$\newcommand{\im}{\operatorname{im}}$ I am looking over Linear Algebra, and am stuck on a question. I want to prove that $$\im(T')=(\ker(T))^0$$ assuming that for any subspace $U$ of a vector space $W$ there exists a subspace $K$ of $W$ such that $W=U \oplus K$. However, we have that V is not necessarily finite dimensional, so we cannot use dimension arguments. I have managed (I think) to prove the inclusion one way, that is:
Considering $v \in \ker(T)$, we see that $$T'(f)(v)=(f\circ T)(v)=f(T(v))=f(0)=0$$ which implies that $$T'(f) \in (\ker(T))^0$$ which implies that $$\im(T')\subseteq(\ker(T))^0$$ However, I don't know how t prove the reverse inclusion, and where to bring in the hint. I'm assuming I want to have that one of the sides of the equality we are trying to prove is a subspace of $V'$ and then use some kind of basis argument? But, as the space is infinite dimensional, I don't know how to construct this.
Any help appreciated-thanks.