It has already been given that:
$T\colon P_3\to P_4 \qquad T(p)=(x+2)p(x)$
$H\colon P_4\to P_3 \qquad H(p)=p'(x)+p'(0)$
It is asked to show that $T$ is one-to-one but not onto, and that $H$ is onto but not one-to-one. How can I show that?
It has already been given that:
$T\colon P_3\to P_4 \qquad T(p)=(x+2)p(x)$
$H\colon P_4\to P_3 \qquad H(p)=p'(x)+p'(0)$
It is asked to show that $T$ is one-to-one but not onto, and that $H$ is onto but not one-to-one. How can I show that?
Just use the definition to check that $T$ is one-to-one. To show that $T$ is not onto, note that for any $0\neq p(x)\in P_3$, $T(p(x))=(x+2)p(x)$ which is a polynomial having degree at least one. Therefore, for any degree zero polynomial $q(x)$, namely the constant, there does not exists any $p(x)\in P_3$ such that $T(p(x))=q(x)$.
Use the definition to check that $H$ is onto. To do it, you need to solve some linear equations. To show that $H$ is not one-to-one, just note that for any degree zero polynomials $p(x)$, we have $H(p(x))=p'(x)+p'(0)=0$.