Prove $W= (W^{0})^{0} $ (where $(W^{0})^{0} $ is annihilator of $W^0$) This question was asked in my Linear Algebra quiz yesterday and I could not solve it.

Edit : V is a finite dimensional vector space and W is a subspace of V.


Question: Prove that $W= {(W^{0})^0} $

Definition of $W^{0} = \{ f \text{ belonging to } V^{*} \mid f(\alpha) =0 , \ \alpha \text{ belongs to } W\}$.
But the problem occurs in the definition of annihilator of $W^{0}$.

How should annihilator of $W^{0} $ be defined?


How should I prove the asked result?


There is another subsequent question which I am asking here as it's related: Why $W$ is always a subspace of $W^{00}$?

Kindly guide.
 A: I believe the question is missing the assumptions $W$ is a subspace and $V$ is finite-dimensional, hence I am going to assume them.
Note that by definition $W^{00}:=(W^{0})^{0}=\{\phi \in V^{**}: \phi(f)=0 \text{ for all } f \in W^0\}.$
We have $$\begin{align*}\dim W+\dim W^0&=\dim V\end{align*}\\
\dim W^0+\dim W^{00}=\dim V^*.$$ Using $\dim V=\dim V^*,$ we obtain $$\dim W=\dim W^{00}$$ but $W$ is a subspace of $W^{00}$ and so $W=W^{00}.$
Edit 1: To see that $W$ is a subspace of $W^{00},$ show that $W \subseteq W^{00}$ by identifying $V^{**}$ with $V.$
Edit 2: In order to show $W$ is a subspace of $W^{00}$, one needs to identify $V^{**}$ with $V$ as follows:
Let $x\in V$ and define $L_x:V^* \to F$ (field) by $L_x(f)=f(x)$ for all $f \in V^*$. Then one can show that $x \mapsto L_x$ is an isomorphism from $V$ onto $V^{**}$ (Theorem 3.6.17 in Hoffman and Kunze, 2nd edition). Therefore for any $x \in V$, the element $L_x$ is its representative in $V^{**}$. Thus to show $W \subseteq W^{00}$, it suffices to show that for any $x \in W$, one has $L_x \in W^{00}$. To this end, let $x \in W$. Then for every $f \in W^0$, we have $L_x(f)=f(x)=0$. This shows that $L_x \in W^{00}$. It follows that $W \subseteq W^{00}$. Moreoever $W$ is a vector space itself and hence $W$ is a subspace of $W^{00}$.
A: I am reading Hoffman and Kunze these days. This confusion got me. Thus, I come to StackExchange for help as usual.
Technically, $(W^0)^0$ is a subspace of $V^{**}$, which is a vector space of linear functionals. Since $V$ is a general vector space, how the heck can a general subspace equal to a subspace of linear functionals? Hoffman and Kunze (2nd Edition) give an explanation on page 108 (the paragraph above Theorem 18).
I think, it is all about definition. Since $V$ and $V^{**}$ are isomorphic, the authors implicitly define $(W^0)^0$ as a subspace of $V$, which is $\{\alpha\in V: f(\alpha)=0\;\forall f\in W^0\}$. If you got this definition, you got $W\subseteq (W^0)^0$ right away.
