| [ QuizWit ] in KIDS 글 쓴 이(By): iLUSiON (화려한집념) 날 짜 (Date): 2003년 8월 9일 토요일 오전 05시 52분 32초 제 목(Title): Re: Linear Algebra 에 관련된 문제 ok let me retry uehehe... b <- b1 , c <- b2 are orthogonal. so b'c=c'b = 0. now... b'Pbc'Pc -------- = 1. b'Pcc'Pb man this is a matrix equation. i think the problem hinges on the fact that P is diagonal. hehehe... let P=diag(p1,....pn). oh based on norm <x,y> = x'Py. <b,b><c,c> = <b,c>^2. f*... didn't i seen this identity somewhere? too familar....swartz inequality is something like <b,c>^2 <= <b,b><c,c> oh f*..joyous moment. i solved !!!!! swartz inequality becomes equal only if b is portional to c. in this case.... b= \gamma Pc or c = \gamma Pb. now you said b, c orthogonal. there is no f* way to make c into b or b into c by multiplying diagonal matrix. so the inequality is always strict. so the matrix is non singular. that is M (\alpha \beta )' = 0 wehre M nonsingular. so we can get M^{-1}. hence all of \alpha \beta should be 0. fucking!..... MY conclusion: a will not belong to span(b1,b2) for any diagonal matrix P. but i think i mised soemthing here. i forgot to prove that <x,y>=xPy is a norm. it is a norm only if P is positive definite. f*... P has to be a digonal matrix with positive elements. yam yam..so happy. yo drake... i solved!!!! YOU OWE me 5 bottles of soju. ehhee... iLUSiON whitepolarcow@hotmail.com |