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Find bases for the image and the kernel of the linear map $f:P_2(x) \to P_1(x)$ given by $f(p(x)) = p'(x).$ Based on your results, indicate whether $f$ is injective or surjective.

This is my answer.

Ker f = p(x) subset P2 given p(x)=0 = p(x) subset P2 given d/dx p(x)=0 = p0 subset R =R

Img f = q(x) subset P1 given q(x)=d/dx p(x), p(x) subset of P = P1

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    $\begingroup$ ...please...? Some self work, some ideas, some effort...? $\endgroup$ – DonAntonio Aug 18 '13 at 12:03
  • $\begingroup$ What are P1 and P2? $\endgroup$ – celtschk Aug 18 '13 at 12:04
  • $\begingroup$ I think $P_2(x)$ means vector spaces of all polynomials of degree atmost $2$ over reals $\endgroup$ – Marso Aug 18 '13 at 12:08
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$\ker f=\{p(x): p'(x)=0\}$ so all constant polynomial. So your $f$ is not Injective.

take any $q(x)\in P_1(x)$ if $f$ is surjective Then there exist $p(x)\in P_2(x)$ such that $f(p(x)=q(x)$ so now can you tell me $f$ is surjective or not?

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  • $\begingroup$ f is surjective. Img f = W $\endgroup$ – ben Aug 19 '13 at 5:11

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