Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 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, [email protected]
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.
Copyright q 2009 J. Liu and J. Zhang. This is an open access article distributed under t he 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 practice see 1–5. Consider the following
linear system see 4:
˙xtAxtBut,x0x
0
, 1.1
with the cost
J
where x
0
∈ R
n
is the initial state of the systems 1.1 and 1.2, K is the positive definite
solution of the following algebraic Riccati equation ARE:
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 to 1.4, 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 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 ARE6–12.
In addition, from 2, 6, we know that an interpretation of trK is that trK/n is the average
value of the optimal cost J
∗
as x
0
varies 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 in 13–15.
,letdAd
1
A,d
2
A, ,d
n
A,λAλ
1
A,λ
2
A, ,λ
n
A,σA
σ
1
A,σ
2
A, ,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
i1
β
i
, 1.6
then it is said that α is controlled by β, which is signed by α ≺ β.
Journal of Inequalities and Applications 3
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 in 16 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. in 13 have observed another result. Let A, B ∈ R
n×n
i1
σ
i
B, 1.9
where S UV
T
is orthogonal.
Liu and He in 14 have obtained the following: let A, B ∈ R
n×n
be arbitrary matrices
with the following singular value decomposition:
B Udiag
σ
1
B,σ
2
B, ,σ
n
B
V
T
, 1.10
where U, V ∈ R
n×n
are orthogonal. Then
min
1≤i≤n
σ
1
B,σ
2
B, ,σ
n
B
V
T
, 1.12
where U, V ∈ R
n×n
are orthogonal. Then
n
i1
σ
i
Bλ
n−i1
AS ≤ trAB ≤
n
i1
σ
i
Bλ
i
i1
y
i
u
i
. 2.1
Lemma 2.2 see 16, page 95, H.3.b. If x
1
≥ ··· ≥ x
n
,y
1
≥ ··· ≥ y
n
and x≺
w
y, then for
any real array u
1
≥···≥ u
n
≥ 0,
n
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 A
T
∈ R
n×n
,then
dA ≺ λA. 2.4
Lemma 2.5 see 16, page 240, F.4.a. Let A ∈ R
n×n
,then
λ
A A
T
2
≺
w
k1
a
k
b
k
≤
n
k1
a
p
k
1/p
n
k1
b
q
k
1/q
≤ c
p,q
n
k1
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
k1
a
2
k
1/2
n
k1
b
2
k
1/2
≤ c
2
n
k1
a
k
b
k
, 2.8
where
c
2
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
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
m
1
M
2
−M
1
m
2
m
1
/M
1
p
1/q
lim
p →∞
q → 1
M
2
M
1/pp/q−p
1
m
1
M
2
lim
p →∞
≤ M
1
n
k1
b
k
. 2.11
Lemma 2.9. If q ≥ 1, a
i
≥ 0 i 1, 2, ,n,then
1
n
n
i1
a
i
q
≤
1
n
n
i1
a
q
i
xqx
q−1
> 0andf
x
qq − 1x
q−2
> 0. Thus, fx is a convex function. As a
i
> 0and1/n
n
i1
a
i
> 0, from the
property of the convex function, we have
1
n
n
i1
a
i
q
f
> 0 i
r 1, ,n. Then from 2, we have
1
n − r
q
n
i1
a
i
q
1
n − r
n
i1
a
i
q
≤
1
n − r
n
n
i1
a
i
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.
i1
σ
i
Bd
i
V
T
AU
. 2.18
Proof. By the matrix theory we have
trABtr
AUdiag
σ
1
B,σ
2
B, ,σ
n
B
V
T
tr
B ≥ σ
2
B ≥ ··· ≥ σ
n
B ≥ 0, without loss of generality, we may assume σB
σ
1
B,σ
2
B, ,σ
n
B. Next, we will prove the left-hand side of 2.18:
n
i1
σ
i
Bd
n−i1
V
T
AU
≤
n
i1
σ
i
, ,d
1
V
T
AU
, 2.21
we obtain the conclusion. Now assume that there exists j<ksuch that d
j
V
T
AU >
d
k
V
T
AU, then
σ
j
Bd
k
V
T
AU
σ
k
k
B
d
k
V
T
AU
− d
j
V
T
AU
≤ 0.
2.22
We use
dV
T
AU to denote the vector of dV
T
AU after changing d
j
V
T
AU and d
AU
. 2.23
After limited steps, we obtain the the left-hand side of 2.18. 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
. 2.24
we obtain the conclusion. Now assume that there exists j>ksuch that d
j
V
T
AU <
d
k
V
T
AU, then
σ
j
Bd
k
V
T
AU
σ
k
Bd
j
V
T
AU
− σ
j
− d
j
V
T
AU
≥ 0.
2.26
8 Journal of Inequalities and Applications
We use
dV
T
AU to denote the vector of dV
T
AU after changing d
j
V
T
AU and d
k
V
T
AU,
then
n
i1
i
Bd
n−i1
V
T
AU
≤ trAB ≤
n
i1
σ
i
Bd
i
V
T
AU
. 2.28
This completes the proof.
Since trABtrBA, 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 Pdiagσ
i
Q
T
BP. 2.30
Now using 2.18 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 Pdiag
σ
1
A,σ
2
A, ,σ
n
A
Q
T
,
B Udiag
σ
1
B,σ
2
B, ,σ
n
n−i1
V
T
AU
≤ trAB ≤ min
n
i1
σ
i
Bd
i
V
T
AU
,
n
i1
σ
i
Ad
i
V
T
AU
≤ trAB ≤
n
i1
σ
i
Bd
i
V
T
AU
≤ max
1≤i≤n
V
T
AU
ii
n
i1
σ
AU
V
T
AU
T
2
≤
n
i1
σ
i
Bd
n−i1
V
T
AU
V
T
AU
T
2
≤ trAB
T
AU
V
T
AU
T
2
.
2.34
In fact, for i 1, 2, ,n, we have
λ
i
V
T
AU V
T
AU
T
2
λ
i
V
T
AUV
Bλ
n−i1
AS ≤
n
i1
σ
i
Bd
n−i1
V
T
AU
≤ trAB
≤
n
i1
σ
i
Bd
i
V
T
AU
≤
i
Bλ
i
AS ≤
n
i1
σ
i
Bσ
i
A. 2.38
Note that λ
i
−AS−λ
n−i1
AS,fromLemma 2.2 and 2.38, we have
−
n
i1
σ
i
Bλ
n−i1
AS
n
i1
σ
n
i1
σ
i
Bσ
i
A ≤
n
i1
σ
i
Bλ
n−i1
AS. 2.40
Both 2.38 and 2.40 show that 1.13 is tighter than 1.7.
Journal of Inequalities and Applications 11
3. Applications of the Results
Wang et al. in 6 have obtained the following: let K be the positive semidefinite solution of
the ARE 1.4. Then the trace of matrix K has the lower and upper bounds given by
λ
n
A
λ
n
A
λ
n
A
λ
n
A
2
n
i1
λ
p
i
R
1/p
trQ
n
i1
λ
p
i
R
trQ
1/c
p,q
n
2−1/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
n
1−1/q
n
i1
λ
p
i
R
1/p
trQ
n
i1
λ
p
i
R
1/p
≤ trK
≤
n
i1
λ
p
i
A
trQ
1/c
p,q
n
2−1/q
n
i1
λ
p
i
R
1/p
.
3.3
12 Journal of Inequalities and Applications
3 If A A
T
/2 ≤ 0, then the trace of matrix K has the lower and upper bounds given by
−
n
i1
n
i1
λ
p
i
R
1/p
trQ
1/c
p,q
n
2−1/q
n
i1
λ
p
i
R
1/p
≤ trK
≤
1/c
p,q
n
1−1/q
n
i1
λ
i
A
p
1/p
2
n
i1
λ
p
m
q
k
p
M
r
m
k
M
q
k
− m
r
M
k
m
q
k
1/p
q
m
r
M
k
M
p,q
M
p
1
M
q
k
− m
p
1
m
q
k
p
M
1
m
k
M
q
k
− m
1
M
k
m
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, λKσK,trK
n
i1
σ
i
K, and from
Lemma 2.9, we have
trK
n
1−1/q
≤
tr
K
q
Journal of Inequalities and Applications 13
By the Cauchy-Schwartz inequality 2.8, it can be shown that
n
i1
σ
i
KK
n
i1
σ
2
i
K ≥
n
i1
σ
i
K
2
n
trK
n
≥
n
i1
λ
i
Rσ
i
K
2
≥
1
c
p,q
n
i1
λ
p
i
R
1/p
n
.
3.11
From 2.34,notethatλ
i
U
T
AUλ
i
A and λ
i
U
T
A
T
Uλ
i
A
T
i 1, ,n, then
we obtain
trAK ≤
n
i1
λ
i
Aσ
i
K ≤ λ
1
3.12
It is easy to see that
trA
T
KtrKA ≤
λ
1
A
T
trKλ
1
AtrK
trK
2λ
1
A
T
A
2
trK2λ
1
AtrK.
3.13
Combine 3.11 and 3.13,weobtain
T
AUλ
i
A and λ
i
U
T
A
T
U
λ
i
A
T
i 1, ,n, by 2.34, 2.6 and 3.7, we have
tr
A
T
K
≤
n
i1
λ
p
i
i1
λ
p
i
A
T
1/p
n
i1
λ
p
i
A
trK
≤ 2
n
i1
λ
p
i
λ
n
A
2
n
i1
λ
2
i
R
1/2
trQ
n
i1
λ
2
i
R
1/2
≤ trK ≤
1
n
3/2
n
i1
λ
2
i
R
1/2
,
3.17
where c
1
M
r
M
k
/m
r
m
k
m
⎠
,
B
⎛
⎜
⎜
⎜
⎝
0.9892 0.1140 0.1233
0.0410 0.3096 0.5125
0.0476 0.7097 0.0962
⎞
⎟
⎟
⎟
⎠
.
4.1
Neither A nor
B is symmetric. In this case, the results of 6–12 are not valid.
Using 1.9 we obtain
0.78 ≤ trAB ≤ 1.97. 4.2
Using 1.11 yields
0.8611 ≤ trAB ≤ 1.9090. 4.3
By 2.18,weobtain
1.0268 ≤ trAB ≤ 1.7524, 4.4
where both lower and upper bounds are better than those of 4.2 and 4.3.
Example 4.2. Let
A
⎛
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
Neither A nor B is symmetric. In this case, the results of 6–12 are not valid.
16 Journal of Inequalities and Applications
Using 1.7 yields
−6.1424 ≤ trAB ≤ 6.1424. 4.6
From 1.9 we have
−1.5007 ≤ trAB ≤ 5.0110. 4.7
Using 1.11 yields
−3.1058 ≤ trAB ≤ 6.0736. 4.8
By 1.13,weobtain
−0.7267 ≤ trAB ≤ 4.3399. 4.9
The bound in 2.18 yields
−0.5375 ≤ tr
AB ≤ 4.2659. 4.10
Obviously, 4.10 is tighter than 4.6, 4.7, 4.8 and 4.9.
Example 4.3. Consider the systems 1.1, 1.2 with
A
⎛
⎜
⎜
⎜
⎝
538 440 266
440 441 321
266 321 296
⎞
⎟
⎟
⎟
⎠
. 4.11
Moreover, the corresponding ARE 1.4 with R BB
T
, A, R is stabilizable and Q, A is
observable.
Using 3.17 yields
8.5498 ≤ trK ≤ 47.9041. 4.12
Using 3.1 we obtain
9.0132 ≤ trK ≤ 19.0099, 4.13
where both lower and upper bounds are better than those of 4.12.
Journal of Inequalities and Applications 17
Example 4.4. Consider the systems 1.1, 1.2 with
A
⎛
⎜
⎜
⎜
⎝
17.57.45 3.465
7.45 9.77.845
3.465 7.845 9.905
⎞
⎟
⎟
⎟
⎠
.
4.14
Moreover, the corresponding ARE 1.4 with R BB
T
, A, R is stabilizable and Q, A is
observable.
Using 3.1 we obtain
1.6039 ≤ trK ≤ 5.6548. 4.15
Using 3.17 yields
1.6771 ≤ trK ≤ 5.5757, 4.16
where both lower and upper bounds are better than those of 4.15.
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 t he 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 China 10671164, Science and
Research Fund of Hunan Provincial Education Department 06A070.
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.
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