# Definition of annihilator is not clear

I have a question regarding the following definition of an annihilator of a finite dimensional vector space. I think I understand the two definitions but I don't really get the link implied by the last sentence.

Definition:

If $A \subset V$, the annihilator of $A, A^{°}$, is the set of all $f$ in $V^*$ such that $f(a) = 0$ for all $a$ in $A$. Similarly, if $A \subset V^*$, then

$$A^{°} = \{a\in V\colon\ f(a) = 0 \text{ for all } f\in A \}.$$

If we view $V$ as $(V^{*})^{*}$, the second definition is included in the first.

So now my questions are why may we view $V$ as $(V^{*})^{*}$, I know they're equivalent by the isomorphism but this doesn't mean that they're equal, does it? And how does the first definition include the second one?

They are not equal, but algebraically isomorphic, because a finite dimensional vector space is reflexive and the second definition is included in the first if You identify every $a\in V$ with it´s image $Ja$ under the canonical injection (in this case isomorphism) $J:V\rightarrow (V^*)^*,a\mapsto(Ja:V^*\rightarrow\mathbb{C},f\mapsto f(a))$,because the first definition then reads for $V^*$:
$A^0=\{Ja\in V^{**}:Ja(f)=f(a)=0\}$
• So it is because of this one to one correspondence (bijection) that you can identify with every element of $V^{**}$ an element of $V$ to reconstruct out of the set of the first definition with $A \subset V^*$ the set in the second definition under the canonical injection? Commented Sep 18, 2015 at 12:41