ_iS^p1/p

_n

i1λ_iS^p2/p

4/λ_nR ⁿ_i1λ^p_iR_1/p tr

QR⁻¹ 2 ⁿ_i1λ^p_iR_1/p

/λnR

≤trK≤ cp,qn^2−1/qλ₁R 2 ⁿ_i1λ^p_iR1/p

×

⎧⎨

⎩ 1 cp,qn^1−1/q

!

−ⁿ

i1

λiS^p"_1/p

#$

$%! 1 cp,qn^1−1/qN

"2

4

cp,qn^2−1/qλ₁RStr QR⁻¹

⎫⎪

⎬

⎪⎭,

3.4

whereNdenotes_n

i1|λiS|^p1/p and Sdenotes_n

i1λ^p_iR^1/p,

We have

cp,q M^prM^q_k−m^prm^q_k p

MrmkM_k^q−mrMkm^q_k1/p q

mrMkM^pr −Mrmkm^pr

_1/q,

Mr λ₁R, mr λ_nR, Mkλ₁K, mkλ_nK,

c_p,q M^psM^q_k−m^psm^q_k p

MsmkM^q_k−msMkm^q_k1/p

q

msMkM^ps−M1mkm^ps

1/q,

Msλ₁S, msλ_nS, SR⁻¹A^TAR⁻¹.

3.5

Proof. 1Multiply1.4on the right and on the left byR^−1/2to get

R^−1/2QR^−1/2K^T₁K1−R^−1/2

A^TKKA

R^−1/2, 3.6

whereK1R^1/2KR^−1/2. Take the trace of all terms in3.6to get

tr K₁^TK1

−tr

R⁻¹A^TKKAR⁻¹

−tr QR⁻¹

0. 3.7

SinceKis positive semidefiniteness,λK ReλK, trK _n

i1λ_iK _n

i1Reλ_iK, and fromLemma 2.7, we have

trK

n^1−1/q ≤trK^q1/q≤trK, 3.8

n i1

λiKK ⁿ

i1

λ²_iK≤

! _n

i1

λiK

"₂

trK². 3.9

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

n i1

λiKK ⁿ

i1

λ²_iK≥ _n

i1λ_iK2

n trK²

n . 3.10

SinceK, Qare positive semidefiniteness,Ris positive definiteness, then by1.5, note that for i1,2, . . . , n, λ_iR⁻¹ λ_iR⁻¹ 1/λ_n−i1R, and considering2.6,3.8, and3.9, we have

tr K^T₁K1

tr

R⁻¹KRK

≤ⁿ

i1

λi

R⁻¹

λiKRK

ⁿ

i1

λiKRK λ_n−i1R ≤ 1

λ_nRtrKRK

≤ 1 λnR

! _n

i1

λ^p_iR

"_1/p

trK².

3.11

Note thatSR⁻¹A^TAR⁻¹,λiS λiS, then from1.5we have

λnStrK≤ⁿ

i1

λn−i1

R⁻¹A^TAR⁻¹ λiK

≤tr

R⁻¹A^TKAR⁻¹K tr

R⁻¹A^TKKAR⁻¹

≤ⁿ

i1

λ_i

R⁻¹A^TAR⁻¹

λ_iK≤λ₁StrK.

3.12

Combining3.11with3.12, we obtain

1 λ_nR

!_n

i1

λ^p_iR

"1/p

trK²−trKλnS−tr QR⁻¹

≥0. 3.13

Solving3.13for trKyields the left-hand side of3.2.

SinceK, Qare positive semidefiniteness,Ris positive definiteness, then by1.5, note that fori 1,2, . . . , n, λn−i1R⁻¹ λn−i1R⁻¹ 1/λiR, and considering2.6,3.8, and3.10, we have

tr K^T₁K1

tr

R⁻¹KRK

≥ⁿ

i1

λ_n−i1 R⁻¹

λ_iKRK

ⁿ

i1

λ_iKRK λiR ≥ 1

λ1RtrKRK

≥ 1

cp,qλ₁R

!_n

i1

λ^p_iR

"1/p!_n

i1

λ^q_iK²

"1/q

≥ 1

cp,qn^2−1/qλ₁R

!_n

i1

λ^p_iR

"1/p

trK².

3.14

Combining3.12with3.14, we obtain

1 cp,qn^2−1/qλ1R

!_n

i1

λ^p_iR

"1/p

trK²−trKλ₁S−tr QR⁻¹

≤0. 3.15

Solving3.15for trKyields the right-hand side of3.2.

2Note that whenS≥0, by1.5,2.6, and3.8, we have

tr

R⁻¹A^TKKAR⁻¹

≥ⁿ

i1

λ_n−i1Sλ_iK

≥ 1 cp,q

!_n

i1

λ^p_iS

"1/q!_n

i1

λ^q_iK

"1/q

≥ 1

cp,qn^1−1/q

! _n

i1

λ^p_iS

"_1/p trK.

3.16

Combining3.11with3.16, we obtain

1 λnR

! _n

i1

λ^p_iR

"_1/p

trK²− 1 cp,qn^1−1/q

! _n

i1

λ^p_iS

"_1/p

trK−tr QR⁻¹

≥0. 3.17

Solving3.17for trKyields the left-hand side of3.3.

By1.5,2.6, and3.8, we have

tr

R⁻¹A^TKKAR⁻¹

≤ⁿ

i1

λiSλiK

≤

! _n

i1

λ^p_iS

"_1/p! _n

i1

λ^q_iK

"_1/q

≤

!_n

i1

λ^p_iS

"_1/p trK.

3.18

Combining3.14with3.18, we obtain

1 cp,qn^2−1/qλ1R

!_n

i1

λ^p_iR

"_1/p

trK²−

!_n

i1

λ^p_iS

"_1/p

trK−tr QR⁻¹

≤0. 3.19

Solving3.19for trKyields the right-hand side of3.3.

3Note that whenS ≤ 0, by3.3, we obtain3.4immediately. This completes the proof.

Remark 3.2. FromRemark 2.6andTheorem 3.1, we have the following results.

1Letp → ∞, q → 1 in3.2, then we have λ_nRλ_nS λ_nR

λ²_nS

4/λ_nR

λ₁Rtr QR⁻¹ 2λ1R

≤trK≤ n

2λ₁S n 2

λ²₁S 4tr QR⁻¹

n .

3.20

2Letp → ∞, q → 1 in3.3, then we obtain3.20.

3Letp → ∞, q → 1 in3.4. Note that whenS≤0,

plim→ ∞

! _n

i1

|λ_iS|^p

"_1/p max

1≤i≤nλ_iS−λ_nS, limp→ ∞

q→1

1 cp,qn^1−1/q

!_n

i1

|λ_iS|^p

"_1/p min

1≤i≤nλ_iS−λ₁S.

3.21

Then we can also obtain3.20.

Note that the right-hand side of 3.20 is 3.1, which implies that Theorem 3.1 improves3.1.

4. Numerical Examples

In this section, firstly, we will give an example to illustrate that our new trace bounds are better than those of the recent results. Then, to illustrate that the application in the algebraic Riccati equations of our results will have different superiority if we choose differentpandq, we will give two examples.

Example 4.1. Let

A

⎛

⎜⎜

⎝

0.2563 0.2588 0.1422 0.2358 2.0451 0.4177 0.8942 0.2547 0.9852

⎞

⎟⎟

⎠,

B

⎛

⎜⎜

⎝

0.2587 0.5236 0.8541 0.5236 1.1254 0.3654 0.8541 0.3654 1.2541

⎞

⎟⎟

⎠.

4.1

Then trAB 4.9933 andBis symmetric. Using1.5yields

0.2173≤trAB≤5.5656. 4.2

Using2.18yields

0.6079≤trAB≤5.1255, 4.3

where both lower and upper bounds are better than those of the main result of18, that is, 1.5.

Example 4.2. Consider the system1.1,1.2with

A

⎛

⎜⎜

⎝

−15 −23 27 26 −9 4 35 72 18

⎞

⎟⎟

⎠, BB^T

⎛

⎜⎜

⎝ 6 1 3 1 7 4 3 4 8

⎞

⎟⎟

⎠, Q

⎛

⎜⎜

⎝

485 49 38 49 64 −92 38 −92 192

⎞

⎟⎟

⎠ 4.4

and consider the corresponding ARE1.4withR BB^T;A, Ris stabilizable andQ, Ais observable. Using3.20yields

39.0104≤trK≤682.1538. 4.5

Using3.2, whenpq2, then we obtain

201.9801≤trK≤271.4, 4.6

where the upper bound is better than that of the main result of16, that is,3.1.

Example 4.3. Consider the system1.1,1.2with

A

⎛

⎜⎜

⎝

20 3 7.5 5 7 9 2 0 4

⎞

⎟⎟

⎠, BB^T

⎛

⎜⎜

⎝ 9 1 3 1 5 2 3 2 6

⎞

⎟⎟

⎠, Q

⎛

⎜⎜

⎝

455 332 209 332 304 127.5 209 127.5 125

⎞

⎟⎟

⎠ 4.7

and consider the corresponding ARE1.4withR BB^T;A, Ris stabilizable andQ, Ais observable. Using3.2, whenpq2, then we obtain

5.2895≤trK≤97.2209. 4.8

Using3.20yields

5.6559≤trK≤25.9683, 4.9

where the lower and upper bounds are better than those of4.8.

5. Conclusions

In this paper, we have proposed lower and upper bounds for the trace of the product of two real square matrices in which one is diagonalizable. We have shown that our bounds for the trace are the tightest among the parallel trace bounds in symmetric case. Then, we have obtained 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 bounds are better than those of the previous results.

Acknowledgments

The authors would like to thank Professor Jozef Banas and the referees for the very helpful comments and suggestions to improve the contents and presentation of this paper. The work was also supported in part by the National Natural Science Foundation of China10971176.

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