Inner products on the Dual Space exercise 
Let $V$ be a finite dimensional vector space over the field $K$, with a non-degenerate scalar product. Let $W$ be a subspace. Show that $W^{\perp\perp}=W$.
  Show that the same conclusion as in the preceding exercise is valid if by $W^{\perp}$ we mean the orthogonal complement of $W$ in the dual space $V^*$.

I am self-studying, and questions and doubts seem to be exponentially increasing while I am going through the functionals.
Questions:
1) Does $W$ lie in the dual space?
2) How can I deal with ortoghonality in the dual space? What about its inner product?
Thanks in advance!
 A: You have a perfect pairing $$V^\ast \times V \ni (f,v)\mapsto f(v) \in \Bbb K$$that works like a inner product. In this light, the annihilator of a subspace is precisely the orthogonal complement.
Bear the relevant definitions in mind: $W^\perp = \{x \in V \mid \langle x,y\rangle = 0,\, \forall\,y \in W  \}$ is a subspace of $V$, while ${\rm Ann}(W) \equiv W^0 = \{ f \in V^\ast \mid f\big|_W = 0 \}$ is a subspace of $V^\ast$. They live in different worlds.
With this in mind, $W^{\perp\perp}$ is again a subspace of $V$, and so it makes sense to wonder if $W^{\perp\perp}$ is equal to $W$. On the other hand, the double annihilator $W^{00}$ is a subspace of $V^{\ast\ast}$, so you must understand the abuse of notation: what the exercise wants you to prove is that $J[W] = W^{00}$, where $J$ is the natural map from $V$ to $V^{\ast \ast}$ (which you should know what it is, since you got this far). 
Alternatively, if $X \subseteq V^\ast$ is a subspace, you can think of $X_0 = \{ v \in V \mid f(v) = 0,\,\forall\,f \in X \}$. With this notion, you can prove that $W^0_{\,\,\,0} = W$, your pick.
