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 $W^{\perp\perp}=W$.
I have no idea how to solve this one. Since I am self-studying I have no way to know if I get this right, that is the reason I ask for a resolution.
Questions:
Can someone provide me a resolution?
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Just show set equality, i.e $W^{\perp \perp} \subseteq W$ and,$W \subseteq W^{\perp \perp}$ .
For example Let $x \in W$, Let $y\in W^{\perp}$, then $\langle x,y \rangle =0$, hence since $y$ arbitrary, $x\in W^{\perp \perp}$.
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Hint: Let $U\subseteq V$ be a subspace and $u_1, \dots, u_n$ be a orthogonal basis of $U$ and define
$$ P: V \rightarrow V, \ P(v)=\sum_{j=1}^n \langle v, u_j \rangle u_j.$$
Show that $\operatorname{Ker}(P)=U^{\perp}$ and $\operatorname{Im}(P)=U$. Thus, by the rank-nulity theorem we have
$$ \dim(V)=\dim(U)+\dim(U^{\perp}).$$
Using this equality for $U=W, W^{\perp}$, we get
$$ \dim(W)=\dim(W^{\perp \perp}).$$
Therefore, it suffices to check that $W\subseteq W^{\perp \perp}$. Let $w\in W$ and $u\in W^{\perp}$, then we have
$$ \langle w, v\rangle =0.$$
As $v\in W^{\perp}$ arbitrary, we get $w\in W^{\perp \perp}$. As $w\in W$ arbitrary, we conclude
$$ W\subseteq W^{\perp \perp}.$$
As they have the same dimension, they are equal.
First of all, according to the definition; $$ W^{\perp}=\{u\in V:u\cdot v=0, \forall v\in W\}\Longrightarrow W^{\perp \perp}=\{u\in V:u\cdot v=0, \forall v\in W^{\perp}\} $$ Let $w \in W$, then we have that $w \cdot v=0, \forall v\in W^{\perp} \Longrightarrow w\in W^{\perp \perp}\Longrightarrow W \subset W^{\perp \perp}$ On the other hand, we have that; $$ V=W^{\perp}\oplus W=W^{\perp \perp}\oplus W^{\perp}\Longrightarrow \dim(W)=\dim(W^{\perp \perp}) $$ Taking everything into account; $$ (\dim(W)=\dim(W^{\perp \perp})) \wedge (W \subset W^{\perp \perp}) \Longleftrightarrow W^{\perp \perp}=W $$