How do I derive a surjective transformation if annihilator of range T is zero

Question

In Linear Algebra Done Right 3rd ed. Axler tries to prove a transformation $T\in \mathcal{L}(V,W)$ is surjective iff the dual map $T^*$ is injective. His reasoning starts with the following equation:$$\operatorname{range}(T)=W\iff (\operatorname{range}(T))^0=\{0\}.$$ How do I derive this? Is there something special about the fact that the only transformation that takes $\operatorname{range}(T)$ to the zero vector is the zero transformation?

Answer 1

Let $\Bbbk$ be the field over which you are working. If $U$ is a subspace of $W$, then you have two possibilities:

1. $U=W$: then the only linear form $\alpha\in W^*$ such that $\alpha(W)=\{0\}$ is the null form.
2. $U\varsubsetneq W$: then, if $v\in W\setminus U$, if $U'$ is a subspace of $V$ such that $U\subset U'$ and that $U'\oplus\Bbbk v=W$, then if $\alpha\colon W\longrightarrow\Bbb k$ is defined by $\alpha(u+\lambda v)=\lambda$ ($u\in U'$ and $\lambda\in\Bbb k$), then $\alpha\in W^*$, $\alpha\neq0$, and $\alpha\in U^0$.

So, this proves that$$U=W\quad\text{if and only if}\quad U^0=\{0\}.$$