| [ QuizWit ] in KIDS 글 쓴 이(By): iLUSiON (화려한집념) 날 짜 (Date): 2003년 8월 8일 금요일 오후 03시 57분 13초 제 목(Title): Re: Linear Algebra 에 관련된 문제 hehe.. i am always the first to respond. call me .. i a man without fear. ^^ Let me restate your problem.. hehe... b<-b1, c<-b2. given b'c=0 a'Pb=0 a'Pc=0 what is the f* condition to give a = \alpha b + \beta c. how the f* should i know. but i am bored... can't sleep... so let me go thru algebra. man.... this smells like a highschool algebra. :( subsitituting a = \alpha b + \beta c into the above three equations and solve it will ya. note that we have 2 unknown papramters \alpha and \beta plus n parameters in P. (just assume for a moment that a, b,c, fixed). so we have n+2 parameters while there are only 3 solutions. hence... n-1 parameters will not be determined. but the first equation b'c = 0 is uselesss.. hehe so for fixed a,b,c,.... n parameters will not be determined. oh ho.. does that mean P will not be determined? ehehe.... why don't we put it into equation.. \alpha b'Pb + \beta c'Pb=0 \alpha b'Pc + \beta c'Pc =0 b'c = 0. oh ho.... note that b'Pc = c'Pb rarara.... now what...? it would be embarrasing wouldn't it if i can't solve this from here. uheehe... who cares.... note that one of \alpha \beta sould not be 0. further asssume that b c positive. i am suer you can rotate b and c such taht all the component to be positive. eheh... oh i got it has to be singular. so you need b'Pbc'Pc - b'Pcc'Pb = 0 with b'c=0. what the f* is this? ok.. sinced everything is scalar..take trace.. i think tr(AB)=tr(BA)... hope i remember it correctly eheheh... if not..well shove it into your a*. hehe... tr(b'Pbc'Pc) = tr(bc'Pcb'P) = 0. so you don't need to worry about the fist term. oh we are going somewhere. uhehe... now... the only condition is b'Pcc'Pb = 0 with b'c=0. note that b'Pcc'Pb = (b'Pc)^2 =0. so b'Pc=0 with b'c=0. fucking! the only solution is Pc=c. projection or Pb=b. ya.. If P is the projection matrix of c or b.. you got it. now you said P is diagonal.... tehrea are very limited number of diagonal projection matrix.. is't it? eheheh.... P is projection if P^2 = P. idempotent. so... man... am i going somewhere? so going to sam-chun-po? if P=diag(p1,p2,....pn) pi^2 = pi. fucking cool... pi= 0 or 1. if P consists of either 0 or 1 nontrivial diagonal matrix. you get your answer. fucking boring.... now somehow i feel didn't i seeen this somewhere? man i feel this has something to do with the cochran's theorem in the mathemtical statistics ..cochran's theorem can be proved in somewhat similar fashion... iLUSiON whitepolarcow@hotmail.com |