[線代] one-to-one and onto
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!!
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