Orthogonal complement of $span(M)$ Let $M$ be a non-empty subset of a Euclidean space $V$. How would I prove this:
$(span\ M)^\bot=M^\bot$
 A: $x\in {\rm span}\ M$ then $$ x= \sum_{i=1}^n a_i m_i,\ m_i\in M $$
$({\rm span}\ M)^\perp \supset M^\perp $ : Hence for $v\in M^\perp$,
then
$$ x\cdot v= \sum_{i=1}^n a_i m_i\cdot v =0 $$
That is, $$v\in ({\rm span}\ M)^\perp $$
$({\rm span}\ M)^\perp \subset M^\perp $ : Let $ v\in {\rm span}\
(M)^\perp $ so that
$$ x\cdot v= \sum_{i=1}^n a_i m_i\cdot v =0 $$
Since $x\neq 0$ then we can assume that $a_1=0$. By definition of
$v$, since $x-a_1m_1\in {\rm span }\ (M)$, then $$ (x-a_1 m_1)\cdot
v= 0
$$
This implies that $$ a_1m_1\cdot v =0 $$
Since $m_1$ is an arbitrary element in $M$, then $v\in M^\perp $. 
A: To show two sets $A, B$ are equal you need to show whenever $x \in A \implies x \in B$ and whenever $x \in B \implies x \in A$. 
Now suppose $x \in M^\perp$. Then by definition this means $\langle x, y \rangle = 0  \, \,   \forall \, \,  y \in M$.
On the other hand, if $x \in (\text{span } M)^\perp$ then similarly $x$ is orthogonal to every element in $\text{span } M$. In other words, if $M = \{v_1, v_2, v_3, \ldots\}$  then $$\langle x, a_1v_1+ a_2v_2 + a_3v_3 \ldots \rangle = 0  $$
In particular, see what happens when the coefficients are all one? Can you finish from here?
