Re: [線代] 一題bilinear form
※ 引述《jimlucky (......)》之銘言:
: Let V be a vector space with a bilinear form B,
: assume that B(x,y)=0 => B(y,x)=0
: prove that B(x,y)=B(y,x) for all x, y belong V
: or B(x,x)=0 for all x belong V.
Lemma:
xy + ax + by + c = 0 => xy + dx + ey + f=0 for all x, y in R
則 a=d,b=e,c=f
proof:
設 D=d-a, E=e-b, F=c-f
即證若 xy+ax+by+c=0 => Dx+ Ey + F=0 for all x,y in R 則 D=E=F=0
平移 x' = x+a, y=y'+b c'=ab-c, F'= bD + eE - F
即證 x'y'= c' for all x',y' in R => Dx' + Ey' = F' 則D=E=F'=0
若c'=0, 則(1,0) ,(0,0), (0,1)三點不共線且滿足x'y'=c,故D=E=F'=0
若c'=/=0, 則
(1,c'), (-1,-c'),(2,c'/2)三點不共線且滿足x'y'=c,故D=E=F'=0
回到原題:
若有B(a,a)=/=0
對任意b,c 考慮
f(x,y) = B(xa+b, ya+c) = xy B(a,a) + xB(a,c) + yB(b,a) + B(b,c)
g(x,y) = B(ya+c, xa+b) = xy B(a,a) + xB(c,a) + yB(a,b) + B(c,b)
由題目知 f(x,y)=0 => g(x,y)=0
同除B(a,a)之後利用Lemma, 得f,g之對應係數相等
特別的,B(b,c)=B(c,b)
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11/01 14:43, , 1F
11/01 14:43, 1F
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