Trace inequalities for positive semidefinite matrices with centrosymmetric structure

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 < ·, · > _F in 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 ${∥A∥}_F=\sqrt{<A,A{>}_F}=\sqrt{Re\left(tr\left(A^{H}A\right)\right)}=\sqrt{tr\left(A^{H}A\right)}$, 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

$$x^{H}Ax\ge 0,\forall x\in {ℂ}^n.$$

(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

$$A\ge B.$$

Next, we introduce some basic definitions and lemmas.

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

$$A=\left(\begin{array}{cc}\hfill A_{11}\hfill & \hfill A_{12}\hfill \\ \hfill A_{21}\hfill & \hfill A_{22}\hfill \end{array}\right)$$

(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

$$a_{ij}=a_{n-i+1,n-j+1,}1\le i\le n,1\le j\le n,or{J}_{n}A{J}_n=A,$$

where J _n = (e _n , e_n-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 ) ∈ ℂ^{n × n}(n = 2m) be centrosymmetric. Then, A has the following form,

$$A=\left(\begin{array}{cc}\hfill B\hfill & \hfill J_{m}C{J}_m\hfill \\ \hfill C\hfill & \hfill J_{m}B{J}_m\hfill \end{array}\right),\mathsf{\text{and}}{P}^{T}AP=\left(\begin{array}{cc}\hfill B-J_{m}C\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill B+J_{m}C\hfill \end{array}\right),$$

(1.3)

where$B\in {ℂ}^{m\times m},C\in {ℂ}^{m\times m},P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}\hfill I_m\hfill & \hfill I_m\hfill \\ \hfill -J_m\hfill & \hfill J_m\hfill \end{array}\right)$.

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,

$${∥{\left(A\circ B\right)}^m∥}_F^2\le {∥{\left(A^{H}A\right)}^m∥}_F\cdot {∥{\left(B^{H}B\right)}^m∥}_F$$

(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 ${∥{\left(A\circ B\right)}^m∥}_F^2\le {∥A^{2m}∥}_F\cdot {∥B^{2m}∥}_F$.

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

$$\left(\sum _{i=1}^k{∥A_i^{\left(\left(s+t\right)/2\right)m}∥}_F^2\right)\le \left(\sum _{i=1}^k{∥A_i^{sm}∥}_F^2\right)\left(\sum _{i=1}^k{∥A_i^{tm}∥}_F^2\right)$$

(1.5)

Lemma 1.4 (see [3]). Let$A=\left(\begin{array}{cc}\hfill A_{11}\hfill & \hfill A_{12}\hfill \\ \hfill A_{21}\hfill & \hfill A_{22}\hfill \end{array}\right)\ge 0,B=\left(\begin{array}{cc}\hfill B_{11}\hfill & \hfill B_{12}\hfill \\ \hfill B_{21}\hfill & \hfill B_{22}\hfill \end{array}\right)\ge 0$. Then,

$$tr{\left[{\left({Ã}_{22}\right)}^{1/2}{B}_{11}^{1/2}\right]}^{2m}+tr{\left[{\left(A_{22}\right)}^{1/2}{\tilde{B}}_{11}^{1/2}\right]}^{2m}\le tr{\left(AB\right)}^m\le tr\left(A^m{B}^m\right),$$

(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:

$$A=\left(\begin{array}{cc}\hfill A_1\hfill & \hfill J_m{A}_2{J}_m\hfill \\ \hfill A_2\hfill & \hfill J_m{A}_1{J}_m\hfill \end{array}\right),B=\left(\begin{array}{cc}\hfill B_1\hfill & \hfill J_m{B}_2{J}_m\hfill \\ \hfill B_2\hfill & \hfill J_m{B}_1{J}_m\hfill \end{array}\right),A_1,A_2,B_1,B_2\in {ℂ}^{m\times m}$$

(2.1)

Then, A ○ B is a centrosymmetric matrix.

Proof. By the definition of Hadamard product,

$$A\circ B=\left(\begin{array}{cc}\hfill A_1\hfill & \hfill J_m{A}_2{J}_m\hfill \\ \hfill A_2\hfill & \hfill J_m{A}_1{J}_m\hfill \end{array}\right)\circ \left(\begin{array}{cc}\hfill B_1\hfill & \hfill J_m{B}_2{J}_m\hfill \\ \hfill B_2\hfill & \hfill J_m{B}_1{J}_m\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill A_1\circ B_1\hfill & \hfill \left(J_m{A}_2{J}_m\right)\circ \left(J_m{B}_2{J}_m\right)\hfill \\ \hfill A_2\circ B_2\hfill & \hfill \left(J_m{A}_1{J}_m\right)\circ \left(J_m{B}_1{J}_m\right)\hfill \end{array}\right)$$

(2.2)

We shall prove the following

$$\left\{\begin{array}{c}\hfill \left(J_m{A}_1{J}_m\right)\circ \left(J_m{B}_1{J}_m\right)=J_m\left(A_1\circ B_1\right)J_m,\hfill \\ \hfill \left(J_m{A}_2{J}_m\right)\circ \left(J_m{B}_2{J}_m\right)=J_m\left(A_1\circ B_2\right)J_m.\hfill \end{array}\right.$$

(2.3)

From (2.1),

$$A_1\circ B_1=\left(\begin{array}{ccc}{a}_{11}\hfill & \cdots \hfill & a_{1m}\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ a_{m1}\hfill & \cdots \hfill & a_{mm}\hfill \end{array}\right)\circ \left(\begin{array}{ccc}{b}_{11}\hfill & \cdots \hfill & b_{1m}\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ b_{m1}\hfill & \cdots \hfill & b_{mm}\hfill \end{array}\right)=\left(\begin{array}{ccc}{a}_{11}{b}_{11}\hfill & \cdots \hfill & a_{1m}{b}_{1m}\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ a_{m1}{b}_{m1}\hfill & \cdots \hfill & a_{mm}{b}_{mm}\hfill \end{array}\right),$$

(2.4)

and

$$J_m\left(A_1\circ B_1\right)J_m=\left(\begin{array}{ccc}{a}_{mm}{b}_{mm}\hfill & \cdots \hfill & a_{m1}{b}_{m1}\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ a_{1m}{b}_{1m}\hfill & \cdots \hfill & a_{11}{b}_{11}\hfill \end{array}\right).$$

(2.5)

Since

$$J_m{A}_1{J}_m=\left(\begin{array}{ccc}{a}_{mm}\hfill & \cdots \hfill & a_{m1}\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ a_{1m}\hfill & \cdots \hfill & a_{11}\hfill \end{array}\right),\mathsf{\text{and}}{J}_m{B}_1{J}_m=\left(\begin{array}{ccc}{b}_{mm}\hfill & \cdots \hfill & b_{m1}\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ b_{1m}\hfill & \cdots \hfill & b_{11}\hfill \end{array}\right),$$

(2.6)

we have the following

$$\left(J_m{A}_1{J}_m\right)\circ \left(J_m{B}_1{J}_m\right)=\left(\begin{array}{ccc}{a}_{mm}{b}_{mm}\hfill & \cdots \hfill & a_{m1}{b}_{m1}\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ a_{1m}{b}_{1m}\hfill & \cdots \hfill & a_{11}{b}_{11}\hfill \end{array}\right).$$

(2.7)

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

$$\left(J_m{A}_1{J}_m\right)\circ \left(J_m{B}_1{J}_m\right)=J_m\left(A_1\circ B_1\right)J_m.$$

(2.8)

Similarly, we can prove that

$$\left(J_m{A}_2{J}_m\right)\circ \left(J_m{B}_2{J}_m\right)=J_m\left(A_2\circ B_2\right)J_m.$$

(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

$$A=\left(\begin{array}{cc}\hfill B\hfill & \hfill J_{m}C{J}_m\hfill \\ \hfill C\hfill & \hfill J_{m}B{J}_m\hfill \end{array}\right),B,C\in {ℂ}^{m\times m}.$$

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 ∈ 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

$$X=A_1\circ B_1-J_m\left(A_2\circ B_2\right),Y=A_1\circ B_1+J_m\left(A_2\circ B_2\right).$$

Then, the following inequality holds:

$${∥X^m∥}_F^2+{∥Y^m∥}_F^2\le \sqrt{\left({∥M_A^{2m}∥}_F^2+{∥N_A^{2m}∥}_F^2\right)\cdot \left({∥M_B^{2m}∥}_F^2+{∥N_B^{2m}∥}_F^2\right)}.$$

(2.10)

Proof. Since A, B are centrosymmetric matrices, by Lemma 2.1, A ○ B is centrosymmetric and

$$A\circ B=\left(\begin{array}{cc}\hfill A_1\circ B_1\hfill & \hfill \left(J_m{A}_2{J}_m\right)\circ \left(J_m{B}_2{J}_m\right)\hfill \\ \hfill A_2\circ B_2\hfill & \hfill \left(J_m{A}_1{J}_m\right)\circ \left(J_m{B}_1{J}_m\right)\hfill \end{array}\right).$$

From Lemma 1.1,

$$P^T\left(A\circ B\right)P=\left(\begin{array}{cc}\hfill A_1\circ B_1-J_m\left(A_2\circ B_2\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill A_1\circ B_1-J_m\left(A_2\circ B_2\right)\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill X\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill Y\hfill \end{array}\right).$$

P is orthogonal, then

$$P^T{\left(A\circ B\right)}^{m}P=\left(\begin{array}{cc}\hfill X^m\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill Y^m\hfill \end{array}\right),$$

and

$${∥{\left(A\circ B\right)}^m∥}_F^2={∥P^T{\left(A\circ B\right)}^{m}P∥}_F^2={∥\left(\begin{array}{cc}\hfill X^m\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill Y^m\hfill \end{array}\right)∥}_F^2={∥X^m∥}_F^2+{∥Y^m∥}_F^2.$$

Similarly from Lemma 1.1,

$$\begin{array}{c}{P}^{T}AP=\left(\begin{array}{cc}\hfill A_1-J_1{A}_2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill A_1+J_1{A}_2\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill M_A\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_A\hfill \end{array}\right),\\ P^{T}BP=\left(\begin{array}{cc}\hfill B_1-J_1{B}_2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill B_1+J_1{B}_2\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill M_B\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_B\hfill \end{array}\right).\end{array}$$

Then,

$$P^T{A}^{2m}P=\left(\begin{array}{cc}\hfill M_A^{2m}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_A^{2m}\hfill \end{array}\right),P^T{B}^{2m}P=\left(\begin{array}{cc}\hfill M_B^{2m}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_B^{2m}\hfill \end{array}\right),$$

and

$$\begin{array}{c}{∥A^{2m}∥}_F\cdot {∥B^{2m}∥}_F={∥\left(\begin{array}{cc}\hfill M_A^{2m}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_A^{2m}\hfill \end{array}\right)∥}_F\cdot {∥\left(\begin{array}{cc}\hfill M_B^{2m}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_B^{2m}\hfill \end{array}\right)∥}_F\\ =\sqrt{\left({∥M_A^{2m}∥}_F^2+{∥N_A^{2m}∥}_F^2\right)\cdot \left({∥M_B^{2m}∥}_F^2+{∥N_B^{2m}∥}_F^2\right)}.\end{array}$$

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

$${∥\left(A\circ B\right)∥}_F^2\le {∥A^{2m}∥}_F\cdot {∥B^{2m}∥}_F.$$

Thus,

$${∥X^m∥}_F^2+{∥Y^m∥}_F^2\le \sqrt{\left({∥M_A^{2m}∥}_F^2+{∥N_A^{2m}∥}_F^2\right)\cdot \left({∥M_B^{2m}∥}_F^2+{∥N_B^{2m}∥}_F^2\right)}.$$

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

$$A=\left(\begin{array}{cc}\hfill B\hfill & \hfill J_{m}C{J}_m\hfill \\ \hfill C\hfill & \hfill J_{m}B{J}_m\hfill \end{array}\right),B,C\in {ℂ}^{m\times m}.$$

Let M = B - J _m C, N = B + J _m C. Then, for positive integers s, m, t, the following two equalities hold

$${∥A^{\left(\left(s+t\right)/2\right)m}∥}_F^2={∥M^{\left(\left(s+t\right)/2\right)m}∥}_F^2+{∥N^{\left(\left(s+t\right)/2\right)m}∥}_F^2,$$

(2.11)

and

$${∥M^{\left(\left(s+t\right)/2\right)m}∥}_F^2+{∥N^{\left(\left(s+t\right)/2\right)m}∥}_F^2\le \left({∥M^{sm}∥}_F^2+{∥N^{sm}∥}_F^2\right)\cdot \left({∥M^{tm}∥}_F^2+{∥N^{tm}∥}_F^2\right).$$

(2.12)

Proof. By Lemma 1.1,

$$P^{T}AP=\left(\begin{array}{cc}\hfill B-J_{m}C\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill B+J_{m}C\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill M\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N\hfill \end{array}\right).$$

Since A is positive semidefinite, we have

$$P^T{A}^{\left(\left(s+t\right)/2\right)m}P=\left(\begin{array}{cc}\hfill M^{\left(\left(s+t\right)/2\right)m}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N^{\left(\left(s+t\right)/2\right)m}\hfill \end{array}\right).$$

Thus,

$$\begin{array}{ll}\hfill {∥A^{\left(\left(s+t\right)/2\right)m}∥}_F^2={∥P^T{A}^{\left(\left(s+t\right)/2\right)m}∥}_F^2& ={∥\left(\begin{array}{cc}\hfill M^{\left(\left(s+t\right)/2\right)m}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N^{\left(\left(s+t\right)/2\right)m}\hfill \end{array}\right)∥}_F^2\\ ={∥M^{\left(\left(s+t\right)/2\right)m}∥}_F^2+{∥N^{\left(\left(s+t\right)/2\right)m}∥}_F^2.\end{array}$$

From Lemma 2.2, M and N are positive semidefinite. Combining Lemma 1.3, we have

$${∥M^{\left(\left(s+t\right)/2\right)m}∥}_F^2+{∥N^{\left(\left(s+t\right)/2\right)m}∥}_F^2\le \left({∥M^{sm}∥}_F^2+{∥N^{sm}∥}_F^2\right)\cdot \left({∥M^{tm}∥}_F^2+{∥N^{tm}∥}_F^2\right).$$

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

$$tr{\left(M_A^{1/2}{M}_B^{1/2}\right)}^{2m}+tr{\left(N_A^{1/2}{N}_B^{1/2}\right)}^{2m}\le tr{\left(AB\right)}^m\le tr\left(A^m{B}^m\right).$$

(2.13)

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

$$\begin{array}{c}{P}^{T}AP=\left(\begin{array}{cc}\hfill A_1-J_m{A}_2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill A_1+J_m{A}_2\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill M_A\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_A\hfill \end{array}\right),\\ P^{T}BP=\left(\begin{array}{cc}\hfill B_1-J_m{B}_2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill B_1+J_m{B}_2\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill M_B\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill N_B\hfill \end{array}\right).\end{array}$$

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,

$$tr{\left(M_A^{1/2}{M}_B^{1/2}\right)}^{2m}+tr{\left(N_A^{1/2}{N}_B^{1/2}\right)}^{2m}\le tr{\left(P^{T}AP{P}^{T}BP\right)}^m=tr{\left(P^{T}ABP\right)}^m.$$

Since P is orthogonal, we have tr(P^TABP)^m= tr(AB)^m. By Lemma 1.4, the following holds

$$tr{\left(M_A^{1/2}{M}_B^{1/2}\right)}^{2m}+tr{\left(N_A^{1/2}{N}_B^{1/2}\right)}^{2m}\le tr{\left(AB\right)}^m\le tr\left(A^m{B}^m\right).$$