Volume 2009, Article ID 620758, 18 pagesdoi:10.1155/2009/620758 Research Article New Trace Bounds for the Product of Two Matrices and Their Applications in the Algebraic Riccati Equation
Trang 1Volume 2009, Article ID 620758, 18 pages
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
Copyrightq 2009 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 practicesee 1 5 Consider the following linear systemsee 4:
˙xt Axt But, x0 x0, 1.1 with the cost
J
∞
0
x T Qx u T u
Moreover, the optimal control rate u∗and the optimal cost J∗of1.1 and 1.2 are
u∗ Px, P B T K,
J∗ x T
0Kx0,
1.3
Trang 2where x0 ∈ R n is the initial state of the systems1.1 and 1.2, K is the positive definite
solution of the following algebraic Riccati equationARE:
A T K KA − KRK −Q, 1.4
with R BB T and Q 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, 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 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 trK is that trK/n is the average value of the optimal cost J∗as x0varies 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, let R n×n denote the set of n × n real matrices Let x x1, x2, , xn be
a real n-element array which is reordered, and its elements are arranged in nonincreasing order That is, x1 ≥ x2 ≥ · · · ≥ x n Let |x| |x1|, |x2|, , |x n | For A a ij ∈
R n×n , let dA d1A, d2A, , d n A, λA λ1A, λ2A, , λ n A, σA
σ1A, σ2A, , σ n A denote the diagonal elements, the eigenvalues, the singular values
of A, respectively, Let trA, A T denote the trace, the transpose of A, respectively We define
A ii a ii d i A, A A A T /2 The notation A > 0 A ≥ 0 is used to denote that A is a
symmetric positive definitesemidefinite matrix
Let α, β be two real n-element arrays If they satisfy
k
i1
α i≤k
i1
β i , k 1, 2, , n, 1.5
then it is said that α is controlled weakly by β, which is signed by α≺ wβ.
If α≺ wβ and
n
i1
α in
i1
then it is said that α is controlled by β, which is signed by α ≺ β.
Trang 3Therefore, 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 in16 have showed that for any A, B ∈ R n×n , then
−n
i1
σ i Aσ i B ≤ trAB ≤n
i1
σ i Aσ i B. 1.7
Xing et al in13 have observed another result Let A, B ∈ R n×nbe arbitrary matrices with the following singular value decomposition:
B Udiag
σ1B, σ2B, , σ n BV T , 1.8
where U, V ∈ R n×nare orthogonal Then
λ n ASn
i1
σ i B ≤ trAB ≤ λ1ASn
i1
σ i B, 1.9
where S UV T is orthogonal
Liu and He in14 have obtained the following: let A, B ∈ R n×nbe arbitrary matrices with the following singular value decomposition:
B Udiag
σ1B, σ2B, , σ n BV T , 1.10
where U, V ∈ R n×nare orthogonal Then
min
1≤i≤n
V T AU
ii n
i1
σ i B ≤ trAB ≤ max
1≤i≤n
V T AU
ii n
i1
σ i B. 1.11
F Zhang and Q Zhang in15 have obtained the following: let A, B ∈ R n×nbe arbitrary matrices with the following singular value decomposition:
B Udiag
σ1B, σ2B, , σ n BV T , 1.12
where U, V ∈ R n×nare orthogonal Then
n
i1
σ i Bλ n−i1 AS ≤ trAB ≤n
i1
σ i Bλ i AS, 1.13
where S UV T is orthogonal They show that1.13 has improved 1.9
Trang 42 Main Results
The following lemmas are used to prove the main results
Lemma 2.1 see 16, 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
Lemma 2.2 see 16, page 95, H.3.b If x1 ≥ · · · ≥ x n , y1 ≥ · · · ≥ y n and x≺wy, then for any real array u1≥ · · · ≥ u n ≥ 0,
n
i1
x i u i≤n
i1
Remark 2.3 Note that if x≺wy, then for k 1, 2, , n, x1, , x k≺w y1, , y k Thus fromLemma 2.2, we have
k
i1
x i u i≤k
i1
y i u i , k 1, 2, , n. 2.3
Lemma 2.4 see 16, page 218, B.1 Let A AT ∈ R n×n , then
Lemma 2.5 see 16, page 240, F.4.a Let A ∈ Rn×n , then
λ
A A T
2
≺w
λ
A A T
2
Lemma 2.6 see 17 Let 0 < m1≤ a k ≤ M1, 0 < m2≤ b k ≤ M2, k 1, 2, , n, 1/p 1/q 1 Then
n
k1 akbk≤
n
k1
a p k
1/p n
k1
b q k
1/q
≤ c p,q n
k1 akbk, 2.6
where
p
1M2q − m p
1m q2
p
M1m2M q2− m1M2m q2 1/p
q
m1M2M p1− M1m2m p1 1/q 2.7
Trang 5Note that if m1 0, m2/ 0 or m2 0, m1/ 0, obviously, 2.6 holds If m1 m2 0, choose
cp,q ∞, then 2.6 also holds.
Remark 2.7 If p q 2, then we obtain Cauchy-Schwartz inequality
n
k1 akbk≤
n
k1
a2
k
1/2 n
k1
b2
k
1/2
≤ c2
n
k1 akbk, 2.8
where
c2 M1M2
m1m2
m1m2
Remark 2.8 Note that
lim
p → ∞
a p1 a p
2 · · · a p
n
1/p
max
1≤k≤n
ak
,
lim
p → ∞
q → 1
cp,q lim
p → ∞
q → 1
M p1M2q − m p
1m q2
p
M1m2M q2− m1M2m q2 1/p
q
m1M2M p1− M1m2m p1 1/q
lim
p → ∞
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/M1p 1/q
lim
p → ∞
q → 1
M2
M 1/pp/q−p1 m1M2
lim
p → ∞
q → 1
1
M11/p−1 m1
M1
m1.
2.10
Let p → ∞, q → 1 in 2.6, then we obtain
m1
n
k1
bk≤n
k1 akbk ≤ M1
n
k1
Lemma 2.9 If q ≥ 1, a i ≥ 0 i 1, 2, , n, then
1
n
n
i1 ai
q
≤ 1
n
n
i1
Trang 6Proof 1 Note that q 1, or a i 0 i 1, 2, , n,
1
n
n
i1 ai
q
1
n
n
i1
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 ai > 0 and 1/nn
i1 ai > 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
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
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
q n
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 2.10 Let A, B ∈ R n×n be arbitrary matrices with the following singular value decomposition:
B Udiagσ1B, σ2B, , σ n BV T , 2.17
where U, V ∈ R n×n are orthogonal Then
n
i1
σ i Bd n−i1V T AU
≤ trAB ≤n
i1
σ i Bd iV T AU
. 2.18
Proof By the matrix theory we have
trAB tr AUdiag
σ1B, σ2B, , σ n BV T
tr V T AUdiag
σ1B, σ2B, , σ n B
n
i1
σi BV T AU
ii
2.19
Trang 7Since σ1B ≥ σ2B ≥ · · · ≥ σ n B ≥ 0, without loss of generality, we may assume σB
σ1B, σ2B, , σ n B Next, we will prove the left-hand side of 2.18:
n
i1
σ i Bd n−i1V T AU
≤n
i1
σ i Bd i
V T AU
If
d
V T AU
d n
V T AU
, d n−1
V T AU
, , d1
V T AU
, 2.21
we obtain the conclusion Now assume that there exists j < k such that d j V T AU >
dk V T AU, then
σ j Bd k
V T AU
σ k Bd j
V T AU
− σ j Bd j
V T AU
− σ k Bd k
V T AU
σ j B − σ k B dk
V T AU
− d j
V T AU
We use dV T AU to denote the vector of dV T AU after changing dj V T AU and dk V T AU,
then
n
i1
σ i B di
V T AU
≤n
i1
σ i Bd i
V T AU
After limited steps, we obtain the the left-hand side of2.18 For the right-hand side of 2.18,
n
i1
σ i Bd i
V T AU
≤n
i1
σ i Bd i
V T AU
If
d
V T AU
d1V T AU
, d2
V T AU
, , d n
V T AU
, 2.25
we obtain the conclusion Now assume that there exists j > k such that d j V T AU <
dk V T AU, then
σ j Bd k
V T AU
σ k Bd j
V T AU
− σ j Bd j
V T AU
− σ k Bd k
V T AU
σ j B − σ k B dk
V T AU
− d j
V T AU
Trang 8We use dV T AU to denote the vector of dV T AU after changing dj V T AU and dk V T AU,
then
n
i1
σ i Bd i
V T AU
≤n
i1
σ i B di
V T AU
After limited steps, we obtain the right-hand side of2.18 Therefore,
n
i1
σ i Bd n−i1V T AU
≤ trAB ≤n
i1
σ i Bd iV T AU
. 2.28
This completes the proof
Since trAB trBA, applying 2.18 with B in lieu of A, we immediately have the
following corollary
Corollary 2.11 Let A, B ∈ R n×n be arbitrary matrices with the following singular value decomposition:
A P diagσ1A, σ2A, , σ n AQ T , 2.29
where P, Q ∈ R n×n are orthogonal Then
n
i1
σ i Ad n−i1
Q T BP
≤ trAB ≤n
i1
σ i Ad i Q T BP . 2.30
Now using2.18 and 2.30, one finally has the following theorem
Theorem 2.12 Let A, B ∈ R n×n be arbitrary matrices with the following singular value decompositions, respectively:
A P diag
σ1A, σ2A, , σ n AQ T ,
B Udiag
σ1B, σ2B, , σ n BV T ,
2.31
where P, Q, U, V ∈ R n×n are orthogonal Then
max
n
i1
σ i Ad n−i1
Q T BP
, n
i1
σ i Bd n−i1
V T AU
≤ trAB ≤ min
n
i1
σ i Bd i
V T AU
, n
i1
σ i Ad i
Q T BP
.
2.32
Trang 9Remark 2.13 We point out that2.18 improves 1.11 In fact, it is obvious that
min
1≤i≤n
V T AU
ii
n
i1
σ i B ≤n
i1
σ i Bd n−i1V T AU
≤ trAB ≤n
i1
σ i Bd i
V T AU
≤ max
1≤i≤n
V T AU
ii n
i1
σ i B.
2.33
This implies that2.18 improves 1.11
Remark 2.14 We point out that2.18 improves 1.13 Since for i 1, , n, σ i B ≥ 0 and
di V T AU di V T AU V T AU T /2, from Lemmas2.1and2.4, then2.18 implies
n
i1
σ i Bλ n−i1
V T AU
V T AUT
2
≤n
i1
σ i Bd n−i1
V T AU
V T AUT
2
≤ trAB
≤n
i1
σ i Bd i
V T AU
V T AUT
2
≤n
i1
σ i Bλ i
V T AU
V T AUT
2.34
In fact, for i 1, 2, , n, we have
λi
V T AU V T AU T
V T AUV
T AUV TT
λ i
AUV T AUV TT
2
λ i AS.
2.35
Trang 10Then2.34 can be rewritten as
n
i1
σ i Bλ n−i1 AS ≤n
i1
σ i Bd n−i1
V T AU
≤ trAB
≤n
i1
σ i Bd iV T AU
≤n
i1
σ i Bλ i AS.
2.36
This implies that2.18 improves 1.13
Remark 2.15 We point out that1.13 improves 1.7 In fact, fromLemma 2.5, we have
λAS≺wσAS. 2.37
Since S is orthogonal, σAS σA Then 2.37 is rewritten as follows: λAS≺ wσA By
using σ1B ≥ σ2B ≥ · · · ≥ σ n B ≥ 0 andLemma 2.2, we obtain
n
i1
σ i Bλ i AS ≤n
i1
σ i Bσ i A. 2.38
Note that λ i −AS −λ n−i1 AS, fromLemma 2.2and2.38, we have
−n
i1
σ i Bλ n−i1 AS n
i1
σ i Bλ i −AS
≤n
i1
σ i Bλ i AS ≤n
i1
σ i Bσ i A.
2.39
Thus, we obtain
−n
i1
σ i Bσ i A ≤n
i1
Both2.38 and 2.40 show that 1.13 is tighter than 1.7
Trang 113 Applications of the Results
Wang et al in6 have obtained the following: let K be the positive semidefinite solution of
the ARE1.4 Then the trace of matrix K has the lower and upper bounds given by
λ n A
λ n A 2 λ1RtrQ
λ1R ≤ trK ≤
λ1A
λ1A 2 λ n R/ntrQ
λ n R/n .
3.1
In this section, we obtain the application in the algebraic Riccati equation of our results including3.1 Some of our results and 3.1 cannot contain each other
Theorem 3.1 If 1/p 1/q 1 and K is the positive semidefinite solution of the ARE 1.4, then
1 the trace of matrix K has the lower and upper bounds given by
λ n A
λ n A 2 n
i1 λ p i R 1/ptrQ
n
i1 λ p i R 1/p
≤ trK ≤ λ1A
λ1A 2
1/c p,qn2−1/q n
i1 λ p i R 1/ptrQ
1/c p,qn2−1/q n
i1 λ p i R 1/p .
3.2
2 If A A T /2 ≥ 0, then the trace of matrix K has the lower and upper bounds given by
1/c p,q n1−1/q n
i1 λ p i A 1/p
1/c p,qn1−1/q n
i1 λ p i A 1/p 2 n
i1 λ p i R 1/ptrQ n
i1 λ p i R 1/p
≤ trK
≤
n
i1 λ p i A 1/p
n
i1 λ p i A 2/p 1/c p,qn2−1/q n
i1 λ p i R 1/p trQ
1/c p,qn2−1/q n
i1 λ p i R 1/p .
3.3
Trang 123 If A A T /2 ≤ 0, then the trace of matrix K has the lower and upper bounds given by
− n
i1λ n−i1 Ap 1/p
n
i1λ n−i1 Ap 2/p
1/c p,qn2−1/q n
i1 λ p i R 1/p trQ
1/c p,qn2−1/q n
i1 λ p i R 1/p
≤ trK
≤
− 1/c
p,q n1−1/q n
i1λ n−i1 Ap 1/p
n
i1 λ p i R 1/p
1/c p,qn1−1/q n
i1λ i Ap 1/p 2 n
i1 λ p i R 1/ptrQ
n
i1 λ p i R 1/p ,
3.4
where
p
r M q k − m p r m q k
p
Mr mkM k q − m rMkm q k 1/p
q
mrMkM p r − M r mkm p r
1/q ,
Mr λ1R, m r λ n R, M k λ1K, m k λ n K,
c p,q M
p
1M q k − m p
1m q k
p
M1mkM q k − m1Mkm q k 1/p
q
m1MkM p1− M1mkm p1 1/q ,
M1 λ1A, m1 λ n A.
3.5
Proof. 1 Take the trace in both sides of the matrix ARE 1.4 to get
tr
A T K
trKA − trKRK −trQ. 3.6
Since K is symmetric positive definite matrix, λK σK, trK n
i1 σ i K, and from
Lemma 2.9, we have
trK
n1−1/q ≤ tr
K q 1/q
n
i1
σ i KK n
i1
σ i2 K ≤
n
i1
σ i K
2
trK 2
... ResultsWang et al in 6 have obtained the following: let K be the positive semidefinite solution of< /i>
the ARE1.4 Then the trace of matrix K has the lower and upper bounds given by...
In this section, we obtain the application in the algebraic Riccati equation of our results including3.1 Some of our results and 3.1 cannot contain each other
Theorem...
Trang 123 If A A T /2 ≤ 0, then the trace of matrix K has the lower and upper bounds given