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 rateu∗and the optimal costJ∗of1.1and1.2are u∗P x, PBTK,
J∗xT0Kx0,
1.3
where x0 ∈ Rn is the initial state of the systems1.1 and1.2,K is the positive definite solution of the following algebraic Riccati equationARE:
ATKKA−KRK−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 costJ∗asx0varies 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, x1 ≥ x2 ≥ · · · ≥ xn. Let |x| |x1|,|x2|, . . . ,|xn|. For A aij ∈ Rn×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
αi≤k
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 byα≺β.
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, B∈Rn×n,then
−n
i1
σiAσiB≤trAB≤n
i1
σiAσiB. 1.7
Xing et al. in13have observed another result. LetA, B∈Rn×nbe arbitrary matrices with the following singular value decomposition:
BUdiag
σ1B, σ2B, . . . , σnB
VT, 1.8
whereU, V ∈Rn×nare orthogonal. Then
λnASn
i1
σiB≤trAB≤λ1ASn
i1
σiB, 1.9
whereSUVT is orthogonal.
Liu and He in14have obtained the following: letA, B∈Rn×nbe arbitrary matrices with the following singular value decomposition:
BUdiag
σ1B, σ2B, . . . , σnB
VT, 1.10
whereU, V ∈Rn×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, B∈Rn×nbe arbitrary matrices with the following singular value decomposition:
BUdiag
σ1B, σ2B, . . . , σnB
VT, 1.12
whereU, V ∈Rn×nare orthogonal. Then n
i1
σiBλn−i1AS≤trAB≤n
i1
σiBλiAS, 1.13
whereSUVT is orthogonal. They show that1.13has improved1.9.
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 andx ≺ y, then for any real arrayu1≥ · · · ≥un,
n i1
xiui≤n
i1
yiui. 2.1
Lemma 2.2see16, page 95, H.3.b. Ifx1 ≥ · · · ≥ xn, y1 ≥ · · · ≥ynandx≺wy, then for any real arrayu1≥ · · · ≥un≥0,
n i1
xiui≤n
i1
yiui. 2.2
Remark 2.3. Note that ifx≺wy, then fork 1,2, . . . , n,x1, . . . , xk≺wy1, . . . , yk. Thus fromLemma 2.2, we have
k i1
xiui≤k
i1
yiui, k1,2, . . . , n. 2.3
Lemma 2.4see16, page 218, B.1. LetAAT ∈Rn×n, then
dA≺λA. 2.4
Lemma 2.5see16, page 240, F.4.a. LetA∈Rn×n, then
λ
AAT 2
≺w
λ
AAT 2
≺wσA. 2.5
Lemma 2.6see17. Let 0< m1≤ak≤M1,0< m2≤bk≤M2, k1,2, . . . , n,1/p1/q1.
Then
n k1
akbk≤ n
k1
apk
1/p n
k1
bqk 1/q
≤cp,q
n k1
akbk, 2.6
where
cp,q Mp1M2q−mp1mq2 p
M1m2Mq2−m1M2mq21/p q
m1M2Mp1−M1m2mp11/q. 2.7
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
akbk≤ n
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→ ∞
q→1
cp,qlim
p→ ∞
q→1
Mp1M2q−mp1mq2 p
M1m2Mq2−m1M2mq21/p q
m1M2Mp1−M1m2mp11/q
lim
p→ ∞
q→1
Mp1
Mq2−m1/M1pmq2 M1/p1
p
m2M2q−m1/M1M2mq21/p
Mq/p1 q
m1M2−M1m2m1/M1p1/q
lim
p→ ∞
q→1
M2
M1/pp/q−p1 m1M2
lim
p→ ∞
q→1
1 M11/p−1m1
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 2.9. Ifq≥1,ai≥0 i1,2, . . . , n, then
1 n
n i1
ai
q
≤ 1 n
n i1
aqi. 2.12
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 n−r
q n
i1
ai
q
1 n−r
n i1
ai
q
≤ 1 n−r
n i1
aqi. 2.15
Sincen−r/nq≤n−r/n, thus 1
n n
i1
ai
q
n−r n
q 1 n−r
q n
i1
ai
q
≤ n−r n
1 n−r
n i1
aqi 1 n
n i1
aqi. 2.16
This completes the proof.
Theorem 2.10. Let A, B ∈ Rn×n be arbitrary matrices with the following singular value decomposition:
BUdiagσ1B, σ2B, . . . , σnBVT, 2.17
whereU, V ∈Rn×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
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
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, B ∈ Rn×n be arbitrary matrices with the following singular value decomposition:
APdiagσ1A, σ2A, . . . , σnAQT, 2.29
whereP, Q∈Rn×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, B ∈ Rn×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, V ∈Rn×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
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
σiBλn−i1 VTAU
VTAUT
2
≤n
i1
σiBdn−i1 VTAU
VTAUT
2
≤trAB
≤n
i1
σiBdi VTAU
VTAUT
2
≤n
i1
σiBλi 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
Then2.34can be rewritten as n i1
σiBλn−i1AS≤n
i1
σiBdn−i1
VTAU
≤trAB
≤n
i1
σiBdi
VTAU
≤n
i1
σiBλiAS.
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
σiBλiAS≤n
i1
σiBσiA. 2.38
Note thatλi−AS −λn−i1AS, fromLemma 2.2and2.38, we have
−n
i1
σiBλn−i1AS n
i1
σiBλi−AS
≤n
i1
σiBλiAS≤n
i1
σiBσiA.
2.39
Thus, we obtain
−n
i1
σiBσiA≤n
i1
σiBλn−i1AS. 2.40
Both2.38and2.40show that1.13is tighter than1.7.
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
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 MprMqk−mprmqk p
MrmkMkq−mrMkmqk1/p q
mrMkMpr −Mrmkmpr
1/q,
Mr λ1R, mrλnR, Mkλ1K, mkλnK,
cp,q Mp1Mqk−mp1mqk p
M1mkMqk−m1Mkmqk1/p q
m1MkMp1−M1mkmp11/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
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
λiRσi 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
λiAσiK≤λ1An
i1
σiK,
trATK≤n
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.
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
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.
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.
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
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 time-varying systems,” IEEE Transactions on Automatic Control, vol. 13, no. 2, pp. 150–159, 1968.
3 R. Davies, P. Shi, and R. Wiltshire, “New upper solution bounds for perturbed continuous algebraic Riccati equations applied to automatic control,” Chaos, Solitons & Fractals, vol. 32, no. 2, pp. 487–495, 2007.
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.
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.
7 J. B. Lasserre, “Tight bounds for the trace of a matrix product,” IEEE Transactions on Automatic Control, vol. 42, no. 4, pp. 578–581, 1997.
8 Y. Fang, K. A. Loparo, and X. Feng, “Inequalities for the trace of matrix product,” IEEE Transactions on Automatic Control, vol. 39, no. 12, pp. 2489–2490, 1994.
9 J. Saniuk and I. Rhodes, “A matrix inequality associated with bounds on solutions of algebraic Riccati and Lyapunov equations,” IEEE Transactions on Automatic Control, vol. 32, no. 8, pp. 739–740, 1987.
10 T. Mori, “Comments on “A matrix inequality associated with bounds on solutions of algebraic Riccati and Lyapunov equation”,” IEEE Transactions on Automatic Control, vol. 33, no. 11, pp. 1088–1091, 1988.
11 J. B. Lasserre, “A trace inequality for matrix product,” IEEE Transactions on Automatic Control, vol. 40, no. 8, pp. 1500–1501, 1995.
12 P. Park, “On the trace bound of a matrix product,” IEEE Transactions on Automatic Control, vol. 41, no.
12, pp. 1799–1802, 1996.
13 W. Xing, Q. Zhang, and Q. Wang, “A trace bound for a general square matrix product,” IEEE Transactions on Automatic Control, vol. 45, no. 8, pp. 1563–1565, 2000.
14 J. Liu and L. He, “A new trace bound for a general square matrix product,” IEEE Transactions on Automatic Control, vol. 52, no. 2, pp. 349–352, 2007.
15 F. Zhang and Q. Zhang, “Eigenvalue inequalities for matrix product,” IEEE Transactions on Automatic Control, vol. 51, no. 9, pp. 1506–1509, 2006.
16 A. W. Marshall and I. Olkin, Inequalities: Theory of Majorization and Its Applications, vol. 143 of Mathematics in Science and Engineering, Academic Press, New York, NY, USA, 1979.
17 C.-L. Wang, “On development of inverses of the Cauchy and H ¨older inequalities,” SIAM Review, vol.
21, no. 4, pp. 550–557, 1979.