Hyperplane - explanation of theorem needed For any subset $W$ of $\mathbb{R^n}$ the following are equivalent:


*

*W is a hyperplane in $\mathbb{R^n}$;

*There is a non-zero $a$ in $\mathbb{R^n}$ such that $W = \{x ∈ \mathbb{R^n} : x\cdot a = 0\}$; 

*There are scalars $a_1, . . . , a_n$, not all zero, such that
$W = \{(x_1, . . . , x_n) ∈ \mathbb{R^n} : x_1a_1 + ··· + x_na_n = 0$


I understood the proof of the theorem, but I don't really understand the theorem itself. Why would we need second and third points of it to have separately? Even the proof states, that that they are equivalent, so why do we bother? 
Also, could somebody explain the hyperplane itself according to this theorem? I mean, the hyperplane is a subspace of dimension 1 lower than the space it is in. So why do we show that $W = \{x ∈ \mathbb{R^n} : x\cdot a = 0\}$?
I need just a small 'click' because I'm almost there! Thanks :)
 A: I agree with you. Statements (2) and (3) are identical. I don't know how much you know at this stage, but if $W$ has dimension $n-1$, then its orthogonal complement has dimension $1$, and $a$ is a vector that spans it.
A: A partial answer arrives from your question. Point 1) and 2) in the theorem are non trivial because, as you said, an hyperplane is "a subspace of dimension 1 lower than the space it is in". Think about the line $y=x$ in the plane. It is a subspace of codimension 1...can you prove that there exists an $a$ in $\mathbb R^{2}$ orthogonal to it? (it is clear from the geometry of the problem) Well, this is a non trivial characterization. Points 2) and 3) are the same.
A: Every subspace contain $\{0\}$ so all hyperplanes pass through $\{0\}$. Now, just like in euclidean geometry a plane (containing the origin) is uniquely determined by a normal vector, a hyperplane is uniquely determined by a normal vector $a$ (it is the set of points $x$ verifying $x.a=0$ where $.$ is a scalar product). (3) is the standard scalar product on $\Bbb{R}^n$; (2) and (3) are equivalent only if the scalar product used is the standard one: $a.b=\sum\limits_{i=1}^na_ib_i$.
The plane in euclidean geometry (3 dimensions) has 2 dimensions (=3-1); in general, a hyperplane has $(n-1)$ dimensions.
A: Like any logical deduction in maths, it is necessary to prove that two (maybe seemingly obvious to you) statements are equivalent if you want to be able to swap between the two statements. Having several characterisations of the same property of an object can be useful, especially when one is trying to make future proofs, which rely on the equivalence, shorter and more readable.
For example, suppose I asked you to prove that if $n\geq 2$, then there are an infinite number of hyperplanes in $\mathbb{R}^n$. It might not be totally clear from the definition because this just says that a hyperplane has codimension $1$ in $\mathbb{R}^n$. But now that you have an equivalent characterisation of a hyperplane (It is a subset $W$ of $\mathbb{R}^n$ such that $W=\{x\in\mathbb{R}^n:x\cdot a=0\}$ for some vector $a$) it should be immediately clear to you how to tackle this problem. As hyperplanes can be classified by vectors which are orthogonal to them, we can show there are an infinite number of them by instead showing that there are an infinite number of unit-vectors which each correspond to a different hyperplane (a little more work is needed to show that two vectors which are linearly independant define different hyperplanes).
A: Definition: If $\,V\,$ is a vector space then $\,H\le V\,$ is a hyperplane if
$$\forall\,v\in V\setminus H\;,\;\;\langle\;H\,,\,v\;\rangle = V$$
Claim: $\,H\le V\;$ is a hyperplane iff there exists $\,0\neq \phi\in V^*\;\;s.t.\;\;H=\ker\phi\;$
Another Claim: If $\,\dim V<\infty\;$ , then $\,H\le V\;$ is a hyperplane iff $\,\dim H=\dim V-1\;$
So now the equivalence $\,(1)\iff (2)\;$ follows from the fact that
$$\forall\,0\neq\phi\in\left(\Bbb R^n\right)^*\;,\;\;n=\dim\text{Im}(\phi)+\dim\ker(\phi)=1+(n-1)$$
and from the fact that for any 
$$\,\phi\in\left(\Bbb R^n\right)^*\;\exists\;a_1,...,a_n\in\Bbb R\;\;s.t.\;\;\phi(x_1,...,x_n)=\sum_{k=1}^na_kx_k$$ 
Point (3) is exactly the same as (2).
