Trace Inequalities for Matrix Products and Trace Bounds for the Solution of the Algebraic Riccati Equations

Abstract

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.

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]):

(11)

with the cost

(12)

The optimal control rate the optimal cost of (1.1) and (1.2) are

(13)

where is the initial state of system (1.1) and (1.2) and is the positive semidefinite solution of the following algebraic Riccati equation (ARE):

(14)

with and 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 is stabilizable, and the pair 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 is that is the average value of the optimal cost as 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 then the following.

(15)

where .

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. In Section 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 denote the set of real matrices and let denote the subset of consisting of symmetric matrices. For , we assume that , denote the trace, the inverse, the transpose, the diagonal elements, the singular values of , respectively, and define . If is an arbitrary symmetric matrix, then and denote the eigenvalues and the real part of eigenvalues of . Suppose is a real -element array such as which is reordered, and its elements are arranged in nonincreasing order; that is, . The notation () is used to denote that is a symmetric positive definite (semidefinite) matrix.

Let be two real -element arrays. If they satisfy

(21)

then it is said that is controlled weakly by , which is signed by .

If and

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then it is said that is controlled by , which is signed by .

The following lemmas are used to prove the main results.

Lemma 2.1 (see [21, Page 92, H.2.c]).

If and , then for any real array ,

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Lemma 2.2 (see [21, Page 218, B.1]).

Let , then

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Lemma 2.3 (see [21, Page 240, F.4.a]).

Let , then

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Lemma 2.4 (see [22]).

Let , then

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where

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Note that if or , obviously, (2.6) holds. If , choose , then (2.6) also holds.

Remark 2.5.

If , then we obtain Cauchy-Schwarz inequality:

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where

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Remark 2.6.

Note that

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Let in (2.6), then we obtain

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Lemma 2.7.

If , , then

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Proof.

() Note that for , or ,

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 () If , , for , choose , then and . Thus, is a convex function. As and , from the property of the convex function, we have

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 () If , without loss of generality, we may assume . Then from (), we have

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Since , thus

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This completes the proof.

Theorem 2.8.

Let , and let be diagonalizable with the following decomposition:

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where is nonsingular. Then

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Proof.

Note that is real; by the matrix theory we have

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Since , without loss of generality, we may assume . Next, we will prove the left-hand side of (2.18):

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If

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then we obtain the conclusion. Now assume that there exists such that then

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We use to denote the vector of after changing and , then

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After a limited number of steps, we obtain the left-hand side of (2.18). For the right-hand side of (2.18)

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If

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then we obtain the conclusion. Now assume that there exists such that then

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We use to denote the vector of after changing and , then

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After a limited number of steps, we obtain the right-hand side of (2.18). Therefore, we have

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Since applying (2.18) with in lieu of we immediately have the following corollary.

Corollary 2.9.

Let , and let be diagonalizable with the following decomposition:

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where is nonsingular. Then

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Theorem 2.10.

Let , be normal. Then

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Proof.

Since is normal, from [23, page 101, Theorem ], we have

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where is orthogonal. Since and , then for , we have

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In terms of Lemmas 2.1 and 2.2, (2.18) implies

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This completes the proof.

Note that if , , then from (2.34) we obtain (1.5) immediately. This implies that (2.18) improves (1.5).

Since applying (2.31) with in lieu of we immediately have the following corollary.

Corollary 2.11.

Let , be normal, then

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3. Trace Bounds for the Solution of the Algebraic Riccati Equations

Komaroff (1994) in [16] obtained the following. Let be the positive semidefinite solution of the ARE (1.4). Then the trace of has the upper bound given by

(31)

where

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.

Theorem 3.1.

If , and is the positive semidefinite solution of the ARE (1.4).

There are both, upper and lower, bounds:

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If , then the trace of has the lower and upper bounds given by

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where denotes and denotes

If , then the trace of has the lower and upper bounds given by

(34)

where ,

We have

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Proof.

() Multiply (1.4) on the right and on the left by to get

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where . Take the trace of all terms in (3.6) to get

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Since is positive semidefiniteness, , and from Lemma 2.7, we have

(38)

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By Cauchy-Schwarz inequality (2.8), it can be shown that

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Since are positive semidefiniteness, is positive definiteness, then by (1.5), note that for , , and considering (2.6), (3.8), and (3.9), we have

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Note that , , then from (1.5) we have

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Combining (3.11) with (3.12), we obtain

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Solving (3.13) for yields the left-hand side of (3.2).

Since are positive semidefiniteness, is positive definiteness, then by (1.5), note that for , , and considering (2.6), (3.8), and (3.10), we have

(314)

Combining (3.12) with (3.14), we obtain

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Solving (3.15) for yields the right-hand side of (3.2).

() Note that when , by (1.5), (2.6), and (3.8), we have

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Combining (3.11) with (3.16), we obtain

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Solving (3.17) for yields the left-hand side of (3.3).

By (1.5), (2.6), and (3.8), we have

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Combining (3.14) with (3.18), we obtain

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Solving (3.19) for yields the right-hand side of (3.3).

() Note that when , by (3.3), we obtain (3.4) immediately. This completes the proof.

Remark 3.2.

From Remark 2.6 and Theorem 3.1, we have the following results.

() Let in (3.2), then we have

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() Let in (3.3), then we obtain (3.20).

() Let in (3.4). Note that when ,

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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 different superiority if we choose different and , we will give two examples.

Example 4.1.

Let

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Then and is symmetric. Using (1.5) yields

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Using (2.18) yields

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where both lower and upper bounds are better than those of the main result of [18], that is, (1.5).

Example 4.2.

Consider the system (1.1), (1.2) with

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and consider the corresponding ARE (1.4) with ; is stabilizable and is observable. Using (3.20) yields

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Using (3.2), when , then we obtain

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where the upper bound is better than that of the main result of [16], that is, (3.1).

Example 4.3.

Consider the system (1.1), (1.2) with

(47)

and consider the corresponding ARE (1.4) with ; is stabilizable and is observable. Using (3.2), when , then we obtain

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Using (3.20) yields

(49)

where the lower and upper bounds are better than those of (4.8).

5. Conclusions

In this paper, we have proposed lower and upper bounds for the trace of the product of two real square matrices in which one is diagonalizable. We have shown that our bounds for the 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.