In this note we generalize the trace inequality derived by [1] to the case
where the number of terms of the sum (denoted by K) is arbitrary

Belmega, E. V.

Jungers, M.

Lasaulce, S.

English

arXiv

The Australian Journal of Mathematical Analysis and Applications (AJMAA)International audienceIn this note we generalize the trace inequality derived in E.V. Belmega, S. Lasaulce and M. Debbah, A Trace Inequality for Positive Definite Matrices, J. Inequal. Pure Appl. Math., 10(1) (2009), to the case where the number of terms of the sum (denoted by K) is arbitrary

Belmega, Elena Veronica

Jungers, Marc

Lasaulce, Samson

HAL-CentraleSupelec

A generalization of a trace inequality for positivedefinite matricesElena Veronica Belmega, Marc Jungers, Samson LasaulceTo cite this version:Elena Veronica Belmega, Marc Jungers, Samson Lasaulce. A generalization of a trace in-equality for positive definite matrices. The Australian Journal of Mathematical Analysis andapplications, 2011, 7 (2), pp.Art. 26, 1-5. <hal-00447007>HAL Id: hal-00447007https://hal.archives-ouvertes.fr/hal-00447007Submitted on 15 Jan 2010HAL is a multi-disciplinary open accessarchive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come fromteaching and research institutions in France orabroad, or from public or private research centers.L’archive ouverte pluridisciplinaire HAL, estdestine´e au de´poˆt et a` la diffusion de documentsscientifiques de niveau recherche, publie´s ou non,e´manant des e´tablissements d’enseignement et derecherche franc¸ais ou e´trangers, des laboratoirespublics ou prive´s.A GENERALIZATION OF A TRACE INEQUALITY FOR POSITIVEDEFINITE MATRICES.ELENA VERONICA BELMEGA, MARC JUNGERS, AND SAMSON LASAULCEAbstract. In this note we generalize the trace inequality derived by [1] to the casewhere the number of terms of the sum (denoted by K) is arbitrary. More precisely weprove that TK = TrK∑k=1(Ak −Bk)( k∑`=1B`)−1−(k∑`=1A`)−1 ≥ 0 for any set ofpositive definite matrices.1. IntroductionTrace inequalities are useful in many areas like multiple input multiple output (MIMO)systems in control theory and communications. Proving the trace inequality under inves-tigation in this note is a sufficient condition to ensure the uniqueness of a Nash equilibriumin certain MIMO communications game [2] where Rosen’s diagonally strict concavity con-dition [3] is valid. The considered inequality has been proven by e.g., [4] for K = 1 andby [1] for K = 2. Here we generalize it to K ∈ N∗. The main result of this note is asfollows.Theorem 1.1. Let K ∈ N∗. Assume that(i): A1 = AH1 Â 0, B1 = BH1 Â 0;(ii): ∀k ∈ {2, . . . , K}, Ak = AHk º 0 and Bk = BHk º 0.Then, we have that(1.1) TK , TrK∑k=1(Ak −Bk)( k∑`=1B`)−1−(k∑`=1A`)−1 ≥ 0.2. Auxiliary ResultsIn order to prove Theorem 1.1, we will use two auxiliary lemmas which are stated herefor the sake of clarity.Lemma 2.1. [1] Let A, B be two positive definite matrices, C, D, two positive semidef-inite matrices whereas X is only assumed to be Hermitian. Then(2.1) Tr{XA−1XB−1}− Tr{X(A+C)−1X(B+D)−1} ≥ 0.The proof is given in [1].Lemma 2.2. Let A, B be two positive definite matrices, C, D, two positive semi-definitematrices. Then(2.2)Tr {(A−B)(B+D)−1(C−D)(A+C)−1} =Tr {(C−D)(B+D)−1(A−B)(A+C)−1} ∈ R.2000 Mathematics Subject Classification. 15A45.Key words and phrases. Trace inequality, positive definite matrices, positive semidefinite matrices.12 ELENA VERONICA BELMEGA, MARC JUNGERS, AND SAMSON LASAULCEProof. To prove the desired result, let us define E by(2.3) E = Tr{(C−D) [(B+D)−1 − (A+C)−1]}and write it in two different ways.(2.4)E = Tr {(C−D)(B+D)−1[A+C−B−D](A+C)−1}= Tr {(C−D)(B+D)−1(C−D)(A+C)−1}+Tr {(C−D)(B+D)−1(A−B)(A+C)−1} .(2.5)E = Tr {(C−D)(A+C)−1[A+C−B−D](B+D)−1}= Tr {(C−D)(A+C)−1(C−D)(B+D)−1}+Tr {(C−D)(A+C)−1(A−B)(B+D)−1} .By using the commutation property of the trace and the two expressions of E , we findthe desired result. The only thing which needs to be proven is that E is real. For thispurpose, observe that if we denote by M = (C −D)(B +D)−1(A − B)(A +C)−1 thenMH = (A +C)−1(A−B)(B +D)−1(C−D) and from the result just proven we obtainthat Tr(MH) = Tr(M) and thus we have that Tr(M) ∈ R. ¤3. Proof of Theorem 1.1Define Xk =k∑i=1Ai, Yk =k∑i=1Bi, for all k ≥ 1 which are both positive definitematrices. Notice that TK can be re-written recursively as follows:(3.1)T1 = Tr{(A1 −B1)Y−11 (A1 −B1)X−11}TK = TK−1 + Tr{(AK −BK)Y−1K (AK −BK)X−1K}+Tr{(AK −BK)Y−1K (XK−1 −YK−1)X−1K}We proceed in two steps. First, we find a lower bound for TK and then we prove that thisbound is positive. First, let us prove that, for all K ≥ 1:(3.2)TK ≥ 12K∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK −YK)Y−1K (XK −YK)X−1K}To this end we proceed by induction. For all K ∈ N∗, define the proposition:(3.3)PK : TK ≥ 12K∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK −YK)Y−1K (XK −YK)X−1K}.It is easy to check that, for K = 1, P1 is true:(3.4)T1 = Tr{(A1 −B1)Y−11 (A1 −B1)X−11}= 12Tr{(A1 −B1)Y−11 (A1 −B1)X−11}+ 12Tr{(X1 −Y1)Y−11 (X1 −Y1)X−11}.Now, let us assume that PK−1 is true and then prove that this implies that PK is alsotrue. We have that:A TRACE INEQUALITY 3(3.5)TK−1 ≥ 12K−1∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK−1 −YK−1)Y−1K−1(XK−1 −YK−1)X−1K−1}.From the recursive formula (3.1) we obtain:(3.6)TK ≥ 12K−1∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK−1 −YK−1)Y−1K−1(XK−1 −YK−1)X−1K−1}+Tr{(AK −BK)Y−1K (AK −BK)X−1K}+ Tr{(AK −BK)Y−1K (XK−1 −YK−1)X−1K}≥ 12K−1∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK−1 −YK−1)Y−1K (XK−1 −YK−1)X−1K}+Tr{(AK −BK)Y−1K (AK −BK)X−1K}+ Tr{(AK −BK)Y−1K (XK−1 −YK−1)X−1K}= 12K−1∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK−1 −YK−1)Y−1K (XK−1 −YK−1)X−1K}+Tr{(AK −BK)Y−1K (AK −BK)X−1K}+ 12Tr{(AK −BK)Y−1K (XK−1 −YK−1)X−1K}+12Tr{(XK−1 −YK−1)Y−1K (AK −BK)X−1K}= 12K∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK−1 −YK−1)Y−1K (XK−1 −YK−1)X−1K}+12Tr{(AK −BK)Y−1K (AK −BK)X−1K}+ 12Tr{(AK −BK)Y−1K (XK−1 −YK−1)X−1K}+12Tr{(XK−1 −YK−1)Y−1K (AK −BK)X−1K}4 ELENA VERONICA BELMEGA, MARC JUNGERS, AND SAMSON LASAULCE= 12K∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK−1 +AK −YK−1 −BK)Y−1K (XK−1 +AK −YK−1 −BK)X−1K}= 12K∑i=1Tr{(Ai −Bi)Y−1i (Ai −Bi)X−1i}+12Tr{(XK −YK)Y−1K (XK −YK)X−1K}The second inequality follows from applying Lemma 2.1 for the second term on theright and considering that XK = XK−1 + AK , YK = YK−1 + BK . The third equalityfollows from Lemma 2.2. Thus, we have proven the desired result.The second step of the proof is straightforward. From (3.2), it is easy to check thatTK ≥ 0 (all the terms of the form Tr {XB−1XA−1} with X = XH , A Â 0, B Â 0 can bere-written as Tr(NNH) ≥ 0 with N = A−1/2XB−1/2).References[1] E. V. BELMEGA, S. LASAULCE, and M. DEBBAH, “A Trace Inequality for Positive DefiniteMatrices”, Journal of inequalities in pure and applied mathematics (JIPAM), Vol. 10, No. 1, Article5, 4 pp., (2009).[2] E. V. BELMEGA, S. LASAULCE, M. DEBBAH, M. JUNGERS, and J. DUMONT, “Power alloca-tion games in wireless networks of multi-antenna terminals”, Springer Telecommunications SystemsJournal, VALUETOOLS special issue, revised (minor revisions), (2009).[3] J. ROSEN, “Existence and uniqueness of equilibrium points for concave n-person games”, Economet-rica, Vol. 33, pp. 520–534, (1965).[4] K. M. ABADIR and J. R. MAGNUS, “Matrix Algebra”, Cambridge University Press, New York,USA, pp. 329, pp. 338, (2005).Universite´ Paris-Sud XI, SUPELEC, Laboratoire des Signaux et Syste`mes, Gif-sur-Yvette, France.E-mail address: belmega@lss.supelec.frURL: http://veronica.belmega.lss.supelec.frCNRS, ENSEM, CRAN, Vandoeuvre, France.E-mail address: marc.jungers@cran.uhp-nancy.frURL: http://perso.ensem.inpl-nancy.fr/Marc.Jungers/CNRS, SUPELEC, Laboratoire des Signaux et Syste`mes, Gif-sur-Yvette, France.E-mail address: lasaulce@lss.supelec.frURL: http://samson.lasaulce.lss.supelec.fr

A generalization of a trace inequality for positive definite matrices

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A generalization of a trace inequality for positive definite matrices

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