$S$ is a subspace of $V$, then does $S$ perp contain $V$ perp? My question is very simple.
If $S$ is a subspace of the vector space $V$, would that make $V^{\perp}$ contained by $S^\perp$? 
I am asked to prove this theorem, but I couldn't move a pencil :(
 A: Assume $S\subset V\subset X$, where $X$ is a vector space and $\subset$ is understood to mean vector subspace. Now, apply the definitions:
$$S^\perp=\{x\in X\colon \left<x,s\right>=0 \text{ for all } s\in S \}$$
$$V^\perp=\{x\in X\colon \left<x,v\right>=0 \text{ for all } v\in V \}$$
Hint: suppose $x\in V^\perp$. What can you say about $\left<x,s\right>$ where $s\in S$?
Remark: I have assumed $V\subset X$ since if $V$ is the universe, then $V^\perp=\{0\}$ and the claim is trivial.
A: This is more of a note, to just show that par's answer will also hold in the case of a vector space with a more general bilinear form...there isn't enough space to do this in the comments

I suppose the question is concerned with inner product spaces - this is usually assumed when no more info is given. It is good to state that explicitly though, as orthogonality could also be defined in terms of an arbitrary bilinear form in a similar way - i.e. $w$ is orthogonal to $v$ if $f(v,w)=0$ - though now it could be that $v$ is not orthogonal to $w$ in the case where $f$ is not symmetric. So now, to denote exactly with regard to which form we are stating orthogonality we denote: $v \perp_f w$. 
Following this definition we can define orthogonal complement in a similar way as for an inner product, but again we have to distinguish between a right- and left- orthogonal complement (right is more common), i.e. the right $f$-orthogonal complement of the subspace $S$ (of the vector space $\mathcal{V}$) is
$$S^{\perp_f}=\{x \in \mathcal{V} : f(s,x) = 0 \text{ for all } s \in S\}.$$
As you can see the result in the question still holds as a direct consequence of the definition, and can be proved in the same way as one would prove it for an inner product space. (What does not still hold in the general case though is that dimensions of the subspace and its complement will add up to the dimension of the vector space). 
