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Wysłany: Czw 22:57, 10 Lut 2011
Temat postu: Fan Ky Au Theorem and promotion _2966
Fan Ky Au Theorem and promotion
P. 445) gives the following theorem: FanKy concavity Theorem Let A, B is a real positive definite matrix, then for 0 ≤ ≤ 1, with l + (1 - X) Bl ≥ lAllBl. [1ib ~ the proof of the generalized use of multiple points. In this paper, known as the Young inequality ka + (1 a A) b ~ ab (where n, b is a positive number, 0 ≤ ≤ 1) of the concavity theorem FanKy a more elementary proof of this theorem and m the case of a positive definite matrix. To this end the first given the following lemma: Lemma on the same order real positive definite matrix A, B, there exists invertible matrix T, so T,】 『, BT also become diagonal, and diagonal elements are all positive . A certificate for a real symmetric matrix, so there is real invertible matrix P, so P'AP-J (J is the identity matrix.) And because P'BP is real symmetric matrix, so there exists an orthogonal matrix Q, so Q, (P ~ BP) Q = D. Where D is the diagonal matrix, and diagonal elements are real positive definite matrix P'BP the eigenvalues are all positive. Order T-PQ, is easy to know T-J, T'BT = D, and T and TBT diagonal matrix the diagonal elements are positive. FanKy concavity theorem Lemma we know that can find invertible matrix T, so T-diag (a1,
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, a2, ..., a) a T'BT = diag (b1, b2, ..., b) a b1 above n, > 0,6> 0 (: 1,2, ...,). According to inequality Aa, + (1 a A) b, ≥ nA1 a known lTll + (1 a) Bl-l + (1 A x) m'grl-ldiag [Aa1 + (1 a A) b, Aa2 + (1 a A) b2 , ...,
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, Aan + (1 a A) b] a set i = 1 [Aa, + (1 - A) bi] ≥ fi (a (Li i = 1n,) (Li i = 1) a [Received ] 2005-04-28124 Mathematics Volume 22 lATlITBTl one by one ITIIAIIBI one. on both sides of elimination that I D I have to I + (1 - 2) BI ≥ IAII cited. In this theorem to the case of a positive definite matrix, before concavity of the first to apply the theorem FanKy two inequalities .1. For it / order real positive definite matrix A, B, are IA + BI {≥ 2 (A) i.2. Let M 1 there IMI> O when, IM + M-1I ≥ 2; MIdO time, I + M-1I ≤ a 2. Certificate 1. If the application Minkowski inequality [Aq-BI ÷ ≥ IAI {+ IBI, the results of IA + BI tone ≥ 2 (A) i is easy to know, and now the inequality I + (1 - 2) BI ≥ IAII BU in order: Temple, that was I lost A + lost BI ≥ IAI ÷ IBI ÷, that is IA + BI ≥ Dong So IA + BI tone ≥ 2 (A) i.2. Since M is invertible, so M'M is a real positive definite matrix, matrix J of course is a real positive definite matrix. of A ~-M 'M and B-I and 2 a ÷ application FanKy concavity theorem, IMM + lost II ≥ II ÷ III ÷ a IIMI [, ie lM + Ml1Ml ≥ l1MI1. So IMI> O when, IM + M-1I ≥ 2; IMId0 time, I + MI ≤ a 2.FanKy concavity theorem in the form set A., A, ..., A is a real positive definite matrix, oK2 ≤ 1 (a 1,2, ...,) and. + + ... + a 1, a l. A. +. A + ... + swollen Al ≥ lA. l-lAlz ... lAl. license to use mathematical induction. When a 2, is the FanKy Au theorem. Suppose inequality m a a real positive definite matrix set up, now consider the m- the case of real positive definite matrix. we may assume that 0 <i <1 (a 1,
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,2, ...,), and noted that A + ... + A is real positive definite matrix,
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, so 1111IlAl +2 A2 + ... + a lA A l + AI-IA- + (1 - 2,) (rAz + ... + fAm) l ≥ lAlIA +..-+ Ar. ≥ IAIhI (IAI ... IAI Yu) - (induction hypothesis) an IAI-JAIz ... IAI. which proves the theorem. concavity of FanKy other forms of promotion of Theorem Ⅲ. [
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