※ 引述《mqazz1 (無法顯示)》之銘言：
: let D be the differentiation operator on P[3], and let
: S = {p屬於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
: 請問這題該怎麼證呢?
: 謝謝!!
(a).
因為向量空間P3書上有兩種定義,在這邊使用定義P3=span{x^2,x,1}
令f(x)=ax^2 + bx + c,f屬於P3
range(D)=D(f(x))= 2ax + b = span{x,1} = P2 --(1)
當f屬於N{D}，D(f(x))= 2ax+b = 0，a=b=0,c任意實數時成立,得N{D}=span{1}≠{0}--(2)
由(1)、(2)敘述(a)得證。
(b).
令g(x)=ax^2+bx+c,因為g(0)=c=0,得當g(x)=ax^2+bx,g(x)屬於S
range(S)=D(g(x))=2ax+b=span{x,1}=P2≠P3 ==> D is not onto 得證
且當g(x)屬於N{D},D(g(x))=2ax+b=0，a=b=0時成立，得N{D}={0}==>D is one-to-one
如果有錯，麻煩提醒，謝謝
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