$T(p(x)) = p'(x)$ is an isomorphism

Question

I proved that $T$ is Injective, how can I prove that $T$ is Surjective? Thought about using induction

Answer 1

Consider any polynomial $p(x)$ in $\mathbb R[x]$, and we need to find another polynomial $q(x)$ in $V$ such that $q'(x) = p(x)$, but this is easy, just consider $$q(x) = \int_0^x p(t)dt \in V.$$

Answer 2

How can I prove that T is Surjective?

Take $p(x)=a_0+a_1x+\cdots+a_nx^{n}\in \mathbb R[x]$.

You have to show that there exists $q(x)\in V$ such that $T(q(x))=p(x)$, that is, $$q'(x)=a_0+a_1x+\cdots+a_nx^n.$$ From the basic rules of differentiation, a polynomial $q(x)\in V$ satisfying this equality does exist and is given by $$q(x)=a_0x+\tfrac{1}{2}a_1x^2+\cdots+\tfrac{1}{n+1}a_nx^{n+1}.$$