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.



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 $V$.

  • $\begingroup$ Never saw this definition on V* but only on V. This makes perfect sense now, TIL! Thank you very much. $\endgroup$ – Tegernako Feb 5 '19 at 11:42

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