Every symmetric bilinear form has a orthogonal basis Task:

Assume $K$ is a field, $V$ is a vetor space over $K$ and $\langle -, - \rangle$ a symmetric bilinear form on $V$. Show that $V$ has a orthogonal basis.

Solution:
Define
$A := \{v \in V | \langle v, u \rangle = 0 \forall u \in V\}$,
and let $U$ be a direct complement such that
$V = A \oplus U.$
The restriction of the bilinear form on $U$ is not degenerated. Let $w_1, ..., w_n$ be a basis of $A$ and $v_1, ..., v_n$ be a basis of $U$. The vectors    $w_i$ can be taken as a part of the orthogonal basis since they are orthogonal to every other vector. So we just have to take care of $U$. This means that we can assume from the very beginning that we have a not-degenerated, symmetric bilinear form. Because of the Polarization identity there are also $v \in V$ with  $\langle v, v \rangle \neq 0$.
Because of this and since the bilinear form is not-degenerated, we can conclude that the orthogonal complement of $v$ has the dimension $\dim(V) - 1$. This orthogonal complement is not-degenerated too. That is why we can conclude by induction over the dimension that there is a basis $v_1, ..., v_n$ with  $\langle v_i, v_i \rangle \neq 0$.
We can orthogonalize a basis like this the following way:
We can find a orthogonal basis $v_1', ..., v_n'$ such that
$\langle v_1, ..., v_i \rangle =  \langle v_1', ..., v_n' \rangle$ for every $i$. 
We show this by induction. Let $v_1', ..., v_n'$ already be constructed. We define
$c_j := \langle v_j, v_{i+1} \rangle$
and
$v_{i+1}' = v_{i+1} - \sum_{j=1}^i \frac{c_j}{\langle v_j, v_j \rangle} v_j.$
Then we receive for $k \le i$:
$\langle v_{i+1}', v_k  \rangle = \langle v_{i+1} - \sum_{j=1}^i \frac{c_j}{\langle v_j, v_j \rangle} v_j, v_k \rangle = \langle v_{i+1}, v_k \rangle - \frac{c_k}{\langle v_k, v_k \rangle} \langle v_k, v_k \rangle = 0$
Questions:
I have tons of questions about this proof. 


*

*Why is he arguing with the Polarization identity? How this it give him the result above?

*What is this orthogonal complement he is talking about all of the sudden? And why is it relevant in this case?

*How did he come to the idea to define $c_j$ and $v_{i+1}'$ the way he did? It looks like it simply came out of nowhere. 

*How did he conclude the last two steps? Why does the sum disappear nearly completely? I guess it has something to do with the orthogonality. And why does this whole thing equal $0$?
 A: The answer to your first question is connected to the third and fourth question. Let me takle the second one first: He is taking a vector $v$ s.t. $\langle v,v\rangle\neq 0$, the orthogonal complement is now defined as:
$$V^\perp:=\{u\in V| \langle v,u\rangle=0 \}. $$
Here you can already see that it is important that $\langle v,v \rangle\neq 0$ because otherwise $v$ would be an element of $V^\perp$ and you don't want this.
The answer to your third and forth question is probably that whoever wrote this was making it more complicated than it actually needs to be. Here is a simple proof by induction on the dimension of $V$:
The base case is $\dim(V)=0$ where it is obvious. Now assume $\dim(V)=n$. Pick a vector $w$ s.t. $\langle w,w\rangle\neq 0$. You get that $V=span(w)\oplus W^\perp$. And since $\dim(W^\perp)=n-1$ you can use the induction hypothesis to get a basis $v_1,\ldots,v_{n-1}$ with $\langle v_i,v_j\rangle=0$ for $i\neq j$. Since the $v_i$ are elements in $W^\perp$ you have that $\langle v_i,w\rangle=0$ and therefore $v_1,\ldots,v_{n-1},w$ is your desired basis for $V$.
It remains the question: why is there always a $w$ with $\langle w,w\rangle\neq 0$? And this is the answer to your first question. Since your bilinearform is nondegenerate by assumption you find two vectors $\langle u,v\rangle \neq 0$. Now the polarization identity for bilinear forms states that 
$$2\langle u,v \rangle=\langle u+v,u+v \rangle-\langle u,u \rangle-\langle v,v \rangle $$
and assuming your field $K$ does not have characteristic $2$ you have that the left hand side is non zero and therefore at least one of the terms on the right hand side has to be non zero which gives you a $w$ as desired namely either $w=u$ or $w=v$ or $w=u+v$.
