Open Access

Trace inequalities for positive semidefinite matrices with centrosymmetric structure

Journal of Inequalities and Applications20122012:62

https://doi.org/10.1186/1029-242X-2012-62

Received: 4 July 2011

Accepted: 12 March 2012

Published: 12 March 2012

Abstract

In this article, we present some results on the Hadamard product of positive semidefinite matrices with centrosymmetric structure. Based on these results, several trace inequalities on positive semidefinite centrosymmetric matrices are obtained.

1 Introduction and preliminaries

We will use the following notation. Let n × nand n × nbe the space of n × n complex and real matrices, respectively. The identity matrix in n × nis denoted by I = I n . Let A T , Ā, A H , and tr(A) denote the transpose, the conjugate, the conjugate transpose, and the trace of a matrix A, respectively. Let Re(a) represent the real part of a. The Frobenius inner product < ·, · > F in m × nover the complex field is defined as follows: < A, B > F = Re(tr(B H A)), for A, B m × n, i.e., < A, B > is the real part of the trace of B H A. The induced matrix norm is A F = < A , A > F = R e ( t r ( A H A ) ) = t r ( A H A ) , which is called the Frobenius (Euclidean) norm.

A matrix A n × nis Hermitian if A H = A, and A is called positive semidefinite, written as A ≥ 0 (see [1, p. 159]), if
x H A x 0 , x n .
(1.1)

A is further called positive definite, symbolized A > 0, if the strict inequality in (1.1) holds for all non-zero x n . An equivalent condition for A n to be positive definite is that A is Hermitian and all eigenvalues of A are positive.

Let A and B be two Hermitian matrices of the same size. If A - B is positive semidefinite, we write
A B .

Next, we introduce some basic definitions and lemmas.

Definition 1.1 (see [1]). Let A be a square complex matrix partitioned as
A = A 11 A 12 A 21 A 22
(1.2)

where A11is a square submatrix of A. If A11is nonsingular, we call à 11 = A 22 - A 21 A 11 - 1 A 12 the Schur complement of A11in A.

Note. If A is a positive definite matrix, then A11 is nonsingular and A22Ã11 ≥ 0.

Definition 1.2 (see [2]). A = (a ij ) n × nis called a centrosymmetric matrix, if
a i j = a n - i + 1 , n - j + 1 , 1 i n , 1 j n , o r J n A J n = A ,

where J n = (e n , en-1,..., e1), e i denotes the unit vector with the ith entry 1.

If a matrix is both positive semidefinite and centrosymmetric, we call this matrix positive semidefinite centrosymmetric.

Using the partition of matrix, the central symmetric character of a square centrosymmetric matrix can be described as follows [2]:

Lemma 1.1 (see [2]). Let A = (a ij ) n × n(n = 2m) be centrosymmetric. Then, A has the following form,
A = B J m C J m C J m B J m , and P T A P = B - J m C 0 0 B + J m C ,
(1.3)

where B m × m , C m × m , P = 1 2 I m I m - J m J m .

Note. In this article, we mainly discuss the case n = 2m. For n is odd, i.e., n = 2m + 1, similar results can be obtained by taking similar steps.

Recently, in [3], Ulukö k and Tü rkmen proved some matrix trace inequalities for positive semidefinite matrices:

Lemma 1.2 (see [3]). Let A, B n × n. Then,
( A B ) m F 2 A H A m F B H B m F
(1.4)

where m is an positive integer, A B stands for the Hadamard product of A and B.

Note. Particularly, if A and B in Lemma 1.2 are both semidefinite matrices, then A B m F 2 A 2 m F B 2 m F .

Lemma 1.3 (see [3]). Let A i n × n, (i = 1,2,...,k) be semidefinite matrices. Then, for positive real numbers s, m, t
i = 1 k A i ( ( s + t ) / 2 ) m F 2 i = 1 k A i s m F 2 i = 1 k A i t m F 2
(1.5)
Lemma 1.4 (see [3]). Let A = A 11 A 12 A 21 A 22 0 , B = B 11 B 12 B 21 B 22 0 . Then,
t r à 22 1 / 2 B 11 1 / 2 2 m + t r ( A 22 ) 1 / 2 B ̃ 11 1 / 2 2 m t r ( A B ) m t r ( A m B m ) ,
(1.6)

where m is an integer.

2 Main results

Lemma 2.1. Let A = (a ij )n × n, B = (b ij )n × n(n = 2m) be two centrosymmetric matrices with the following form:
A = A 1 J m A 2 J m A 2 J m A 1 J m , B = B 1 J m B 2 J m B 2 J m B 1 J m , A 1 , A 2 , B 1 , B 2 m × m
(2.1)

Then, A B is a centrosymmetric matrix.

Proof. By the definition of Hadamard product,
A B = A 1 J m A 2 J m A 2 J m A 1 J m B 1 J m B 2 J m B 2 J m B 1 J m = A 1 B 1 ( J m A 2 J m ) ( J m B 2 J m ) A 2 B 2 ( J m A 1 J m ) ( J m B 1 J m )
(2.2)
We shall prove the following
( J m A 1 J m ) ( J m B 1 J m ) = J m ( A 1 B 1 ) J m , ( J m A 2 J m ) ( J m B 2 J m ) = J m ( A 1 B 2 ) J m .
(2.3)
From (2.1),
A 1 B 1 = a 11 a 1 m a m 1 a m m b 11 b 1 m b m 1 b m m = a 11 b 11 a 1 m b 1 m a m 1 b m 1 a m m b m m ,
(2.4)
and
J m ( A 1 B 1 ) J m = a m m b m m a m 1 b m 1 a 1 m b 1 m a 11 b 11 .
(2.5)
Since
J m A 1 J m = a m m a m 1 a 1 m a 11 , and J m B 1 J m = b m m b m 1 b 1 m b 11 ,
(2.6)
we have the following
( J m A 1 J m ) ( J m B 1 J m ) = a m m b m m a m 1 b m 1 a 1 m b 1 m a 11 b 11 .
(2.7)
From (2.5) and (2.7), it is clear that
J m A 1 J m J m B 1 J m = J m A 1 B 1 J m .
(2.8)
Similarly, we can prove that
J m A 2 J m J m B 2 J m = J m A 2 B 2 J m .
(2.9)

From (2.8) and (2.9), we can see that (2.3) holds. By Lemma 1.1, A B is a centrosymmetric matrix.

Lemma 2.2 (see [4]) Let A = (a ij )n × n(n = 2m) a positive semidefinite centrosymmetric matrix with the following form
A = B J m C J m C J m B J m , B , C m × m .

Let M = B - J m C and N = B + J m C. Then, M, N are positive semidefinite matrices.

Theorem 2.1 Let A, B n × n(n = 2m) be two positive semidefinite centrosymmetric matrices with the same form as in (2.1). Let
M A = A 1 - J m A 2 , N A = A 1 + J m A 2 , M B = B 1 - J m B 2 , N B = B 1 + J m B 2 ,
and
X = A 1 B 1 - J m A 2 B 2 , Y = A 1 B 1 + J m A 2 B 2 .
Then, the following inequality holds:
X m F 2 + Y m F 2 M A 2 m F 2 + N A 2 m F 2 M B 2 m F 2 + N B 2 m F 2 .
(2.10)
Proof. Since A, B are centrosymmetric matrices, by Lemma 2.1, A B is centrosymmetric and
A B = A 1 B 1 ( J m A 2 J m ) ( J m B 2 J m ) A 2 B 2 ( J m A 1 J m ) ( J m B 1 J m ) .
From Lemma 1.1,
P T A B P = A 1 B 1 - J m A 2 B 2 0 0 A 1 B 1 - J m A 2 B 2 = X 0 0 Y .
P is orthogonal, then
P T A B m P = X m 0 0 Y m ,
and
A B m F 2 = P T A B m P F 2 = X m 0 0 Y m F 2 = X m F 2 + Y m F 2 .
Similarly from Lemma 1.1,
P T A P = A 1 - J 1 A 2 0 0 A 1 + J 1 A 2 = M A 0 0 N A , P T B P = B 1 - J 1 B 2 0 0 B 1 + J 1 B 2 = M B 0 0 N B .
Then,
P T A 2 m P = M A 2 m 0 0 N A 2 m , P T B 2 m P = M B 2 m 0 0 N B 2 m ,
and
A 2 m F B 2 m F = M A 2 m 0 0 N A 2 m F M B 2 m 0 0 N B 2 m F = M A 2 m F 2 + N A 2 m F 2 M B 2 m F 2 + N B 2 m F 2 .
Since both A and B are positive semidefinite matrices, by Lemma 1.2
A B F 2 A 2 m F B 2 m F .
Thus,
X m F 2 + Y m F 2 M A 2 m F 2 + N A 2 m F 2 M B 2 m F 2 + N B 2 m F 2 .
Theorem 2.2. Let A n × n(n = 2m) be postive semidefinite centroysymmetric with the form:
A = B J m C J m C J m B J m , B , C m × m .
Let M = B - J m C, N = B + J m C. Then, for positive integers s, m, t, the following two equalities hold
A ( ( s + t ) / 2 ) m F 2 = M ( ( s + t ) / 2 ) m F 2 + N ( ( s + t ) / 2 ) m F 2 ,
(2.11)
and
M ( ( s + t ) / 2 ) m F 2 + N ( ( s + t ) / 2 ) m F 2 M s m F 2 + N s m F 2 M t m F 2 + N t m F 2 .
(2.12)
Proof. By Lemma 1.1,
P T A P = B - J m C 0 0 B + J m C = M 0 0 N .
Since A is positive semidefinite, we have
P T A ( ( s + t ) / 2 ) m P = M ( ( s + t ) / 2 ) m 0 0 N ( ( s + t ) / 2 ) m .
Thus,
A ( ( s + t ) / 2 ) m F 2 = P T A ( ( s + t ) / 2 ) m F 2 = M ( ( s + t ) / 2 ) m 0 0 N ( ( s + t ) / 2 ) m F 2 = M ( ( s + t ) / 2 ) m F 2 + N ( ( s + t ) / 2 ) m F 2 .
From Lemma 2.2, M and N are positive semidefinite. Combining Lemma 1.3, we have
M ( ( s + t ) / 2 ) m F 2 + N ( ( s + t ) / 2 ) m F 2 M s m F 2 + N s m F 2 M t m F 2 + N t m F 2 .
Theorem 2.3 Let A, B n × n(n = 2m) are positive semidefinite centrosymmetric matrices with the same form as in (2.1). Let
M A = A 1 - J m A 2 , N A = A 1 + J m A 2 , M B = B 1 - J m B 2 , N B = B 1 + J m B 2 .
Then, the following inequality holds
t r M A 1 / 2 M B 1 / 2 2 m + t r N A 1 / 2 N B 1 / 2 2 m t r ( A B ) m t r A m B m .
(2.13)
Proof. From Lemma 1.1, there exists an orthogonal matrix P such that
P T A P = A 1 - J m A 2 0 0 A 1 + J m A 2 = M A 0 0 N A , P T B P = B 1 - J m B 2 0 0 B 1 + J m B 2 = M B 0 0 N B .
According to Definition 1.1,
Ñ A = M A - 0 N A - 1 0 = M A , M ̃ B = N B - 0 M B - 1 0 = N B .
From Lemma 1.4,
t r M A 1 / 2 M B 1 / 2 2 m + t r N A 1 / 2 N B 1 / 2 2 m t r ( P T A P P T B P ) m = t r P T A B P m .
Since P is orthogonal, we have tr(P T ABP) m = tr(AB) m . By Lemma 1.4, the following holds
t r M A 1 / 2 M B 1 / 2 2 m + t r N A 1 / 2 N B 1 / 2 2 m t r ( A B ) m t r A m B m .

Declarations

Acknowledgements

We would like to thank the reviewers for providing valuable comments and suggestions to improve the manuscript. This study was supported by the National Natural Science Foundation of China (No. 60831001).

Authors’ Affiliations

(1)
LMIB, School of Mathematics and System Science, Beihang University
(2)
Faculty of Science and Technology, University of Macau

References

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Copyright

© Zhao et al; licensee Springer. 2012

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