Zhou Journal of Inequalities and Applications 2014, 2014:64 http://www.journalofinequalitiesandapplications.com/content/2014/1/64 RESEARCH Open Access On some trace inequalities for positive definite Hermitian matrices Houqing Zhou* *Correspondence: [email protected] Abstract Department of Mathematics, Shaoyang University, Shaoyang City, Let A be a positive definite Hermitian matrix, we investigate the trace inequalities Hunan 422000, China of A. By using the equivalence of the deformed matrix, according to some properties of positive definite Hermitian matrices and some elementary inequalities, we extend some previous works on the trace inequalities for positive definite Hermitian matrices, and we obtain some valuable theory. MSC: Primary 15A45; 15A57 Keywords: Hermitian matrix; positive definite; trace inequality 1 Introduction In mathematics, a Hermitian matrix (or self-adjoint matrix) is a square matrix with com- plex entries that is equal to its own conjugate transpose. That is, the elements in the ith row and jth column are equal to the complex conjugates of the elements in the jth row T and ith column. In other words, the matrix A is Hermitian if and only if A = A ,where T A denotes the conjugate transpose of matrix A. Hermitian matrices play an important role in statistical mechanics [], engineering; in casessuchascommunication,todescriben-dimensional signal cross-correlation proper- ties, like conjugate symmetry, we can use Hermitian matrices. The earliest study of matrix inequality work in the literature was []. In , Bellman [] proved some trace inequalities for positive definite Hermitian matrices: () tr(AB) ≤ tr(AB); () tr(AB) ≤ tr A + tr B; () tr(AB) ≤ (tr A) (tr B) . Since then, the problems of the trace inequality for positive definite (semidefinite) Her- mitian matrices have caught the attention of scholars, getting a lot of interesting results. There exists a vast literature that studies the trace (see [–]). In the paper, using the identical deformation of matrix, and combined with some elementary inequalities, our purpose is to derive some new results on the trace inequality for positive definite Hermi- tian matrices. The rest of this paper is organized as follows. In Section , we will give the relevant defi- nitions and properties of Hermitian matrices. In Section , we will quote some lemmas; in Section , which is the main part of the paper, using the properties of Hermitian matrices, we investigate the trace inequalities for positive definite Hermitian matrices. ©2014 Zhou; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribu- tion License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Zhou Journal of Inequalities and Applications 2014, 2014:64 Page 2 of 6 http://www.journalofinequalitiesandapplications.com/content/2014/1/64 2Preliminaries By Mn,m(F)wedenotethen-by-m matrices over a field F, usually the real numbers R or the complex numbers C. Most often, the facts discussed are valid in the setting of the complex- entried matrices, in which case Mn,m(C) is abbreviated as Mn,m.Incaseofsquarematrices we replace Mn,n by Mn. Let A =(aij) ∈ Mn; we may denote the eigenvalues of A by λ, λ,...,λn,withoutloss of generality, where we let λ ≥ λ ≥ ··· ≥ λn.Letσ (A) denote the singular value, and σ ≥ σ ≥···≥σn.ItiswellknownthatifA is Hermitian, then all eigenvalues of A are real numbers and if A is unitary, then every eigenvalue of A has modulus . The sum of two Hermitian matrices of the same size is Hermitian. If A is Hermitian, then Ak is Hermitian for all k =,,....IfA is invertible as well, then A– is Hermitian. A Hermitian matrix A ∈ Mn is said to be positive semidefinite, denoted by A ≥ , if (Ax, x) ≥ for all x ∈ Cn, and it is called positive definite, denoted by A >,if(Ax, x)> for all nonzero x ∈ Cn (Cn denotes complex vector spaces), where (·) denotes the Euclidean inner product on Cn. Remark . () The sum of any two positive definite matrices of the same size is positive definite. () Each eigenvalue of a positive definite matrix is a non-negative (positive) real number. () The trace and the determinant of a positive definite matrix are non-negative (positive) real numbers. () Any principal submatrix of a positive definite matrix is positive definite. A Hermitian matrix is positive definite if and only if all of its eigenvalues are non- negative (positive) real numbers. We will use this fact several times. ∈ n Let A Mn. Then the trace of A is given by tr A = i= aii.Thetracefunctionhasthe following properties. Let A, B ∈ Mn, α ∈ C.Then () tr(A + B)=tr A + tr B; () tr αA = α tr A; () tr(AB)=tr(BA). 3 Some lemmas The following lemmas play a fundamental role in this paper. Lemma . [] Let Ai ∈ Mn (i =,,...,m). Then n n tr(AA ···Am) ≤ σi(AA ···Am) ≤ σi(A)σi(A) ···σi(Am), i= i= where σ(Ai) ≥ σ(Ai) ≥···≥σn(Ai). n ≥ Lemma . [] Let αi >(i =,,...,n), and i= αi . Let aij >(j =,,...,m). Then m m α m α m αn α α ··· αn ≤ ··· aj aj anj aj aj anj . j= j= j= j= Zhou Journal of Inequalities and Applications 2014, 2014:64 Page 3 of 6 http://www.journalofinequalitiesandapplications.com/content/2014/1/64 α α α Lemma . [] Let A, A,...,An be same size positive definite matrices, and , ,..., n n are positive real numbers, and i= αi =.Then n n αi ≤ tr Ai αi tr Ai. i= i= 4 Main results Now, we prove the following theorem. Theorem . Let Ai, Bi (i =,,...,m) be same size positive definite matrices, p >,and p + q =.Then m m p m q ≤ p q tr (AiBi) tr Ai tr Bi . i= i= i= Proof Since the eigenvalues and traces of positive definite matrices are all positive real numbers, the eigenvalues are equal to the singular values. Then, according to Lemma . and the spectral mapping theorem, we have m p q tr Ai Bi i= m n ≤ p q σj Ai Bi i= j= m n ≤ p q σj Ai σj Bi i= j= m n p q = λj Ai λj Bi i= j= m n p q = λj (Ai)λj (Bi). (.) i= j= By using Lemma .,weget m p q tr Ai Bi i= m n p n q ≤ λj(Ai) λj(Bi) i= j= j= m p q ≤ (tr Ai) (tr Bi) i= m p m q ≤ tr Ai tr Bi .(.) i= i= Zhou Journal of Inequalities and Applications 2014, 2014:64 Page 4 of 6 http://www.journalofinequalitiesandapplications.com/content/2014/1/64 p q Let Ai = Ai , Bi = Bi .Thenweobtain m m p m q ≤ p q tr(AiBi) tr Ai tr Bi . i= i= i= By tr(A + B)=tr A + tr B,wehave m m p m q ≤ p q tr (AiBi) tr Ai tr Bi . i= i= i= Thus, the proof is completed. Next, we give a trace inequality for positive definite matrices. n Theorem . Let αi >(i =,,...,n) and i= αi =.Ai, Bi, Ci (i =,,...,n) are same size positive definite matrices. Then n n n n αi αi αi ≤ αi tr Ai + Bi + Ci tr(Ai + Bi + Ci) . i= i= i= i= Proof Since the trace of a matrix is a linear operation, by using Lemma ., it follows that n αi n αi n αi tr( i= Ai + i= Bi + i= Ci ) n αi i=[tr(Ai + Bi + Ci)] n Aαi Bαi Cαi = tr i + i + i tr(A + B + C )αi tr(A + B + C )αi tr(A + B + C )αi i= i i i i i i i i i n A αi n B αi = tr i + tr i tr(A + B + C ) tr(A + B + C ) i= i i i i= i i i n C αi + tr i tr(A + B + C ) i= i i i n α tr A n α tr B n α tr C ≤ i i + i i + i i tr(A + B + C ) tr(A + B + C ) tr(A + B + C ) i= i i i i= i i i i= i i i n α (tr A + tr B + tr C ) = i i i i tr(A + B + C ) i= i i i n = αi =. (.) i= Then we have n n n n αi αi αi ≤ αi tr Ai + Bi + Ci tr(Ai + Bi + Ci) . i= i= i= i= Now we use mathematical induction to deduce our third result. Zhou Journal of Inequalities and Applications 2014, 2014:64 Page 5 of 6 http://www.journalofinequalitiesandapplications.com/content/2014/1/64 Theorem . Let Ai (i =,,...,n) be same size positive definite matrices. Then we have the inequality tr( n A ) n tr A i= i ≤ i . tr( n A ) tr A i= i i= i Proof When n = , according to () on the first page, we have tr A · tr A · tr(AA) ≤ · · tr A tr A tr A tr A = tr A tr A tr A tr A ≤ tr A tr A + tr A tr A =(tr A) tr A +(tr A) tr A . (.) That is, · · tr A tr A tr A tr A tr(AA) ≤ + .(.) tr A tr A On the other hand, tr(A + A) = tr A +tr(AA)+tr A. (.) Note that tr A tr A + tr(A + A) tr A tr A tr A tr A = + (tr A + tr A) tr A tr A · · tr A tr A tr A tr A = tr A + tr A + + .(.) tr A tr A So, according to (.), (.), we deduce tr A tr A + tr(A + A) tr A tr A ≥ tr A +tr(AA)+tr A = tr(A + A) .(.) Hence, tr(A + A ) tr A tr A ≤ + .(.) tr(A + A) tr A tr A Then the inequality holds.