作者a81288653 (Bow)
看板Math
标题Re: [线代] one-to-one and onto
时间Mon May 16 22:44:34 2011
※ 引述《mqazz1 (无法显示)》之铭言：
: let D be the differentiation operator on P[3], and let
: S = {p属於P[3] | p(0)=0}
: show that
: (a) D maps P[3] onto the subspace P[2], but
: D: P[3] -> P[2] is not one-to-one
: (b) D: S -> P[3] is one-to-one but not onto
: 请问这题该怎麽证呢?
: 谢谢!!
(a).
因为向量空间P3书上有两种定义,在这边使用定义P3=span{x^2,x,1}
令f(x)=ax^2 + bx + c,f属於P3
range(D)=D(f(x))= 2ax + b = span{x,1} = P2 --(1)
当f属於N{D}，D(f(x))= 2ax+b = 0，a=b=0,c任意实数时成立,得N{D}=span{1}≠{0}--(2)
由(1)、(2)叙述(a)得证。
(b).
令g(x)=ax^2+bx+c,因为g(0)=c=0,得当g(x)=ax^2+bx,g(x)属於S
range(S)=D(g(x))=2ax+b=span{x,1}=P2≠P3 ==> D is not onto 得证
且当g(x)属於N{D},D(g(x))=2ax+b=0，a=b=0时成立，得N{D}={0}==>D is one-to-one
如果有错，麻烦提醒，谢谢
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