Exercise 3.F.15

Proof. Suppose $T^{\prime }=0$. Let $w\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$. Suppose by contradiction that $w\ne 0$. By _Exercise 3_, there is a linear functional $\psi \in W^{\prime }$ such that $\psi (w)=1$. Let $v$ be a vector in $V$ such that $\mathit{Tv}=w$. We have

But $T^{\prime }=0$, we have a contradiction. Therefore $w=0$.

Conversely, suppose $T=0$. Let $v\in V$ and $\psi \in W^{\prime }$. We have

Since both $v$ and $\psi$ were arbitrary, we get that $T^{\prime }$ must be zero. □