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On Some Matrix Trace Inequalities

Journal of Inequalities and Applications20102010:201486

Received: 23 December 2009

Accepted: 14 March 2010

Published: 6 April 2010


We first present an inequality for the Frobenius norm of the Hadamard product of two any square matrices and positive semidefinite matrices. Then, we obtain a trace inequality for products of two positive semidefinite block matrices by using block matrices.


Positive Real NumberComplex MatriceReal VectorComplex MatrixPositive Definite Matrix

1. Introduction and Preliminaries

Let denote the space of complex matrices and write . The identity matrix in is denoted . As usual, denotes the conjugate transpose of matrix . A matrix is Hermitian if . A Hermitian matrix is said to be positive semidefinite or nonnegative definite, written as , if
is further called positive definite, symbolized , if the strict inequality in (1.1) holds for all nonzero . An equivalent condition for to be positive definite is that is Hermitian and all eigenvalues of are positive real numbers. Given a positive semidefinite matrix and , denotes the unique positive semidefinite power of .
Let and be two Hermitian matrices of the same size. If is positive semidefinite, we write

Denote and eigenvalues and singular values of matrix , respectively. Since is Hermitian matrix, its eigenvalues are arranged in decreasing order, that is, and if is any matrix, its singular values are arranged in decreasing order, that is, The trace of a square matrix (the sum of its main diagonal entries, or, equivalently, the sum of its eigenvalues) is denoted by .

Let be any matrix. The Frobenius (Euclidean) norm of matrix is
It is also equal to the square root of the matrix trace of that is,

A norm on is called unitarily invariant for all and all unitary .

Given two real vectors and in decreasing order, we say that is weakly log majorized by , denoted , if , and we say that is weakly majorized by , denoted , if . We say is majorized by denoted by , if

As is well known, yields (see, e.g., [1, pages 17–19]).

Let be a square complex matrix partitioned as
where is a square submatrix of . If is nonsingular, we call
the Schur complement of in (see, e.g., [2, page 175]). If is a positive definite matrix, then is nonsingular and
Recently, Yang [3] proved two matrix trace inequalities for positive semidefinite matrices and ,


Also, authors in [4] proved the matrix trace inequality for positive semidefinite matrices and ,

where is a positive integer.

Furthermore, one of the results given in [5] is

for and positive definite matrices, where is any positive integer.

2. Lemmas

Lemma 2.1 (see, e.g., [6]).

For any and .

Lemma 2.2 (see, e.g., [7]).

Let then

Lemma 2.3 (Cauchy-Schwarz inequality).

Let and be real numbers. Then,

Lemma 2.4 (see, e.g., [8, page 269]).

If and are poitive semidefinite matrices, then,

Lemma 2.5 (see, e.g., [9, page 177]).

Let and are matrices. Then,

Lemma 2.6 (see, e.g., [10]).

Let and are positive semidefinite matrices. Then,

where is a positive integer.

3. Main Results

Horn and Mathias [11] show that for any unitarily invariant norm on
Also, the authors in [12] show that for positive semidefinite matrix , where

for all and all unitarily invariant norms .

By the following theorem, we present an inequality for Frobenius norm of the power of Hadamard product of two matrices.

Theorem 3.1.

Let and be -square complex matrices. Then
where is a positive integer. In particular, if and are positive semidefinite matrices, then


From definition of Frobenius norm, we write
Also, for any and , it follows that (see, e.g., [13])
Since for and from inequality (3.7), we write
From Lemma 2.1 and Cauchy-Schwarz inequality, we write
By combining inequalities (3.7), (3.8), and (3.9), we arrive at
Thus, the proof is completed. Let and be positive semidefinite matrices. Then

where .

Theorem 3.2.

Let be positive semidefinite matrices. For positive real numbers


We know that , then by using the definition of Frobenius norm, we write

Thus, by using Theorem 3.1, the desired is obtained.

Now, we give a trace inequality for positive semidefinite block matrices.

Theorem 3.3.


where is an integer.


with . Then (see, e.g., [14]). Let
with , , . Then (see, e.g., [14]). We know that
By using Lemma 2.2, it follows that
Therefore, we get
As result, we write

Example 3.4.

Then From inequality (1.11), for we get
Also, for , since and , we get
Thus, according to this example from (3.24) and (3.25), we get



This study was supported by the Coordinatorship of Selçuk University's Scientific Research Projects (BAP).

Authors’ Affiliations

Department of Mathematics, Science Faculty, Selçuk University, Konya, Turkey


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© Z. Ulukök and R. Türkmen. 2010

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