Trace inequalities for positive semidefinite matrices with centrosymmetric structure

Di Zhao; Hongyi Li; Zhiguo Gong

doi:10.1186/1029-242X-2012-62

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Trace inequalities for positive semidefinite matrices with centrosymmetric structure

Di Zhao¹,
Hongyi Li¹Email author and
Zhiguo Gong²

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 ℂ^{n × n}and ℝ^{n × n}be the space of n × n complex and real matrices, respectively. The identity matrix in ℂ^{n × n}is 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 < ·, · >_Fin ℂ^{m × n}over the complex field is defined as follows: < A, B >_F= Re(tr(B^HA)), for A, B ∈ ℂ^{m × n}, i.e., < A, B > is the real part of the trace of B^HA. The induced matrix norm is ${∥A∥}_{F} = \sqrt{< A, A >_{F}} = \sqrt{R e (t r (A^{H} A))} = \sqrt{t r (A^{H} A)}$ , which is called the Frobenius (Euclidean) norm.

A matrix A ∈ ℂ^{n × n}is Hermitian if A^H= A, and A is called positive semidefinite, written as A ≥ 0 (see [1, p. 159]), if

x^{H} A x \geq 0, \forall x \in ℂ^{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 ∈ ℂⁿ. An equivalent condition for A ∈ ℂⁿto 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 \geq B .

Next, we introduce some basic definitions and lemmas.

Definition 1.1 (see [1]). Let A be a square complex matrix partitioned as

A = (\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix})

(1.2)

where A₁₁is a square submatrix of A. If A₁₁is nonsingular, we call $Ã_{11} = A_{22} - A_{21} A_{11}^{- 1} A_{12}$ the Schur complement of A₁₁in A.

Note. If A is a positive definite matrix, then A₁₁ is nonsingular and A₂₂ ≥ Ã₁₁ ≥ 0.

Definition 1.2 (see [2]). A = (a_ij) ∈ ℂ^{n × n}is called a centrosymmetric matrix, if

a_{i j} = a_{n - i + 1, n - j + 1,} 1 \leq i \leq n, 1 \leq j \leq n, o r J_{n} A J_{n} = A,

where J_n= (e_n, e_n-1,..., e₁), e_idenotes 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) ∈ ℂ^{n × n}(n = 2m) be centrosymmetric. Then, A has the following form,

A = (\begin{matrix} B & J_{m} C J_{m} \\ C & J_{m} B J_{m} \end{matrix}), and P^{T} A P = (\begin{matrix} B - J_{m} C & 0 \\ 0 & B + J_{m} C \end{matrix}),

(1.3)

where $B \in ℂ^{m \times m}, C \in ℂ^{m \times m}, P = \frac{1}{\sqrt{2}} (\begin{matrix} I_{m} & I_{m} \\ - J_{m} & J_{m} \end{matrix})$ .

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 ∈ ℂ^{n × n}. Then,

{∥{(A \circ B)}^{m}∥}_{F}^{2} \leq {∥{(A^{H} A)}^{m}∥}_{F} \cdot {∥{(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 ${∥{(A \circ B)}^{m}∥}_{F}^{2} \leq {∥A^{2 m}∥}_{F} \cdot {∥B^{2 m}∥}_{F}$ .

Lemma 1.3 (see [3]). Let A_i∈ ℂ^{n × n}, (i = 1,2,...,k) be semidefinite matrices. Then, for positive real numbers s, m, t

(\sum_{i = 1}^{k} {∥A_{i}^{((s + t) / 2) m}∥}_{F}^{2}) \leq (\sum_{i = 1}^{k} {∥A_{i}^{s m}∥}_{F}^{2}) (\sum_{i = 1}^{k} {∥A_{i}^{t m}∥}_{F}^{2})

(1.5)

Lemma 1.4 (see [3]). Let

A = (\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}) \geq 0, B = (\begin{matrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{matrix}) \geq 0

. Then,

t r {[{(Ã_{22})}^{1 / 2} B_{11}^{1 / 2}]}^{2 m} + t r {[{(A_{22})}^{1 / 2} {\tilde{B}}_{11}^{1 / 2}]}^{2 m} \leq t r {(A B)}^{m} \leq 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 = (\begin{matrix} A_{1} & J_{m} A_{2} J_{m} \\ A_{2} & J_{m} A_{1} J_{m} \end{matrix}), B = (\begin{matrix} B_{1} & J_{m} B_{2} J_{m} \\ B_{2} & J_{m} B_{1} J_{m} \end{matrix}), A_{1}, A_{2}, B_{1}, B_{2} \in ℂ^{m \times m}

(2.1)

Then, A ○ B is a centrosymmetric matrix.

Proof. By the definition of Hadamard product,

A \circ B = (\begin{matrix} A_{1} & J_{m} A_{2} J_{m} \\ A_{2} & J_{m} A_{1} J_{m} \end{matrix}) \circ (\begin{matrix} B_{1} & J_{m} B_{2} J_{m} \\ B_{2} & J_{m} B_{1} J_{m} \end{matrix}) = (\begin{matrix} A_{1} \circ B_{1} & (J_{m} A_{2} J_{m}) \circ (J_{m} B_{2} J_{m}) \\ A_{2} \circ B_{2} & (J_{m} A_{1} J_{m}) \circ (J_{m} B_{1} J_{m}) \end{matrix})

(2.2)

We shall prove the following

\{\begin{matrix} (J_{m} A_{1} J_{m}) \circ (J_{m} B_{1} J_{m}) = J_{m} (A_{1} \circ B_{1}) J_{m}, \\ (J_{m} A_{2} J_{m}) \circ (J_{m} B_{2} J_{m}) = J_{m} (A_{1} \circ B_{2}) J_{m} . \end{matrix}

(2.3)

From (2.1),

A_{1} \circ B_{1} = (\begin{matrix} a_{11} & \dots & a_{1 m} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m m} \end{matrix}) \circ (\begin{matrix} b_{11} & \dots & b_{1 m} \\ ⋮ & ⋱ & ⋮ \\ b_{m 1} & \dots & b_{m m} \end{matrix}) = (\begin{matrix} a_{11} b_{11} & \dots & a_{1 m} b_{1 m} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} b_{m 1} & \dots & a_{m m} b_{m m} \end{matrix}),

(2.4)

and

J_{m} (A_{1} \circ B_{1}) J_{m} = (\begin{matrix} a_{m m} b_{m m} & \dots & a_{m 1} b_{m 1} \\ ⋮ & ⋱ & ⋮ \\ a_{1 m} b_{1 m} & \dots & a_{11} b_{11} \end{matrix}) .

(2.5)

Since

J_{m} A_{1} J_{m} = (\begin{matrix} a_{m m} & \dots & a_{m 1} \\ ⋮ & ⋱ & ⋮ \\ a_{1 m} & \dots & a_{11} \end{matrix}), and J_{m} B_{1} J_{m} = (\begin{matrix} b_{m m} & \dots & b_{m 1} \\ ⋮ & ⋱ & ⋮ \\ b_{1 m} & \dots & b_{11} \end{matrix}),

(2.6)

we have the following

(J_{m} A_{1} J_{m}) \circ (J_{m} B_{1} J_{m}) = (\begin{matrix} a_{m m} b_{m m} & \dots & a_{m 1} b_{m 1} \\ ⋮ & ⋱ & ⋮ \\ a_{1 m} b_{1 m} & \dots & a_{11} b_{11} \end{matrix}) .

(2.7)

From (2.5) and (2.7), it is clear that

(J_{m} A_{1} J_{m}) \circ (J_{m} B_{1} J_{m}) = J_{m} (A_{1} \circ B_{1}) J_{m} .

(2.8)

Similarly, we can prove that

(J_{m} A_{2} J_{m}) \circ (J_{m} B_{2} J_{m}) = J_{m} (A_{2} \circ 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 = (\begin{matrix} B & J_{m} C J_{m} \\ C & J_{m} B J_{m} \end{matrix}), B, C \in ℂ^{m \times m} .

Let M = B - J_mC and N = B + J_mC. Then, M, N are positive semidefinite matrices.

Theorem 2.1 Let A, B ∈ ℂ^{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} \circ B_{1} - J_{m} (A_{2} \circ B_{2}), Y = A_{1} \circ B_{1} + J_{m} (A_{2} \circ B_{2}) .

Then, the following inequality holds:

{∥X^{m}∥}_{F}^{2} + {∥Y^{m}∥}_{F}^{2} \leq \sqrt{({∥M_{A}^{2 m}∥}_{F}^{2} + {∥N_{A}^{2 m}∥}_{F}^{2}) \cdot ({∥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 \circ B = (\begin{matrix} A_{1} \circ B_{1} & (J_{m} A_{2} J_{m}) \circ (J_{m} B_{2} J_{m}) \\ A_{2} \circ B_{2} & (J_{m} A_{1} J_{m}) \circ (J_{m} B_{1} J_{m}) \end{matrix}) .

From Lemma 1.1,

P^{T} (A \circ B) P = (\begin{matrix} A_{1} \circ B_{1} - J_{m} (A_{2} \circ B_{2}) & 0 \\ 0 & A_{1} \circ B_{1} - J_{m} (A_{2} \circ B_{2}) \end{matrix}) = (\begin{matrix} X & 0 \\ 0 & Y \end{matrix}) .

P is orthogonal, then

P^{T} {(A \circ B)}^{m} P = (\begin{matrix} X^{m} & 0 \\ 0 & Y^{m} \end{matrix}),

and

{∥{(A \circ B)}^{m}∥}_{F}^{2} = {∥P^{T} {(A \circ B)}^{m} P∥}_{F}^{2} = {∥(\begin{matrix} X^{m} & 0 \\ 0 & Y^{m} \end{matrix})∥}_{F}^{2} = {∥X^{m}∥}_{F}^{2} + {∥Y^{m}∥}_{F}^{2} .

Similarly from Lemma 1.1,

\begin{gathered} P^{T} A P = (\begin{matrix} A_{1} - J_{1} A_{2} & 0 \\ 0 & A_{1} + J_{1} A_{2} \end{matrix}) = (\begin{matrix} M_{A} & 0 \\ 0 & N_{A} \end{matrix}), \\ P^{T} B P = (\begin{matrix} B_{1} - J_{1} B_{2} & 0 \\ 0 & B_{1} + J_{1} B_{2} \end{matrix}) = (\begin{matrix} M_{B} & 0 \\ 0 & N_{B} \end{matrix}) . \end{gathered}

Then,

P^{T} A^{2 m} P = (\begin{matrix} M_{A}^{2 m} & 0 \\ 0 & N_{A}^{2 m} \end{matrix}), P^{T} B^{2 m} P = (\begin{matrix} M_{B}^{2 m} & 0 \\ 0 & N_{B}^{2 m} \end{matrix}),

and

\begin{gathered} {∥A^{2 m}∥}_{F} \cdot {∥B^{2 m}∥}_{F} = {∥(\begin{matrix} M_{A}^{2 m} & 0 \\ 0 & N_{A}^{2 m} \end{matrix})∥}_{F} \cdot {∥(\begin{matrix} M_{B}^{2 m} & 0 \\ 0 & N_{B}^{2 m} \end{matrix})∥}_{F} \\ = \sqrt{({∥M_{A}^{2 m}∥}_{F}^{2} + {∥N_{A}^{2 m}∥}_{F}^{2}) \cdot ({∥M_{B}^{2 m}∥}_{F}^{2} + {∥N_{B}^{2 m}∥}_{F}^{2})} . \end{gathered}

Since both A and B are positive semidefinite matrices, by Lemma 1.2

{∥(A \circ B)∥}_{F}^{2} \leq {∥A^{2 m}∥}_{F} \cdot {∥B^{2 m}∥}_{F} .

Thus,

{∥X^{m}∥}_{F}^{2} + {∥Y^{m}∥}_{F}^{2} \leq \sqrt{({∥M_{A}^{2 m}∥}_{F}^{2} + {∥N_{A}^{2 m}∥}_{F}^{2}) \cdot ({∥M_{B}^{2 m}∥}_{F}^{2} + {∥N_{B}^{2 m}∥}_{F}^{2})} .

Theorem 2.2. Let A ∈ ℂ^{n × n}(n = 2m) be postive semidefinite centroysymmetric with the form:

A = (\begin{matrix} B & J_{m} C J_{m} \\ C & J_{m} B J_{m} \end{matrix}), B, C \in ℂ^{m \times m} .

Let M = B - J_mC, N = B + J_mC. 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} \leq ({∥M^{s m}∥}_{F}^{2} + {∥N^{s m}∥}_{F}^{2}) \cdot ({∥M^{t m}∥}_{F}^{2} + {∥N^{t m}∥}_{F}^{2}) .

(2.12)

Proof. By Lemma 1.1,

P^{T} A P = (\begin{matrix} B - J_{m} C & 0 \\ 0 & B + J_{m} C \end{matrix}) = (\begin{matrix} M & 0 \\ 0 & N \end{matrix}) .

Since A is positive semidefinite, we have

P^{T} A^{((s + t) / 2) m} P = (\begin{matrix} M^{((s + t) / 2) m} & 0 \\ 0 & N^{((s + t) / 2) m} \end{matrix}) .

Thus,

\begin{align} {∥A^{((s + t) / 2) m}∥}_{F}^{2} = {∥P^{T} A^{((s + t) / 2) m}∥}_{F}^{2} & = {∥(\begin{matrix} M^{((s + t) / 2) m} & 0 \\ 0 & N^{((s + t) / 2) m} \end{matrix})∥}_{F}^{2} \\ = {∥M^{((s + t) / 2) m}∥}_{F}^{2} + {∥N^{((s + t) / 2) m}∥}_{F}^{2} . \end{align}

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} \leq ({∥M^{s m}∥}_{F}^{2} + {∥N^{s m}∥}_{F}^{2}) \cdot ({∥M^{t m}∥}_{F}^{2} + {∥N^{t m}∥}_{F}^{2}) .

Theorem 2.3 Let A, B ∈ ℂ^{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} \leq t r {(A B)}^{m} \leq t r (A^{m} B^{m}) .

(2.13)

Proof. From Lemma 1.1, there exists an orthogonal matrix P such that

\begin{gathered} P^{T} A P = (\begin{matrix} A_{1} - J_{m} A_{2} & 0 \\ 0 & A_{1} + J_{m} A_{2} \end{matrix}) = (\begin{matrix} M_{A} & 0 \\ 0 & N_{A} \end{matrix}), \\ P^{T} B P = (\begin{matrix} B_{1} - J_{m} B_{2} & 0 \\ 0 & B_{1} + J_{m} B_{2} \end{matrix}) = (\begin{matrix} M_{B} & 0 \\ 0 & N_{B} \end{matrix}) . \end{gathered}

According to Definition 1.1,

Ñ_{A} = M_{A} - 0 \cdot N_{A}^{- 1} \cdot 0 = M_{A}, {\tilde{M}}_{B} = N_{B} - 0 \cdot M_{B}^{- 1} \cdot 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} \leq 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^TABP)^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} \leq t r {(A B)}^{m} \leq 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

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