An Inequality for the trace of matrix products, using absolute values

Transcription

1 arxv: v2 [math-ph] 1 Sep 2011 An Inequalty for the trace of matrx products, usng absolute values Bernhard Baumgartner 1 Fakultät für Physk, Unverstät Wen Boltzmanngasse 5, A-1090 Venna, Austra September 1, 2011 Abstract The absolute value of matrces s used n order to gve nequaltes for the trace of products. An applcaton gves a very short proof of the tracal matrx Hölder nequalty. Keywords: matrx, absolute value, trace nequalty, Hölder nequalty MSC: 39B42, 15A45, 47A50, PACS: Yn 1 Bernhard.Baumgartner@unve.ac.at 1

2 An Inequalty for the trace of matrx products, usng absolute values September 1, The baby nequalty and ts applcaton 1 THEOREM. Usng absolute values: Consder two complex m n matrces A, B and ther absolute values, A = (A A) 1/2, A = (AA ) 1/2. Then TrA B (Tr A B ) 1/2 (Tr A B ) 1/2 (1) Proof. Let e be an orthonormal bass made of normalzed egenvectors of A A, wth egenvalues a 2 0. For a 0, f = a 1 Ae are normalzed egenvectors of AA, obeyng A f = a e. Eventually, to make a full bass, ths set has to be completed by ntroducng egenvectors of AA wth egenvalue 0. Analogously, therearebass-sets ofnormalzed vectors g j andh j, such that Bg j = b j h j, B h j = b j g j, wth b j 0. Wth these vectors we get TrA B = a b j g j,e f,h j. (2) TrA B Applyng the Cauchy-Schwarz nequalty gves ( ) 1/2 ( 1/2 a b j e,g j g j,e a b j f,h j h j,f ) (3) = (Tr A B ) 1/2 (Tr A B ) 1/2 (4) In the last step the denttes A e = a e, A f = a f, B g j = b j g j, and B h j = b j h j have been used. We remark that the nequalty s sharp. It becomes an equalty n case both matrces A and B have rank one. Ths follows from the fact that the sum n (2) conssts of only one term, so the Schwarz nequalty n (3) becomes an equalty. A smple example s gven wth 2 2 matrces: A = wth B = ( 1 0 ) ( ) 1 1, B =, (5) ( ) 1 1 1, B = ( ) ; so TrA B = 1, Tr A B = 1/ 2, Tr A B = 2. Other cases where (3) becomes an equalty are e,g j = α f,h j, for some constant α. The example shows also that the dstncton between the absolute values B and B s necessary. Only f both A and B are normal matrces the absolute values are equal, A = A, B = B. In such a case equaton (1) becomes TrA B Tr A B. The followng applcaton was at the orgn of my nvestgatons, searchng for a smple proof of the Hölder nequalty for matrces and operators. (For other proofs, see f.e. [MBR72, RS75, RB97, EC09])

3 An Inequalty for the trace of matrx products, usng absolute values September 1, THEOREM. Matrx Hölder Inequalty: Consder two m m matrces A, B and ther absolute values, then TrA B (Tr A p ) 1/p (Tr B q ) 1/q, 1 p,q, p 1 +q 1 = 1 (6) Proof. Usng the same notaton as n the proof of Theorem 1, we note that Tr A p = Tr A p = ap. So, the Hölder Inequalty for the left hand sde of (1) s proven, f t holds for each factor on the rght hand sde. There we have normal operators. For these we can use the classcal Hölder Inequalty for weghted sums, followed by usng completeness of the bass sets {e } and {g j }: Tr A B = ( ) 1/p ( ) 1/q a b j e,g j 2 a p e,g j 2 b q j e,g j 2 ( 1/p ( 1/q = a) p bj) q = (Tr A p ) 1/p (Tr B q ) 1/q. Analogously for Tr A B. j If one s nterested n characterzng cases of equalty for the matrx Hölder nequalty, one has to check whether e,g j = α f,h j, as stated above, and also whether the classcal Hölder nequalty used n the proof becomes an equalty. 2 Generalzatons There are possbltes to generalze the baby nequalty. It can grow by: Insertng extra matrces, one m m another one n n; usng dfferent exponents for the dfferent absolute values; gong nto vector spaces wth nfnte dmenson. One can nsert extra matrces M and N: 3 THEOREM. Inequalty wth ntermedate matrces: TrMA NB (TrM A M B ) 1/2 (TrN A N B ) 1/2 (7) Proof. Just nsert the extra matrces n the rght places n the nner products appearng n (2) and (3): Replace g j,e by g j,me and f,h j by f,nh j. There s the possblty to consder dfferent exponents: 4 THEOREM. Inequalty wth exponents: TrA B ( Tr A α B β) 1/2 ( Tr A 2 α B 2 β) 1/2, 0 α,β 2, (8) wth = 0, so that A 0 = projector onto range( A ).

4 An Inequalty for the trace of matrx products, usng absolute values September 1, Proof. Modfy (2) and (3) as ( ) TrA 1/2 ( ) 1/2 B = ϕ ψ ϕ 2 ψ 2, (9) wth ϕ = a α/2 b β/2 j e,g j, ψ = a 1 α/2 b 1 β/2 j h j,f. Observe, that for a = 0 the matrx elements nvolvng e or f are just absent. The same holds for b j,g j,h j. Extensons nto nfnte dmensons can be done n dfferent ways. I present the followng result: 5 THEOREM. Inequalty for two Hlbert Schmdt class operators: Let A and B be operators from Hlbert space H to the Hlbert space K, wth the propertes Tr H A A = Tr K A A < and Tr H B B = Tr K B B <. Then Tr H A B (Tr H A B ) 1/2 (Tr K A B ) 1/2 (10) Proof. As n the proof of Theorem 1 we use the sngular values a and bass sets e, f, here e H, f K, such that Ae = a f, and analogously Bg = b h. In Drac s notaton A = f a e and B = h b g. A beng n the Hlbert-Schmdt class means ( 1/2 Tr H A A = Tr K A A = Tr H A 2 = Tr K A 2 = a) 2 <, and the analogue for B. Introducng the operators wth fnte rank A N = N f a e and B N = N h b g, (11) one can apply Theorem 1 to the matrces whch represent these operators, gvng the nequalty (10) for A N and B N. One observes the convergences n norm: ( 1/2 A A N = A A N = A A N = A A N = a) 2 N 0, and the same for B. The Hlbert-Schmdt nner products are jontly normcontnuous n both factors, so each sde of the nequalty (10) for A N and B N converges as N, gvng the same nequalty wthout the N as an ndex. Applcatons are new proofs for Hölder type nequaltes used n mathematcal physcs. They wll be dscussed n a followng artcle. N+1

5 An Inequalty for the trace of matrx products, usng absolute values September 1, Comparson wth another use of absolute values The product A B can be represented as A B = U A B V, (12) by extendng the m n matrces to N N matrces, where N = max{m,n}, and usng the polar decompostons A = U A, B = V B wth untary operators U, V. (Equvalently, one can stay wth the m n matrces and use sometres U and V nstead of untary operators.) Ths equalty mples that the set of sngular values of A B s dentcal to that of A B. (Eventually, when stayng wth m n, the numbers of zeroes are dfferent.) So, all the untarly nvarant norms, see [RB97], are dentcal. One dentty s for the operator norm another one gves A B = A B, (13) Tr A B = Tr ( A B ). (14) Together wth TrM Tr M, whch holds for each matrx, we get TrA B Tr ( A B ). (15) Ths nequalty s not as sharp as the baby nequalty gven n Theorem 1. As an example use the same matrces as n equaton (5). They gve 1 on the l.h.s. but 2 on the r.h.s. of (15). References [MBR72] Mary Beth Ruska: Inequaltes for Traces on von Neumann Algebras Commun. Math. Phys. 26, , (1972) [RS75] Mchael Reed and Barry Smon: Methods of Modern Mathematcal Physcs II (Academc Press, New York) 1975 [RB97] Rajendra Bhata: Matrx Analyss. Graduate Texts n Mathematcs 169, (Sprnger Verlag, New York) 1997 [EC09] Erc A. Carlen: Trace Inequaltes and Quantum Entropy: An ntroductory course. Lecture Course gven at Entropy and the Quantum - Tucson, Arzona, March 16-20,