I have the following Problem and I am not quite sure if I solve it right, could someone help me?
Investigate if the following map is injective or surjective and prove of linearity .
$l_1$ : $P_2$ ($\mathbb{R}$ ) -> $P_2$ ($\mathbb{R}$ ) ,$l_1$($p$)=:=($p') ^2$ , where $p'$ is the derivative of $p$
I know that in order to prove linearity I have to look if the map fullfilts
$$(p + q)(x) = p(x) + q(x)$$ $$(\alpha p)(x) = \alpha(p(x))$$
So if we say $$l_1(p)=l_2(p) =x$$ then $$(l_1 +l_2)(p) =(p_1')^2 +( p_2')^2=1 +1 = 2 $$ so it is not linear.
and in order to check injectivity we have :
$$f(x)=f(y), x=y$$ $$ l_1(p)= l_1(q)$$ $$(p_1')^2=(q_1')^2$$ and if we take square root then $p_1'=q_1'$, so the map is injective
First of all consider $p=x^{2}$ and $q=-x^{2}$, then $l_{1}(p)=4x^{2}=l_{1}(q)$ so $l_{1}$ is not injective. Also note that there is no polynomial $p$ such that $l_{1}(p)=x(x-1)$, so $l_{1}$ is not surjective. Finally note that $$l_{1}(p+q)=(p')^{2}+(q')^{2}+2p'q'=l_{1}(p)+l_{1}(q)+2p'q'$$ so $l_{1}$ is not linear.