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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!

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  • $\begingroup$ "Let $W$ be a subspace" (of $V$ or $V^*$)? $\endgroup$ – Jack Jul 10 '17 at 16:57
  • $\begingroup$ I dont have an answer to 2), but surely this is explained in the book you are reading. I can ask what book is? This will be essential to ask these questions properly. In 1) I assume that $W$ is a subspace of $V$ in first place. After it make the same question supposing that $W$ is a subpsace of $V^*$. $\endgroup$ – Masacroso Jul 10 '17 at 16:58
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    $\begingroup$ @Masacroso Serge Lang Linear Algebra $\endgroup$ – Pedro Gomes Jul 10 '17 at 17:32
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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.

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  • $\begingroup$ Why is $W^{00}$ a subspace of $V^{**}$? $\endgroup$ – Pedro Gomes Jul 10 '17 at 17:43
  • $\begingroup$ Write the definition $W^{00} =\{\cdots\} $, it will become clear. I want you to have the "aha" moment :P $\endgroup$ – Ivo Terek Jul 10 '17 at 17:45
  • $\begingroup$ $W^{00}=\{f\in V^{**}:f(v)\neq 0\}$? $\endgroup$ – Pedro Gomes Jul 10 '17 at 17:55
  • $\begingroup$ No. Where does $W^0$ come in there? Think a bit. $\endgroup$ – Ivo Terek Jul 10 '17 at 17:57
  • $\begingroup$ Sorry but I am not getting it. $\endgroup$ – Pedro Gomes Jul 10 '17 at 18:00

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