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

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?

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$$U=W\quad\text{if and only if}\quad U^0=\{0\}.$$