# Why is the annilihator of the zero linear functional the vector space itself?

Let V be a vector space of a finite dimension. $$T:V \rightarrow V$$ is a linear transformation. I have to prove that T* is injective iff T is injective.

I know T* is injective iff $$kerT^* = 0$$, and $$kerT^* = (ImT)^0$$. Therefore iff $$(ImT)^0 = 0$$, wherethe zero is the zero linear functional $$\lambda = 0_V*$$ ( $$\forall v \in V: \lambda(v) = 0)$$.

The solution I have uses annihilator again, because V is of a finite dimension, $$((ImT)^0)^0 = ImT$$, and argues that $$(0_V*)^0 = V$$ with no explanation. I looked up the definitions and I cant understand why. Once I have this is easy to show that $$KerT = {0}$$ from the dimension theorem.

Thanks!

If $$S\subset V^*$$, then$$S^o=\{v\in V\,|\,(\forall\alpha\in S):\alpha(v)=0\}.$$So,\begin{align}\{0_{V^*}\}^o&=\{v\in V\,|\,0_{V^*}(v)=0\}\\&=V,\end{align}since the equaliy $$o_{V^*}(v)=0$$ holds for every element of $$V$$.