vol20_pp314-321. 100KB Jun 04 2011 12:06:08 AM
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
ELA
http://math.technion.ac.il/iic/ela
A NEW UPPER BOUND FOR THE EIGENVALUES OF THE
CONTINUOUS ALGEBRAIC RICCATI EQUATION∗
JIANZHOU LIU† , JUAN ZHANG‡ , AND YU LIU§
Abstract. By using majorization inequalities, we propose a new upper bound for summations
of eigenvalues (including the trace) of the solution of the continuous algebraic Riccati equation. The
bound extends some of the previous results under certain conditions. Finally, we give a numerical
example to illustrate the effectiveness of our results.
Key words. Eigenvalue, Majorization inequality, Trace inequality, Algebraic Riccati equation.
AMS subject classifications. 15A24.
1. Introduction. We consider the continuous algebraic Riccati equation(CAR
E)
(1.1)
AT K + KA − KRK = −Q,
where Q ≥ 0, R > 0. The CARE has a maximal positive semidefinite solution K,
which is the solution of practical interest (see Theorem 9.4.4 in Lancaster and Rodman
[1]). Various bounds for this solution have been presented in [2-12]. These include
norm bounds, eigenvalue bounds and matrix bounds. In this paper, we apply majorization inequality methods in Marshall and Olkin [13] to obtain a new upper bound
for summations of eigenvalues (including the trace) of the solution of the CARE. The
bound is a refinement of an upper bound presented in [10] under certain conditions.
2. A new upper bound for summations of eigenvalues for the solution of
the continuous algebraic Riccati equation. Throughout this paper, we use Rn×n
to denote the set of n × n real matrices. Let Rn = {(x1 , . . . , xn ) : xi ∈ R for all i},
Rn++ = {(x1 , . . . , xn ) : xi > 0 for all i}, D = {(x1 , . . . , xn ) : x1 ≥ · · · ≥ xn }, D+ =
∗ Received by the editors January 4, 2010. Accepted for publication April 19, 2010. Handling
Editor: Peter Lancaster.
† Department of Mathematical Science and Information Technology, Hanshan Normal University,
Chaozhou, Guangdong 521041, China, and Department of Mathematics and Computational Science,
Xiangtan University, Xiangtan, Hunan 411105, China ([email protected], corresponding author).
Supported in part by Natural Science Foundation of China (10971176).
‡ Department of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan
411105, China.
§ Department of Mathematical Science and Information Technology, Hanshan Normal University,
Chaozhou, Guangdong 521041, China.
314
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
ELA
http://math.technion.ac.il/iic/ela
Eigenvalues of the Continuous Algebraic Riccati Equation
315
{(x1 , . . . , xn ) : x1 ≥ · · · ≥ xn ≥ 0}. Suppose x = (x1 , x2 , . . . , xn ) is a real n-element
array which is reordered so that its elements are arranged in non-increasing order.
i.e., x[1] ≥ x[2] ≥ · · · ≥ x[n] . Let |x| = (|x1 |, |x2 |, . . . , |xn |). For A = (aij ) ∈ Rn×n ,
denote by d(A) = (d1 (A), d2 (A), . . . , dn (A)) and λ(A) = (λ1 (A), λ2 (A), . . . , λn (A))
the diagonal elements and the eigenvalues of A, respectively. Let tr(A) and AT
denote the trace and the transpose of A, respectively, and define (A)ii = aii = di (A)
T
. The notation A > 0 (A ≥ 0) is used to denote that A is a symmetric
and A = A+A
2
positive definite (semi-definite) matrix.
Let α and β be two real n-element arrays.
If they satisfy
k
X
α[i] ≤
k
X
β[i] , k = 1, 2, . . . , n,
i=1
i=1
then it is said that α is weakly majorized by β, which is denoted by α ≺w β.
If they satisfy
k
X
α[n−i+1] ≥
k
X
β[n−i+1] , k = 1, 2, . . . , n,
i=1
i=1
then it is said that α is weakly submajorized by β, which is denoted by α ≺w β.
If α ≺w β and
n
X
α[i] =
n
X
β[i] ,
i=1
i=1
then it is said that α is majorized by β, which is denoted by α ≺ β.
The following lemmas are used to prove the main results.
Lemma 2.1. [13, p. 141, A.3] If xi , yi , i = 1, 2, . . . , n, are two sets of numbers,
then
n
X
x[i] y[n−i+1] ≤
n
X
xi yi ≤
i=1
i=1
n
X
x[i] y[i] .
i=1
Lemma 2.2. [13, p. 95, H.3.b] If x, y ∈ D and x ≺w y, then for any k =
1, 2, . . . , n,
k
X
i=1
x[i] u[i] ≤
k
X
i=1
y[i] u[i] ,
∀u ∈ D+ .
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
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316
J. Liu, J. Zhang, and Y. Liu
Lemma 2.3. [13, p. 95, H.3.c] If x, y ∈ D+ , a, b ∈ D+ , (x1 , . . . , xn ) ≺w
(y1 , . . . , yn ) and (a1 , . . . , an ) ≺w (b1 , . . . , bn ), then
(a1 x1 , . . . , an xn ) ≺w (b1 y1 , . . . , bn yn ).
Lemma 2.4. [13, p. 218, B.1] If A = AT ∈ Rn×n , then
d(A) ≺ λ(A),
which is equivalent to
d(A) ≺w λ(A) and d(A) ≺w λ(A).
Lemma 2.5. [13, p. 118, A.2.h] If x, y ∈ Rn++ and x ≺w y, then for r < 0,
(xr1 , . . . , xrn ) ≺w (y1r , . . . , ynr ).
Remark 2.6. If A = AT ∈ Rn×n , then we have
d(A) ≺w λ(A)
(2.1)
from Lemma 2.4. Combining (2.1) with Lemma 2.5, we obtain for A > 0,
1
1
1
1
(2.2)
≺w
.
,...,
,...,
d1 (A)
dn (A)
λ1 (A)
λn (A)
Lemma 2.7. (Cauchy-Schwartz inequality) For real numbers ai and bi , i =
1, 2, . . . , n,
! 12
! 21
n
n
n
X
X
X
2
2
ai
ai b i ≤
bi
(2.3)
.
i=1
i=1
i=1
We are now ready to give a new upper bound for summations of eigenvalues
(including the trace) of the solution of the continuous algebraic Riccati equation,
which improves under certain conditions the following bound in Komaroff [10]: Let K
be the positive semi-definite solution of the CARE (1.1). Then for any l = 1, 2, . . . , n,
s
l
λ (R) P
λ[j] (Q)
lλ[1] (A) + l λ2[1] (A) + [n]l
l
X
j=1
(2.4)
.
λ[j] (K) ≤
λ[n] (R)
j=1
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
ELA
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Eigenvalues of the Continuous Algebraic Riccati Equation
317
Theorem 2.8. Let K be the positive semi-definite solution of the CARE (1.1)
and assume that A ≥ 0. Then for any l = 1, 2, . . . , n, we have
v
u l
l
l
l
X λ2[j] (A)
X
X
X
λ[j] (Q)
λ[j] (A)
1u
t
2
+
(2.5)
+l
.
λ[j] (K) ≤
2
(R)
λ
(R)
λ
λ
(R)
j=1 [n−j+1]
j=1 [n−j+1]
j=1 [n−j+1]
j=1
Proof. Since K is the positive semi-definite solution of the CARE (1.1), we have
K = U diag(λ[1] (K), λ[2] (K), . . . , λ[n] (K))U T ,
where U ∈ Rn×n is orthogonal. Thus, (1.1) can be written as
e+A
eT Λk + Q,
e
e k = Λk A
Λk RΛ
(2.6)
e = U T RU, A
e = U T AU, Q
e = U T QU, Λk = diag(λ[1] (K), λ[2] (K), . . . , λ[n] (K)).
where R
Then from (2.6), for j = 1, . . . , l, we have
e+A
eT Λk ) + dj (Q)
e = 2λ[j] (K)dj (A)
e k ) = dj (Λk A
e = dj (Λk RΛ
e + dj (Q).
e
λ2[j] (K)dj (R)
Hence,
(2.7)
l
X
λ2[j] (K) = 2
l
X
λ[j] (K)
j=1
j=1
l
X
e
e
dj (Q)
dj (A)
+
.
e
e
dj (R)
dj (R)
j=1
e ≥ d[n−j+1] (R).
e Furthermore, we have
Obviously, we have d[l−j+1] (R)
!
!
1
1
1
1
≺w
,
,...,
,...,
e
e
e
e
d1 (R)
dl (R)
d[n] (R)
d[n−l+1] (R)
e . . . , dl (A))
e ≺w (d[1] (A),
e . . . , d[l] (A)).
e
(d1 (A),
Applying Lemma 2.1 and Lemma 2.3 yields
(2.8)
l
X
e
dj (Q)
j=1
(2.9)
e
dj (R)
≤
l
X
j=1
l
e
e
X
d[j] (Q)
d[j] (Q)
≤
,
e
e
d[l−j+1] (R)
j=1 d[n−j+1] (R)
l
l
l
e
e
X
X
X
e
d[j] (A)
d[j] (A)
dj (A)
≤
≤
.
e
e
e
j=1 dj (R)
j=1 d[l−j+1] (R)
j=1 d[n−j+1] (R)
According to Lemma 2.2 and (2.9), it is evident that
(2.10)
l
X
j=1
λ[j] (K)
l
e
X
e
d[j] (A)
dj (A)
λ[j] (K)
≤
.
e
e
dj (R)
d[n−j+1] (R)
j=1
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
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318
J. Liu, J. Zhang, and Y. Liu
By (2.8), (2.10), (2.2), Lemma 2.3 and Lemma 2.4, (2.7) leads to
l
X
(2.11)
λ2[j] (K) ≤ 2
l
X
λ[j] (K)
l
X
λ[j] (K)
j=1
j=1
≤2
j=1
e
d[j] (A)
e
d[n−j+1] (R)
+
l
X
j=1
e
d[j] (Q)
e
d[n−j+1] (R)
l
X
λ[j] (A)
λ[j] (Q)
+
.
λ[n−j+1] (R) j=1 λ[n−j+1] (R)
Consequently,
l
X
λ2[j] (K) − 2
l
X
j=1
λ[j] (K)
j=1
j=1
≤
l
X
λ2[j] (A)
λ2[n−j+1] (R)
+
l
X
j=1
l
X
λ2[j] (A)
λ[j] (A)
+
λ[n−j+1] (R) j=1 λ2[n−j+1] (R)
λ[j] (Q)
,
λ[n−j+1] (R)
which is equivalent to
l
X
(2.12)
j=1
λ[j] (A)
λ[j] (K) −
λ[n−j+1] (R)
!2
≤
l
X
j=1
λ2[j] (A)
λ2[n−j+1] (R)
+
l
X
j=1
λ[j] (Q)
.
λ[n−j+1] (R)
By the Cauchy-Schwartz inequality (2.3), it can be shown that
(2.13)
l
X
j=1
λ[j] (A)
λ[j] (K) −
λ[n−j+1] (R)
!2
2
l
λ[j] (A)
1 X
|λ[j] (K) −
≥
|
l j=1
λ[n−j+1] (R)
! 2
X
l
λ[j] (A)
1
.
≥
λ[j] (K) −
l j=1
λ[n−j+1] (R)
Combining (2.12) with (2.13) implies that
v
u l
X
l
l
X
X λ2[j] (A)
X
l
λ[j] (A)
λ[j] (Q)
1u
2t
λ
(K)
−
+
.
≤
l
[j]
2
(R)
λ
(R)
λ
λ
(R)
j=1
j=1 [n−j+1]
j=1 [n−j+1]
j=1 [n−j+1]
Therefore,
l
X
j=1
λ[j] (K) ≤
l
X
j=1
v
u l
l
X λ2[j] (A)
X
λ[j] (Q)
λ[j] (A)
1u
+
+ l2 t
.
2
(R) j=1 λ[n−j+1] (R)
λ[n−j+1] (R)
λ
j=1 [n−j+1]
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
ELA
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Eigenvalues of the Continuous Algebraic Riccati Equation
319
Corollary 2.9. Let K be the positive semi-definite solution of the CARE (1.1)
and assume that A ≥ 0. The trace of matrix K has the bound given by
v
u n
n
n
X
X λ2[j] (A)
X
λ[j] (Q)
λ[j] (A)
1u
+
+ n2 t
.
tr(K) ≤
2
(R) j=1 λ[n−j+1] (R)
λ
(R)
λ
j=1 [n−j+1]
j=1 [n−j+1]
Remark 2.10. We point out that (2.5) improves (2.4) when A ≥ 0. Actually, if
1
1
A ≥ 0, noting that for j = 1, 2, . . . , l, λ[n−j+1]
(R) ≤ λ[n] (R) , then we have
l
X
j=1
v
u l
l
X
X λ2[j] (A)
λ[j] (Q)
λ[j] (A)
1u
2
+
+l t
2
(R)
λ[n−j+1] (R)
λ
λ
(R)
j=1 [n−j+1]
j=1 [n−j+1]
v
u
n
X
λ2[j] (A)
λ[j] (A)
λ[j] (Q)
1u
≤ l max
+
+ l 2 tl max 2
1≤j≤l λ[n−j+1] (R)
1≤j≤l λ
λ
(R)
[n−j+1] (R)
j=1 [n−j+1]
v
!2
u
l
u λ[1] (A)
lλ[1] (A)
1 X λ[j] (Q)
t
+l
+
.
=
λ[n] (R)
λ[n] (R)
l j=1 λ[n−j+1] (R)
v
!2
u
l
u λ[1] (A)
X
lλ[1] (A)
1
t
≤
λ[j] (Q)
+l
+
λ[n] (R)
λ[n] (R)
lλ[n] (R) j=1
v
u
l
X
u
lλ[1] (A)
l
tλ2 (A) + λ[n] (R)
+
=
λ[j] (Q).
[1]
λ[n] (R)
λ[n] (R)
l
j=1
This implies that (2.5) is better than (2.4) when A ≥ 0.
3. A numerical example. In this section, we present a numerical example to
illustrate the effectiveness of the main results.
Example 3.1. Let
11
2 8
1
9
2
A=
−2 −1 2
1
4 3
7
3
, R =
−5
12
10 −1 6
2
−1 9
4
3
,
6
4
16 −5
2
3 −5 12
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
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320
J. Liu, J. Zhang, and Y. Liu
2283 809 1003 1022
809 2693 1170 1423
Q=
1003 1170 1119 374 .
1022 1423 374 1458
Obviously, A ≥ 0.
Case 1: Choose l = 3. Using (2.4) yields
(3.1)
3
X
λ[j] (K) ≤ 183.45.
3
X
λ[j] (K) ≤ 130.85,
j=1
By (2.5), we have
j=1
which is better than that of (3.1).
Case 2: Choose l = n = 4. Using (2.4) yields
(3.2)
tr(K) ≤ 244.82.
By (2.5), we have
tr(K) ≤ 148.75,
which is better than that of (3.2).
Acknowledgment. The authors would like to thank Professor Peter Lancaster
and a referee for the very helpful comments and suggestions to improve the contents
and presentation of this paper.
REFERENCES
[1] P. Lancaster and L. Rodman. Algebraic Riccati Equations. Oxford University Press, 1995.
[2] S.D. Wang, T.S. Kuo, and C.F. Hsu. The bounds on the solution of the algebraic Riccati and
Lyapunov equation. IEEE Trans. Automat. Control, 31:654–656, 1986.
[3] R.V. Patel and M. Toda. On norm bounds on the solution of the algebraic Riccati equation.
IEEE Trans. Automat. Control, 23:87–88, 1978.
[4] K. Yasuda and K. Hirai. Upper and lower bounds on the solution of algebraic Riccati equation.
IEEE Trans. Automat. Control, 24:483–487, 1979.
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
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Eigenvalues of the Continuous Algebraic Riccati Equation
321
[5] T. Mori. A note on bounds for the solution to the algebraic Riccati and Lyapunov matrix
equations. IEEE Trans. Automat. Control, 62:760–761, 1979.
[6] V.R. Karanam. Eigenvalue bounds for the algebraic Riccati and Lyapunov matrix equations.
IEEE Trans. Automat. Control, 28:109–111, 1982.
[7] J.M. Saniuk and I.B. Rhodes. A matrix inequality associated with bounds on solutions of
algebraic Riccati and Lyapunov equations. IEEE Trans. Automat. Control, 32:739–740,
1987.
[8] T. Mori. Comments on a matrix inequality associated with bounds on solutions of algebraic
Riccati and Lyapunov equations. IEEE Trans. Automat. Control, 33:1088–1091, 1988.
[9] N. Komaroff. Bounds on eigenvalues of matrix products with an application to the algebraic
Riccati equation. IEEE Trans. Automat. Control, 35:348–350, 1990.
[10] N. Komaroff. Diverse bounds for the eigenvalues of the continuous algebraic Riccati equation.
IEEE Trans. Automat. Control, 39:532–534, 1994.
[11] W.H. Kwon, Y.S. Moon, and S.C. Ahn. Bounds on solutions of algebraic Riccati and Lyapunov
equations: A survey and new results. Internat. J. Control, 64:377–389, 1996.
[12] C.H. Lee. On the upper and lower bounds of the solution for the continuous Riccati matrix
equation. Internat. J. Control, 66:105–118, 1997.
[13] A.W. Marshall and I. Olkin. Inequalities Theory of Majorisation and Its Applications. Academic Press, New York, 1979.
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
ELA
http://math.technion.ac.il/iic/ela
A NEW UPPER BOUND FOR THE EIGENVALUES OF THE
CONTINUOUS ALGEBRAIC RICCATI EQUATION∗
JIANZHOU LIU† , JUAN ZHANG‡ , AND YU LIU§
Abstract. By using majorization inequalities, we propose a new upper bound for summations
of eigenvalues (including the trace) of the solution of the continuous algebraic Riccati equation. The
bound extends some of the previous results under certain conditions. Finally, we give a numerical
example to illustrate the effectiveness of our results.
Key words. Eigenvalue, Majorization inequality, Trace inequality, Algebraic Riccati equation.
AMS subject classifications. 15A24.
1. Introduction. We consider the continuous algebraic Riccati equation(CAR
E)
(1.1)
AT K + KA − KRK = −Q,
where Q ≥ 0, R > 0. The CARE has a maximal positive semidefinite solution K,
which is the solution of practical interest (see Theorem 9.4.4 in Lancaster and Rodman
[1]). Various bounds for this solution have been presented in [2-12]. These include
norm bounds, eigenvalue bounds and matrix bounds. In this paper, we apply majorization inequality methods in Marshall and Olkin [13] to obtain a new upper bound
for summations of eigenvalues (including the trace) of the solution of the CARE. The
bound is a refinement of an upper bound presented in [10] under certain conditions.
2. A new upper bound for summations of eigenvalues for the solution of
the continuous algebraic Riccati equation. Throughout this paper, we use Rn×n
to denote the set of n × n real matrices. Let Rn = {(x1 , . . . , xn ) : xi ∈ R for all i},
Rn++ = {(x1 , . . . , xn ) : xi > 0 for all i}, D = {(x1 , . . . , xn ) : x1 ≥ · · · ≥ xn }, D+ =
∗ Received by the editors January 4, 2010. Accepted for publication April 19, 2010. Handling
Editor: Peter Lancaster.
† Department of Mathematical Science and Information Technology, Hanshan Normal University,
Chaozhou, Guangdong 521041, China, and Department of Mathematics and Computational Science,
Xiangtan University, Xiangtan, Hunan 411105, China ([email protected], corresponding author).
Supported in part by Natural Science Foundation of China (10971176).
‡ Department of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan
411105, China.
§ Department of Mathematical Science and Information Technology, Hanshan Normal University,
Chaozhou, Guangdong 521041, China.
314
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
ELA
http://math.technion.ac.il/iic/ela
Eigenvalues of the Continuous Algebraic Riccati Equation
315
{(x1 , . . . , xn ) : x1 ≥ · · · ≥ xn ≥ 0}. Suppose x = (x1 , x2 , . . . , xn ) is a real n-element
array which is reordered so that its elements are arranged in non-increasing order.
i.e., x[1] ≥ x[2] ≥ · · · ≥ x[n] . Let |x| = (|x1 |, |x2 |, . . . , |xn |). For A = (aij ) ∈ Rn×n ,
denote by d(A) = (d1 (A), d2 (A), . . . , dn (A)) and λ(A) = (λ1 (A), λ2 (A), . . . , λn (A))
the diagonal elements and the eigenvalues of A, respectively. Let tr(A) and AT
denote the trace and the transpose of A, respectively, and define (A)ii = aii = di (A)
T
. The notation A > 0 (A ≥ 0) is used to denote that A is a symmetric
and A = A+A
2
positive definite (semi-definite) matrix.
Let α and β be two real n-element arrays.
If they satisfy
k
X
α[i] ≤
k
X
β[i] , k = 1, 2, . . . , n,
i=1
i=1
then it is said that α is weakly majorized by β, which is denoted by α ≺w β.
If they satisfy
k
X
α[n−i+1] ≥
k
X
β[n−i+1] , k = 1, 2, . . . , n,
i=1
i=1
then it is said that α is weakly submajorized by β, which is denoted by α ≺w β.
If α ≺w β and
n
X
α[i] =
n
X
β[i] ,
i=1
i=1
then it is said that α is majorized by β, which is denoted by α ≺ β.
The following lemmas are used to prove the main results.
Lemma 2.1. [13, p. 141, A.3] If xi , yi , i = 1, 2, . . . , n, are two sets of numbers,
then
n
X
x[i] y[n−i+1] ≤
n
X
xi yi ≤
i=1
i=1
n
X
x[i] y[i] .
i=1
Lemma 2.2. [13, p. 95, H.3.b] If x, y ∈ D and x ≺w y, then for any k =
1, 2, . . . , n,
k
X
i=1
x[i] u[i] ≤
k
X
i=1
y[i] u[i] ,
∀u ∈ D+ .
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Lemma 2.3. [13, p. 95, H.3.c] If x, y ∈ D+ , a, b ∈ D+ , (x1 , . . . , xn ) ≺w
(y1 , . . . , yn ) and (a1 , . . . , an ) ≺w (b1 , . . . , bn ), then
(a1 x1 , . . . , an xn ) ≺w (b1 y1 , . . . , bn yn ).
Lemma 2.4. [13, p. 218, B.1] If A = AT ∈ Rn×n , then
d(A) ≺ λ(A),
which is equivalent to
d(A) ≺w λ(A) and d(A) ≺w λ(A).
Lemma 2.5. [13, p. 118, A.2.h] If x, y ∈ Rn++ and x ≺w y, then for r < 0,
(xr1 , . . . , xrn ) ≺w (y1r , . . . , ynr ).
Remark 2.6. If A = AT ∈ Rn×n , then we have
d(A) ≺w λ(A)
(2.1)
from Lemma 2.4. Combining (2.1) with Lemma 2.5, we obtain for A > 0,
1
1
1
1
(2.2)
≺w
.
,...,
,...,
d1 (A)
dn (A)
λ1 (A)
λn (A)
Lemma 2.7. (Cauchy-Schwartz inequality) For real numbers ai and bi , i =
1, 2, . . . , n,
! 12
! 21
n
n
n
X
X
X
2
2
ai
ai b i ≤
bi
(2.3)
.
i=1
i=1
i=1
We are now ready to give a new upper bound for summations of eigenvalues
(including the trace) of the solution of the continuous algebraic Riccati equation,
which improves under certain conditions the following bound in Komaroff [10]: Let K
be the positive semi-definite solution of the CARE (1.1). Then for any l = 1, 2, . . . , n,
s
l
λ (R) P
λ[j] (Q)
lλ[1] (A) + l λ2[1] (A) + [n]l
l
X
j=1
(2.4)
.
λ[j] (K) ≤
λ[n] (R)
j=1
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Theorem 2.8. Let K be the positive semi-definite solution of the CARE (1.1)
and assume that A ≥ 0. Then for any l = 1, 2, . . . , n, we have
v
u l
l
l
l
X λ2[j] (A)
X
X
X
λ[j] (Q)
λ[j] (A)
1u
t
2
+
(2.5)
+l
.
λ[j] (K) ≤
2
(R)
λ
(R)
λ
λ
(R)
j=1 [n−j+1]
j=1 [n−j+1]
j=1 [n−j+1]
j=1
Proof. Since K is the positive semi-definite solution of the CARE (1.1), we have
K = U diag(λ[1] (K), λ[2] (K), . . . , λ[n] (K))U T ,
where U ∈ Rn×n is orthogonal. Thus, (1.1) can be written as
e+A
eT Λk + Q,
e
e k = Λk A
Λk RΛ
(2.6)
e = U T RU, A
e = U T AU, Q
e = U T QU, Λk = diag(λ[1] (K), λ[2] (K), . . . , λ[n] (K)).
where R
Then from (2.6), for j = 1, . . . , l, we have
e+A
eT Λk ) + dj (Q)
e = 2λ[j] (K)dj (A)
e k ) = dj (Λk A
e = dj (Λk RΛ
e + dj (Q).
e
λ2[j] (K)dj (R)
Hence,
(2.7)
l
X
λ2[j] (K) = 2
l
X
λ[j] (K)
j=1
j=1
l
X
e
e
dj (Q)
dj (A)
+
.
e
e
dj (R)
dj (R)
j=1
e ≥ d[n−j+1] (R).
e Furthermore, we have
Obviously, we have d[l−j+1] (R)
!
!
1
1
1
1
≺w
,
,...,
,...,
e
e
e
e
d1 (R)
dl (R)
d[n] (R)
d[n−l+1] (R)
e . . . , dl (A))
e ≺w (d[1] (A),
e . . . , d[l] (A)).
e
(d1 (A),
Applying Lemma 2.1 and Lemma 2.3 yields
(2.8)
l
X
e
dj (Q)
j=1
(2.9)
e
dj (R)
≤
l
X
j=1
l
e
e
X
d[j] (Q)
d[j] (Q)
≤
,
e
e
d[l−j+1] (R)
j=1 d[n−j+1] (R)
l
l
l
e
e
X
X
X
e
d[j] (A)
d[j] (A)
dj (A)
≤
≤
.
e
e
e
j=1 dj (R)
j=1 d[l−j+1] (R)
j=1 d[n−j+1] (R)
According to Lemma 2.2 and (2.9), it is evident that
(2.10)
l
X
j=1
λ[j] (K)
l
e
X
e
d[j] (A)
dj (A)
λ[j] (K)
≤
.
e
e
dj (R)
d[n−j+1] (R)
j=1
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By (2.8), (2.10), (2.2), Lemma 2.3 and Lemma 2.4, (2.7) leads to
l
X
(2.11)
λ2[j] (K) ≤ 2
l
X
λ[j] (K)
l
X
λ[j] (K)
j=1
j=1
≤2
j=1
e
d[j] (A)
e
d[n−j+1] (R)
+
l
X
j=1
e
d[j] (Q)
e
d[n−j+1] (R)
l
X
λ[j] (A)
λ[j] (Q)
+
.
λ[n−j+1] (R) j=1 λ[n−j+1] (R)
Consequently,
l
X
λ2[j] (K) − 2
l
X
j=1
λ[j] (K)
j=1
j=1
≤
l
X
λ2[j] (A)
λ2[n−j+1] (R)
+
l
X
j=1
l
X
λ2[j] (A)
λ[j] (A)
+
λ[n−j+1] (R) j=1 λ2[n−j+1] (R)
λ[j] (Q)
,
λ[n−j+1] (R)
which is equivalent to
l
X
(2.12)
j=1
λ[j] (A)
λ[j] (K) −
λ[n−j+1] (R)
!2
≤
l
X
j=1
λ2[j] (A)
λ2[n−j+1] (R)
+
l
X
j=1
λ[j] (Q)
.
λ[n−j+1] (R)
By the Cauchy-Schwartz inequality (2.3), it can be shown that
(2.13)
l
X
j=1
λ[j] (A)
λ[j] (K) −
λ[n−j+1] (R)
!2
2
l
λ[j] (A)
1 X
|λ[j] (K) −
≥
|
l j=1
λ[n−j+1] (R)
! 2
X
l
λ[j] (A)
1
.
≥
λ[j] (K) −
l j=1
λ[n−j+1] (R)
Combining (2.12) with (2.13) implies that
v
u l
X
l
l
X
X λ2[j] (A)
X
l
λ[j] (A)
λ[j] (Q)
1u
2t
λ
(K)
−
+
.
≤
l
[j]
2
(R)
λ
(R)
λ
λ
(R)
j=1
j=1 [n−j+1]
j=1 [n−j+1]
j=1 [n−j+1]
Therefore,
l
X
j=1
λ[j] (K) ≤
l
X
j=1
v
u l
l
X λ2[j] (A)
X
λ[j] (Q)
λ[j] (A)
1u
+
+ l2 t
.
2
(R) j=1 λ[n−j+1] (R)
λ[n−j+1] (R)
λ
j=1 [n−j+1]
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Corollary 2.9. Let K be the positive semi-definite solution of the CARE (1.1)
and assume that A ≥ 0. The trace of matrix K has the bound given by
v
u n
n
n
X
X λ2[j] (A)
X
λ[j] (Q)
λ[j] (A)
1u
+
+ n2 t
.
tr(K) ≤
2
(R) j=1 λ[n−j+1] (R)
λ
(R)
λ
j=1 [n−j+1]
j=1 [n−j+1]
Remark 2.10. We point out that (2.5) improves (2.4) when A ≥ 0. Actually, if
1
1
A ≥ 0, noting that for j = 1, 2, . . . , l, λ[n−j+1]
(R) ≤ λ[n] (R) , then we have
l
X
j=1
v
u l
l
X
X λ2[j] (A)
λ[j] (Q)
λ[j] (A)
1u
2
+
+l t
2
(R)
λ[n−j+1] (R)
λ
λ
(R)
j=1 [n−j+1]
j=1 [n−j+1]
v
u
n
X
λ2[j] (A)
λ[j] (A)
λ[j] (Q)
1u
≤ l max
+
+ l 2 tl max 2
1≤j≤l λ[n−j+1] (R)
1≤j≤l λ
λ
(R)
[n−j+1] (R)
j=1 [n−j+1]
v
!2
u
l
u λ[1] (A)
lλ[1] (A)
1 X λ[j] (Q)
t
+l
+
.
=
λ[n] (R)
λ[n] (R)
l j=1 λ[n−j+1] (R)
v
!2
u
l
u λ[1] (A)
X
lλ[1] (A)
1
t
≤
λ[j] (Q)
+l
+
λ[n] (R)
λ[n] (R)
lλ[n] (R) j=1
v
u
l
X
u
lλ[1] (A)
l
tλ2 (A) + λ[n] (R)
+
=
λ[j] (Q).
[1]
λ[n] (R)
λ[n] (R)
l
j=1
This implies that (2.5) is better than (2.4) when A ≥ 0.
3. A numerical example. In this section, we present a numerical example to
illustrate the effectiveness of the main results.
Example 3.1. Let
11
2 8
1
9
2
A=
−2 −1 2
1
4 3
7
3
, R =
−5
12
10 −1 6
2
−1 9
4
3
,
6
4
16 −5
2
3 −5 12
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J. Liu, J. Zhang, and Y. Liu
2283 809 1003 1022
809 2693 1170 1423
Q=
1003 1170 1119 374 .
1022 1423 374 1458
Obviously, A ≥ 0.
Case 1: Choose l = 3. Using (2.4) yields
(3.1)
3
X
λ[j] (K) ≤ 183.45.
3
X
λ[j] (K) ≤ 130.85,
j=1
By (2.5), we have
j=1
which is better than that of (3.1).
Case 2: Choose l = n = 4. Using (2.4) yields
(3.2)
tr(K) ≤ 244.82.
By (2.5), we have
tr(K) ≤ 148.75,
which is better than that of (3.2).
Acknowledgment. The authors would like to thank Professor Peter Lancaster
and a referee for the very helpful comments and suggestions to improve the contents
and presentation of this paper.
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A publication of the International Linear Algebra Society
Volume 20, pp. 314-321, May 2010
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Eigenvalues of the Continuous Algebraic Riccati Equation
321
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