Báo cáo hóa học: "Research Article Trace Inequalities for Matrix Products and Trace Bounds for the Solution of the Algebraic Riccati Equations" pot

Hindawi Publishing Corporation Journal of Inequalities and Applications 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 Liu 1 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. Copyright q 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 systems see 1–5. For example, consider the following linear system see 5: ˙x  t   Ax  t   Bu  t  ,x  0   x 0 , 1.1 with the cost J   ∞ 0  x T Qx  u T u  dt. 1.2 2 Journal of Inequalities and Applications The optimal control rate u ∗ the optimal cost J ∗ of 1.1 and 1.2 are u ∗  Px, P  B T K, J ∗  x T 0 Kx 0 , 1.3 where x 0 ∈ R n is the initial state of system 1.1 and 1.2 and K is the positive semidefinite solution of the following algebraic Riccati equation ARE: A T K  KA − KRK  −Q, 1.4 with R  BB T and Q being positive definite and positive semidefinite matrices, respectively. To guarantee the existence of the positive definite solution to 1.4, we will make the following assumptions: the pair A, 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 ARE see 6– 16. Moreover, in terms of 2, 6, we know that an interpretation of trK is that trK/n is the average value of the optimal cost J ∗ as x 0 varies 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 products 2, 6–8, 17–20,and18 is the tightest among the parallel results. In 1995, Lasserre showed 18 the following given any matrix A ∈ R n×n ,B∈ S n , then the following. n  i1 λ i  A  λ  n−i1   B  ≤ tr  AB  ≤ n  i1 λ i  A  λ i  B  , 1.5 where A 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.InSection 4, we give numerical examples to demonstrate the effectiveness of our results. Finally, we get conclusions in Section 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 trA,A −1 ,A T , dAd 1 A, ,d n A,σAσ 1 A, ,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.IfA ∈ R n×n is an arbitrary symmetric matrix, then λAλ 1 A, ,λ n A and Re λARe λ 1 A, ,Re λ n A denote the eigenvalues Journal of Inequalities and Applications 3 and the real part of eigenvalues of A. Suppose x x 1 ,x 2 , ,x n  is a real n-element array such as dA,σA,λA, Re λA which is reordered, and its elements are arranged in nonincreasing order; that is, x 1 ≥ x 2 ≥ ··· ≥ 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  i1 α i ≤ k  i1 β i ,k 1, 2, ,n, 2.1 then it is said that α is controlled weakly by β, which is signed by α≺ w β. If α≺ w β and n  i1 α i  n  i1 β i , 2.2 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 x 1 ≥ ··· ≥ x n ,y 1 ≥ ··· ≥ y n and x ≺ y, then for any real array u 1 ≥···≥u n , n  i1 x i u i ≤ n  i1 y i u i . 2.3 Lemma 2.2 see 21, Page 218, B.1. Let A  A T ∈ R n×n ,then d  A  ≺ λ  A  . 2.4 Lemma 2.3 see 21, Page 240, F.4.a. Let A ∈ R n×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 <m 1 ≤ a k ≤ M 1 , 0 <m 2 ≤ b k ≤ M 2 ,k 1, 2, ,n,1/p 1/q  1, then n  k1 a k b k ≤  n  k1 a p k  1/p  n  k1 b q k  1/q ≤ c p,q n  k1 a k b k , 2.6 4 Journal of Inequalities and Applications where c p,q  M p 1 M q 2 − m p 1 m q 2  p  M 1 m 2 M q 2 − m 1 M 2 m q 2  1/p  q  m 1 M 2 M p 1 − M 1 m 2 m p 1  1/q . 2.7 Note that if m 1  0,m 2 /  0orm 2  0,m 1 /  0, obviously, 2.6 holds. If m 1  m 2  0, choose c p,q ∞, then 2.6 also holds. Remark 2.5. If p  q  2, then we obtain Cauchy-Schwarz inequality: n  k1 a k b k ≤  n  k1 a 2 k  1/2  n  k1 b 2 k  1/2 ≤ c 2 n  k1 a k b k , 2.8 where c 2  ⎛ ⎝  M 1 M 2 m 1 m 2   m 1 m 2 M 1 M 2 ⎞ ⎠ . 2.9 Remark 2.6. Note that lim p →∞  a p 1  a p 2  ··· a p n  1/p  max 1≤k≤n { a k } , lim p →∞ q → 1 c p,q  lim p →∞ q → 1 M p 1 M q 2 − m p 1 m q 2  p  M 1 m 2 M q 2 − m 1 M 2 m q 2  1/p  q  m 1 M 2 M p 1 − M 1 m 2 m p 1  1/q  lim p →∞ q → 1 M p 1  M q 2 −  m 1 /M 1  p m q 2  M 1/p 1  p  m 2 M q 2 −m 1 /M 1 M 2 m q 2  1/p M q/p 1  q  m 1 M 2 −M 1 m 2  m 1 /M 1  p  1/q  lim p →∞ q → 1 M 2 M 1/pp/q−p 1 m 1 M 2  lim p →∞ q → 1 1 M 1/p−1 1 m 1  M 1 m 1 . 2.10 Let p →∞,q → 1in2.6, then we obtain m 1 n  k1 b k ≤ n  k1 a k b k ≤ M 1 n  k1 b k . 2.11 Journal of Inequalities and Applications 5 Lemma 2.7. If q ≥ 1, a i ≥ 0 i  1, 2, ,n,then  1 n n  i1 a i  q ≤ 1 n n  i1 a q i . 2.12 Proof. 1 Note that for q  1, or a i  0 i  1, 2, ,n,  1 n n  i1 a i  q  1 n n  i1 a q i . 2.13 2 If q>1, a i > 0, for x>0, choose fxx q , then f  xqx q−1 > 0andf  x qq − 1x q−2 > 0. Thus, fx is a convex function. As a i > 0and1/n  n i1 a i > 0, from the property of the convex function, we have  1 n n  i1 a i  q  f  1 n n  i1 a i  ≤ 1 n n  i1 f  a i   1 n n  i1 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  q  n  i1 a i  q   1 n − r n  i1 a i  q ≤ 1 n − r n  i1 a q i . 2.15 Since n − r/n q ≤ n − r/n,thus  1 n n  i1 a i  q   n − r n  q  1 n − r  q  n  i1 a i  q ≤ n − r n 1 n − r n  i1 a q i  1 n n  i1 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  λ 1  B  ,λ 2  B  , ,λ n  B  U −1 , 2.17 where U ∈ R n×n is nonsingular. Then n  i1 Re λ  i   B  d  n−i1   U −1 AU  ≤ tr  AB  ≤ n  i1 Re λ  i   B  d  i   U −1 AU  . 2.18 6 Journal of Inequalities and Applications Proof. Note that U −1 AU ii is real; by the matrix theory we have tr  AB   Re tr  AB   Re tr  AU diag  λ 1  B  ,λ 2  B  , ,λ n  B  U −1   Re tr  U −1 AU diag  λ 1  B  ,λ 2  B  , ,λ n  B    Re n  i1 λ i  B   U −1 AU  ii  n  i1 Re  λ i  B   U −1 AU  ii   n  i1  U −1 AU  ii Re λ i  B   n  i1 ⎡ ⎣ U −1 AU   U −1 AU  T 2 ⎤ ⎦ ii Re λ i  B   n  i1  U −1 AU  ii Re λ i  B  . 2.19 Since Re λ 1 B ≥ Re λ 2 B ≥···≥Re λ n B ≥ 0, without loss of generality, we may assume Re λBRe λ 1 B, Re λ 2 B, ,Re λ n B. Next, we will prove the left-hand side of 2.18: n  i1 Re λ  i   B  d  n−i1   U −1 AU  ≤ n  i1 Re λ  i   B  d i  U −1 AU  . 2.20 If d  U −1 AU    d  n   U −1 AU  ,d  n−1   U −1 AU  , ,d  1   U −1 AU  , 2.21 then we obtain the conclusion. Now assume that there exists j<ksuch that d j U −1 AU > d k U −1 AU, then Re λ  j   B  d k  U −1 AU   Re λ  k   B  d j  U −1 AU  − Re λ  j   B  d j  U −1 AU  − Re λ  k   B  d k  U −1 AU    Re λ  j   B  − Re λ  k   B   d k  U −1 AU  − d j  U −1 AU  ≤ 0. 2.22 Journal of Inequalities and Applications 7 We use  d U −1 AU to denote the vector of dU −1 AU after changing d j U −1 AU and d k U −1 AU, then n  i1 σ  i   B   d i  U −1 AU  ≤ n  i1 σ  i   B  d i  U −1 AU  . 2.23 After a limited number of steps, we obtain the left-hand side of 2.18. For the right-hand side of 2.18 n  i1 Re λ  i   B  d i  U −1 AU  ≤ n  i1 Re λ  i   B  d  i   U −1 AU  . 2.24 If d  V T AU    d  1   U −1 AU  ,d  2   U −1 AU  , ,d  n   U −1 AU  , 2.25 then we obtain the conclusion. Now assume that there exists j>ksuch that d j U −1 AU < d k U −1 AU, then σ  j   B  d k  U −1 AU   σ  k   B  d j  U −1 AU  − σ  j   B  d j  U −1 AU  − σ  k   B  d k  U −1 AU    σ  j   B  − σ  k   B   d k  U −1 AU  − d j  U −1 AU  ≥ 0. 2.26 We use  d U −1 AU to denote the vector of dU −1 AU after changing d j U −1 AU and d k U −1 AU, then n  i1 σ  i   B  d i  U −1 AU  ≤ n  i1 σ  i   B   d i  U −1 AU  . 2.27 After a limited number of steps, we obtain the right-hand side of 2.18. Therefore, we have n  i1 Re λ  i   B  d  n−i1   U −1 AU  ≤ tr  AB  ≤ n  i1 Re λ  i   B  d  i   U −1 AU  . 2.28 8 Journal of Inequalities and Applications Since trABtrBA, 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  λ 1  A  ,λ 2  A  , ,λ n  A  V −1 , 2.29 where V ∈ R n×n is nonsingular. Then n  i1 Re λ  i   A  d  n−i1   V −1 BV  ≤ tr  AB  ≤ n  i1 Re λ  i   A  d  i   V −1 BV  . 2.30 Theorem 2.10. Let A ∈ R n×n , B ∈ R n×n be normal. Then n  i1 Re λ  i   B  λ  n−i1   A  ≤ tr  AB  ≤ n  i1 Re λ  i   B  λ  i   A  . 2.31 Proof. Since B is normal, from 23, page 101, Theorem 2.5.4, we have B  U diag  λ 1  B  ,λ 2  B  , ,λ n  B  U −1 , 2.32 where U ∈ R n×n is orthogonal. Since U T  U −1 and UU T  I, then for i  1, 2, ,n, we have λ  i   U −1 AU   λ  i   U T AU   λ i ⎛ ⎝ U T AU   U T AU  T 2 ⎞ ⎠  λ  i  ⎛ ⎝ U T ⎛ ⎝ AUU T   AUU T  T 2 ⎞ ⎠ U ⎞ ⎠  λ  i  ⎛ ⎝ AUU T   AUU T  T 2 ⎞ ⎠  λ i  A  . 2.33 Journal of Inequalities and Applications 9 In terms of Lemmas 2.1 and 2.2, 2.18 implies n  i1 Re λ  i   B  λ  n−i1   A   n  i1 Re λ  i   B  λ  n−i1   U −1 AU  ≤ n  i1 Re λ  i   B  d  n−i1   U −1 AU  ≤ tr  AB  ≤ n  i1 Re λ  i   B  d  i   U −1 AU  ≤ n  i1 Re λ  i   B  λ  i   U −1 AU   n  i1 Re λ  i   B  λ  i   A  . 2.34 This completes the proof. Note that if B ∈ S n ,Reλ i Bλ i B, then from 2.34 we obtain 1.5 immediately. This implies that 2.18  improves 1.5. Since trABtrBA, 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  i1 Re λ  i   A  λ  n−i1   B  ≤ tr  AB  ≤ n  i1 Re λ  i   A  λ  i   B  . 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 ARE 1.4. Then the trace of K has the upper bound given by tr  K  ≤ n 2 λ 1  S   n 2  λ 2 1  S   4tr  QR −1  n , 3.1 where S  R −1 A T  AR −1 . In this section, by appling our new trace bounds in Section 2, we obtain some lower trace bounds for the solution of the algebraic Riccati equations. Furthermore, we obtain some upper trace bounds which improve 3.1 under certain conditions. 10 Journal of Inequalities and Applications 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  R  λ n  S   λ n  R    λ n S  2   4/λ n  R     n i1 λ p i R  1/p tr  QR −1  2   n i1 λ p i R  1/p ≤ tr  K  ≤ λ 1  S    λ 2 1  S    4/c p,q n 2−1/q λ 1  R     n i1 λ p i R  1/p tr  QR −1  2   n i1 λ p i R  1/p /c p,q n 2−1/q λ 1  R  . 3.2 2 If S ≥ 0, then the trace of K has the lower and upper bounds given by  1/c  p,q n 1−1/q  H    1/c  p,q n 1−1/q  H  2   4/λ n  R   tr  QR −1  2/λ n  R  ≤ tr  K  ≤ H     n i1 λ p  i  S  2/p   4/c p,q n 2−1/q λ  1   R   tr  QR −1  2/c p,q n 2−1/q λ  1   R  , 3.3 where H denotes   n i1 λ p i S 1/p and  denotes   n i1 λ p i R 1/p . 3 If S ≤ 0, then the trace of K has the lower and upper bounds given by −   n i1   λ i S   p  1/p     n i1   λ  i   S    p  2/p  4/λ n  R    n i1 λ p i R  1/p tr  QR −1  2   n i1 λ p i R  1/p /λ n  R  ≤ tr  K  ≤ c p,q n 2−1/q λ 1  R  2   n i1 λ p i R  1/p × ⎧ ⎨ ⎩ 1 c  p,q n 1−1/q  − n  i1   λ  i  S   p  1/p       1 c  p,q n 1−1/q N  2  4 c p,q n 2−1/q λ  1   R  Str  QR −1  ⎫ ⎪ ⎬ ⎪ ⎭ , 3.4 where N denotes   n i1 |λ i S| p  1/p and S denotes   n i1 λ p i R 1/p , [...]... bounds of the solution of the continuous Riccati equation,” IEEE Transactions on Automatic Control, vol 42, no 9, pp 1268–1271, 1997 14 C.-H Lee, “New results for the bounds of the solution for the continuous Riccati and Lyapunov equations,” IEEE Transactions on Automatic Control, vol 42, no 1, pp 118–123, 1997 15 C.-H Lee, “On the upper and lower bounds of the solution for the continuous Riccati matrix. .. 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... 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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 China 10971176 References 1 K Kwakernaak and R Sivan, Linear Optimal Control Systems, John Wiley & Sons, New York, NY, USA, 1972 2 D L Kleinman and M Athans, The design of suboptimal linear... min λ i S 1≤i≤n −λ 1 S Then we can also obtain 3.20 Note that the right-hand side of 3.20 is 3.1 , which implies that Theorem 3.1 improves 3.1 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... 1994 18 J B Lasserre, “A trace inequality for matrix product,” IEEE Transactions on Automatic Control, vol 40, no 8, pp 1500–1501, 1995 Journal of Inequalities and Applications 17 19 P Park, “On the trace bound of a matrix product,” IEEE Transactions on Automatic Control, vol 41, no 12, pp 1799–1802, 1996 20 J B Lasserre, “Tight bounds for the trace of a matrix product,” IEEE Transactions on Automatic . 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,. λ  i   A  λ  i   B  . 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 ARE 1.4. Then the trace of. Corporation Journal of Inequalities and Applications 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