Question: Let D be the differentiation operator on P3, and let 5 = [p P3| p(0) = 0) Show that (a) D maps P3, onto

Let D be the differentiation operator on P3, and let
5 = [p ∈ P3| p(0) = 0)
Show that
(a) D maps P3, onto the subspace P2, but D: P3 → P2 is not one-to-one.
(b) D: S → P3 is one-to-one but not onto.

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a If p ax 2 bx c P 3 then D p 2 ax b Thus D P 3 Span1 x P 2 The operator is not onetoone for if p 1 ... View full answer

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