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A matrix trace inequality and its application

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A matrix trace inequality and its application
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    a  r   X   i  v  :   1   0   0   1 .   3   8   0   3  v   6   [  m  a   t   h .   F   A   ]   2   0   A  u  g   2   0   1   0 A matrix trace inequality and its application Shigeru Furuichi 1 ∗ and Minghua Lin 2 † 1 Department of Computer Science and System Analysis,College of Humanities and Sciences, Nihon University,3-25-40, Sakurajyousui, Setagaya-ku, Tokyo, 156-8550, Japan 2 Department of Mathematics and Statistics,University of Regina, Regina, Saskatchewan, S4S 0A2, Canada Abstract.  In this short paper, we give a complete and affirmative answer to a conjectureon matrix trace inequalities for the sum of positive semidefinite matrices. We also apply theobtained inequality to derive a kind of generalized Golden-Thompson inequality for positivesemidefinite matrices. Keywords :  Matrix trace inequality, positive semidefinite matrix, majorization andGolden-Thompson inequality 2000 Mathematics Subject Classification :  15A39 and 15A45 1 Introduction We give some notations. The set of all  n × n  matrices on the complex field  C  is represented by M  ( n, C ). The set of all  n  × n  Hermitian matrices is also represented by  M  h ( n, C ). Moreoverthe set of all  n × n  nonnegative (positive semidefinite) matrices is also represented by  M  + ( n, C ).Here  X   ∈  M  + ( n, C ) means we have   φ | X  | φ  ≥  0 for any vector  | φ  ∈ C n .The purpose of this short paper is to give the answer to the following conjecture which wasgiven in the paper [1]. Conjecture 1.1 ([1])  For   X,Y   ∈  M  + ( n, C )  and   p  ∈ R , the following inequalities hold or not? (i)  Tr [( I   + X   + Y   + Y  1 / 2 XY  1 / 2 )  p ]  ≤  Tr [( I   + X   + Y   + XY  )  p ]  for   p  ≥  1 .(ii)  Tr [( I   + X   + Y   + Y  1 / 2 XY  1 / 2 )  p ]  ≥  Tr [( I   + X   + Y   + XY  )  p ]  for   0  ≤  p  ≤  1 . We firstly note that the matrix  I   + X   + Y   + XY   = ( I   + X  )( I   + Y  ) is generally not positivesemidefinite. However, the eigenvalues of the matrix ( I   +  X  )( I   +  Y  ) are same to those of thepositive semidefinite matrix ( I   + X  ) 1 / 2 ( I   + Y  )( I   + X  ) 1 / 2 . Therefore the expression  Tr [( I   + X   + Y   + XY   )  p ] always makes sense.We easily find that the equality for (i) and (ii) in Conjecture 1.1 holds in the case of   p  = 1.In addition, the case of   p  = 2 was proven by elementary calculations in [1].Putting  T   = ( I   + X  ) 1 / 2 and  S   =  Y  1 / 2 , Conjecture 1.1 can be reformulated by the followingproblem, because we have  Tr [( I   +  X   +  Y   +  XY  )  p ] =  Tr [( T  2 +  T  2 S  2 )  p ] =  Tr [( T  2 ( I   + S  2 ))  p ] = Tr [( T  ( I   + S  2 ) T  )  p ] =  Tr [( T  2 + TS  2 T  )  p ] . ∗ E-mail:furuichi@chs.nihon-u.ac.jp † E-mail:lin243@uregina.ca 1  Problem 1.2  For   T,S   ∈  M  + ( n, C )  and   p  ∈ R , the following inequalities hold or not? (i)  Tr [( T  2 + ST  2 S  )  p ]  ≤  Tr [( T  2 + TS  2 T  )  p ]  for   p  ≥  1 .(ii)  Tr [( T  2 + ST  2 S  )  p ]  ≥  Tr [( T  2 + TS  2 T  )  p ]  for   0  ≤  p  ≤  1 . 2 Main results To solve Problem 1.2, we use the concept of the majorization. See [2] for the details on the majorization. Here for  X   ∈  M  h ( n, C ),  λ ↓ ( X  ) =  λ ↓ 1 ( X  ) , ··· ,λ ↓ n ( X  )   represents the eigenvaluesof the Hermitian matrix  X   in decreasing order,  λ ↓ 1 ( X  )  ≥ ··· ≥  λ ↓ n ( X  ). In addition  x  ≺  y  meansthat  x  = ( x 1 , ··· ,x n ) is majorized by  y  = ( y 1 , ··· ,y n ), if we have k   j =1 x  j  ≤ k   j =1 y  j  ( k  = 1 , ··· ,n − 1)and n   j =1 x  j  = n   j =1 y  j . We need the following lemma which can be obtained as a consequence of Ky Fan’s maximumprinciple. Lemma 2.1 (p.35 in [3])  For   A,B  ∈  M  h ( n, C )  and any   k  = 1 , 2 , ··· ,n , we have  k   j =1 λ ↓  j ( A  + B )  ≤ k   j =1 λ ↓  j ( A ) + k   j =1 λ ↓  j ( B ) .  (1)Then we have the following theorem. Theorem 2.2  For   S,T   ∈  M  + ( n, C ) , we have  λ ↓ ( T  2 + ST  2 S  )  ≺  λ ↓ ( T  2 + TS  2 T  ) (2) Proof  : For  S,T   ∈  M  + ( n, C ), we need only to show the following k   j =1 λ ↓  j ( T  2 + ST  2 S  )  ≤ k   j =1 λ ↓  j ( T  2 + TS  2 T  ) (3)for  k  = 1 , 2 , ··· ,n − 1, since we have n   j =1 λ ↓  j ( T  2 + ST  2 S  ) = n   j =1 λ ↓  j ( T  2 + TS  2 T  ) , which is equivalent to  Tr [ T  2 + ST  2 S  ] =  Tr [ T  2 + TS  2 T  ].By Lemma 2.1, we have2 k   j =1 λ ↓  j ( X  )  ≤ k   j =1 λ ↓  j  ( X   + Y  ) + k   j =1 λ ↓  j  ( X   − Y  ) .  (4)2  for  X,Y   ∈  M  h ( n, C ) and any  k  = 1 , 2 , ··· ,n .For  X   ∈  M  ( n, C ), the matrices  XX  ∗ and  X  ∗ X   are unitarily similar so that we have λ ↓  j ( XX  ∗ ) =  λ ↓  j ( X  ∗ X  ). Then we have the following inequality:2 k   j =1 λ ↓  j  T  2 + TS  2 T    = k   j =1 λ ↓  j  T  2 + TS  2 T   + k   j =1 λ ↓  j  T  2 + TS  2 T   = k   j =1 λ ↓  j  (( T   + iTS  )( T   − iST  )) + k   j =1 λ ↓  j  (( T   − iTS  )( T   + iST  ))= k   j =1 λ ↓  j  (( T   − iST  )( T   + iTS  )) + k   j =1 λ ↓  j  (( T   + iST  )( T   − iTS  ))= k   j =1 λ ↓  j  T  2 + ST  2 S   + i  T  2 S   − ST  2  + k   j =1 λ ↓  j  T  2 + ST  2 S   − i  T  2 S   − ST  2  ≥  2 k   j =1 λ ↓  j  T  2 + ST  2 S   , for any  k  = 1 , 2 , ··· ,n − 1, by using the inequality (4) for  X   =  T  2 + ST  2 S   and  Y   =  i ( T  2 S  − ST  2 ).Thus we have the inequality (3) so that the proof is completed.From Theorem 2.2, we have the following corollary. Corollary 2.3  For   T,S   ∈  M  + ( n, C )  and   p  ∈ R , the following inequalities hold.(i)  Tr [( T  2 + ST  2 S  )  p ]  ≤  Tr [( T  2 + TS  2 T  )  p ]  for   p  ≥  1 .(ii)  Tr [( T  2 + ST  2 S  )  p ]  ≥  Tr [( T  2 + TS  2 T  )  p ]  for   0  ≤  p  ≤  1 .Proof :  Since  f  ( x ) =  x  p , (  p  ≥  1) is convex function and  f  ( x ) =  x  p , (0  ≤  p  ≤  1) is con-cave function, we have the present corollary thanks to Theorem 2.2 and a general property of majorization (See p.40 in [3]).As mentioned in Introduction, Corollary 2.3 implies the following corollary by putting  T   =( I   + X  ) 1 / 2 and  S   =  Y  1 / 2 . Corollary 2.4  For   X,Y   ∈  M  + ( n, C )  and   p  ∈ R , the following inequalities hold.(i)  Tr [( I   + X   + Y   + Y  1 / 2 XY  1 / 2 )  p ]  ≤  Tr [( I   + X   + Y   + XY  )  p ]  for   p  ≥  1 .(ii)  Tr [( I   + X   + Y   + Y  1 / 2 XY  1 / 2 )  p ]  ≥  Tr [( I   + X   + Y   + XY  )  p ]  for   0  ≤  p  ≤  1 . Thus Conjecture 1.1 was completely solved with an affirmative answer. 3 An application In this section, we give a kind of one-parameter extension of the famous Golden-Thompson in-equality [4, 5] for positive semidefinite matrices, applying the obtained result in the previous sec-tion. For this purpose, we denote the generalized exponential function by exp ν  ( X  )  ≡  ( I   + νX  ) 1 ν for  ν   ∈  (0 , 1] and  X   ∈  M  ( n, C ) such that  Tr [( I   + νX  ) 1 ν ]  ∈ R . In addition, we use the followinginequalities proved in [6].3  Lemma 3.1 ([6])  For   X,Y   ∈  M  + ( n, C ) , and   ν   ∈  (0 , 1] , we have (i) Tr [exp ν  ( X   + Y  )]  ≤  Tr [exp ν  ( X   + Y   + νY  1 / 2 XY  1 / 2 )] .  (5) (ii) Tr [exp ν  ( X   + Y   + νXY  )]  ≤  Tr [exp ν  ( X  )exp ν  ( Y  )] .  (6)As mentioned in the below of Conjecture 1.1, the expression of the left hand side in (6) makes also sense, since we have  Tr [exp ν  ( X  + Y   + νXY  )] =  Tr [  ( I   + νX  ) 1 / 2 ( I   + νY  )( I   + νX  ) 1 / 2  1 ν ]  ≥ 0.From (i) of Corollary 2.4 and Lemma 3.1, we have the following proposition. Proposition 3.2  For   X,Y   ∈  M  + ( n, C )  and   ν   ∈  (0 , 1] , we have  Tr [exp ν  ( X   + Y  )]  ≤  Tr [exp ν  ( X  )exp ν  ( Y   )] .  (7) Proof  : The right hand side of (5) is bounded from the above by applying (i) of Corollary 2.4 and putting  X  1  =  νX  ,  Y  1  =  νY   and  p  =  1 ν  : Tr  exp ν  ( X   + Y   + νY  1 / 2 XY  1 / 2 )   =  Tr  I   + ν  ( X   + Y   + νY  1 / 2 XY  1 / 2 )  1 ν  =  Tr  ( I   + X  1  + Y  1  + Y  1 / 21  X  1 Y  1 / 21  )  p  ≤  Tr [( I   + X  1  + Y  1  + X  1 Y  1 )  p ]=  Tr  { I   + ν  ( X   + Y   + νXY  ) } 1 ν  =  Tr [exp ν  ( X   + Y   + νXY  )] , which is the left hand side of (6). Thus we have the present proposition thanks to Lemma 3.1. Note that the inequality (7) can be regarded as a kind of one-parameter extension of theGolden-Thompson inequality for positive semidefinite matrices  X   and  Y  . Ackowledgements We would like to thank the anonymous reviewer for providing valuable comments to improvethe manuscript. The first author (S.F.) was supported in part by the Japanese Ministry of Education, Science, Sports and Culture, Grant-in-Aid for Encouragement of Young Scientists(B), 20740067 References [1] S.Furuichi,A mathematical review of the generalized entropies and their matrix trace in-equalities, Proceedings of WEC2007, pp.840-845 (2007).[2] A.W.Marshall and I.Olkin, Inequalities: Theory of majorization and its applications, Aca-demic Press, 1979.[3] R.Bhatia, Matrix Analysis, Springer, 1997.4  [4] S.Golden, Lower bounds for the Helmholtz function, Phys. Rev., Vol.137(1965), pp.B1127-B1128.[5] C.J.Thompson, Inequality with applications in statistical mechanics, J.Math.Phys.,Vol.6(1965), pp.1812-1813.[6] S.Furuichi, Trace inequalities in nonextensive statistical mechanics, Linear Alg.Appl.,Vol.418(2006), pp.821-827.5
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