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U linijnij algebri i teoriyi matric dopovnennya Shura dlya bloku matrici tobto pidmatrici v bilshij matrici viznacheno tak Pripustimo A B C D ye matrici vidpovidno p p p q q p i q q i D oborotna Nehaj M A B C D displaystyle M left begin matrix A amp B C amp D end matrix right tak sho M ce matricya p q p q Todi dopovnennya Shura dlya bloku D matrici M ce matricya p p A B D 1 C displaystyle A BD 1 C Jogo nazvali na chest yakij vikoristav jogo dlya dovedennya lemi Shura hocha jogo vikoristovuvali i do togo PidgruntyaDopovnennya Shura vinikaye yak rezultat zastosuvannya metodu Gausa shodo blokiv cherez mnozhennya na matricyu M na blokovu nizhnotrikutnu matricyu L I p 0 D 1 C I q displaystyle L left begin matrix I p amp 0 D 1 C amp I q end matrix right Tut Ip poznachaye odinichnu matricyu p p Pislya mnozhennya na matricyu L dopovnennya Shura z yavlyayetsya u gorishnomu p p bloku Matricyu dobutku taka M L A B C D I p 0 D 1 C I q A B D 1 C B 0 D I p B D 1 0 I q A B D 1 C 0 0 D displaystyle begin aligned ML amp left begin matrix A amp B C amp D end matrix right left begin matrix I p amp 0 D 1 C amp I q end matrix right left begin matrix A BD 1 C amp B 0 amp D end matrix right amp left begin matrix I p amp BD 1 0 amp I q end matrix right left begin matrix A BD 1 C amp 0 0 amp D end matrix right end aligned Ce analogichno do LDU rozkladu matrici Tobto mi shojno pokazali sho A B C D I p B D 1 0 I q A B D 1 C 0 0 D I p 0 D 1 C I q displaystyle begin aligned left begin matrix A amp B C amp D end matrix right amp left begin matrix I p amp BD 1 0 amp I q end matrix right left begin matrix A BD 1 C amp 0 0 amp D end matrix right left begin matrix I p amp 0 D 1 C amp I q end matrix right end aligned otzhe obernena do M mozhna predstaviti za uchastyu D 1 i obernenogo dopovnennya Shura yaksho vono isnuye yak A B C D 1 I p 0 D 1 C I q A B D 1 C 1 0 0 D 1 I p B D 1 0 I q A B D 1 C 1 A B D 1 C 1 B D 1 D 1 C A B D 1 C 1 D 1 D 1 C A B D 1 C 1 B D 1 displaystyle begin aligned amp quad left begin matrix A amp B C amp D end matrix right 1 left begin matrix I p amp 0 D 1 C amp I q end matrix right left begin matrix A BD 1 C 1 amp 0 0 amp D 1 end matrix right left begin matrix I p amp BD 1 0 amp I q end matrix right 12pt amp left begin matrix left A BD 1 C right 1 amp left A BD 1 C right 1 BD 1 D 1 C left A BD 1 C right 1 amp D 1 D 1 C left A BD 1 C right 1 BD 1 end matrix right end aligned Yaksho M simetrichna dodatnooznachena matricya to j tak samo bude dopovnennya Shura dlya D u M Yaksho p i q dorivnyuyut 1 toyuto A B C i D ye skalyarami to mi otrimuyemo formulu dlya obernennya matrici 2 na 2 M 1 1 A D B C D B C A displaystyle M 1 frac 1 AD BC left begin matrix D amp B C amp A end matrix right za umovi sho AD BC ne nul Bilshe togo takozh chitko vidno sho viznachnik M zadayetsya formuloyu det M det D det A B D 1 C displaystyle det M det D det A BD 1 C yaka uzagalnyuye formulu viznachnika u vipadku matric 2 na 2 Umovi na dodatnyu viznachenist i dodatnyu napivviznachenistNehaj X ce simetrichna matricya zadana tak X A B B T C displaystyle X left begin matrix A amp B B T amp C end matrix right Nehaj X A bude dopovnennyam Shura dlya A v X tobto X A C B T A 1 B displaystyle X A C B T A 1 B i X C bude dopovnennyam Shura dlya C v X tobto X C A B C 1 B T displaystyle X C A BC 1 B T Todi X dodatno viznachena todi i tilki todi koli A i X A dodatno viznacheni X 0 A 0 X A C B T A 1 B 0 displaystyle X succ 0 Leftrightarrow A succ 0 X A C B T A 1 B succ 0 X dodatno viznachena todi i tilki todi koli C i X C dodatno viznacheni X 0 C 0 X C A B C 1 B T 0 displaystyle X succ 0 Leftrightarrow C succ 0 X C A BC 1 B T succ 0 Yaksho A dodatno viznachena todi X dodatno napivviznachena todi i tilki todi koli X A ye dodatno napivviznachenoyu If displaystyle text If A 0 displaystyle A succ 0 then displaystyle text then X 0 X A C B T A 1 B 0 displaystyle X succeq 0 Leftrightarrow X A C B T A 1 B succeq 0 Yaksho C ye dodatno viznachenoyu todi X dodatno napivviznachenoyu todi i tilki todi koli X C ye dodatno napivviznachenoyu If displaystyle text If C 0 displaystyle C succ 0 then displaystyle text then X 0 X C A B C 1 B T 0 displaystyle X succeq 0 Leftrightarrow X C A BC 1 B T succeq 0 Pershe i tretye tverdzhennya mozhna otrimati cherez rozglyad minimizatora velichini u T A u 2 v T B T u v T C v displaystyle u T Au 2v T B T u v T Cv yak funkciyi vid v dlya fiksovanogo u Dali oskilki A B B T C 0 C B T B A 0 displaystyle left begin matrix A amp B B T amp C end matrix right succ 0 Longleftrightarrow left begin matrix C amp B T B amp A end matrix right succ 0 i podibno dlya dodatno napivviznachenih matric druge chetverte tverdzhennya negajno viplivaye z pershogo vidpovidno tretogo tverdzhennya Takozh isnuye neobhidna i dostatnya umova na dodatnyu napivviznachennist X v terminah uzagalnenogo dopovnennya Shura A same X 0 A 0 C B T A g B 0 I A A g B 0 displaystyle X succeq 0 Leftrightarrow A succeq 0 C B T A g B succeq 0 I AA g B 0 i X 0 C 0 A B C g B T 0 I C C g B T 0 displaystyle X succeq 0 Leftrightarrow C succeq 0 A BC g B T succeq 0 I CC g B T 0 de A g displaystyle A g poznachaye dlya A displaystyle A Div takozhMatrichna totozhnist Vudburi Lema obernennya matriciPrimitkiZhang Fuzhen 2005 The Schur Complement and Its Applications Springer doi 10 1007 b105056 ISBN 0 387 24271 6 Schur Complement Lemma 10 serpnya 2016 u Wayback Machine na berkeley edu Boyd S and Vandenberghe L 2004 Convex Optimization Cambridge University Press Appendix A 5 5
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