Show that $(S^{\perp})^{\perp} \subseteq L(S) $.

Question

Let $V$ be a finite dimensional inner product space, and $S \subseteq V$. Then, show that :

$(S^{\perp})^{\perp} \subseteq L(S) $

I actually have to show that $(S^{\perp})^{\perp} = L(S) $. I have shown the reverse inclusion, but I am not able to do the other way. Please help.

NOTE: $L(S)$ is the linear span of $S$

Answer 1

Firstly it should be easy to show that $ L(S)^\perp = S^\perp$.

Use the following theorem: Let $W$ be a subspace of a finite dimensional vsp $V$. Then $dim(W^\perp)+dim(W)=dim(V)$.

Applying theorem to $W = S^\perp$ yields: $dim(S^\perp)+dim(S^\perp)^\perp=dim(V)$.

But we can also apply to $W=L(S)$ and so: $dim(L(S))+dim(L(S))^\perp=dim(V)$.

All in all, $dim(L(S))+dim(L(S))^\perp=dim(S^\perp)+dim(S^\perp)^\perp$

Then $dim(L(S))=dim(S^\perp)^\perp$.

Then because you already have one inclusion and so $L(S)=(S^\perp){^\perp}$