I would like to know how to prove that if if the scalar product is non degenerate then the metric matrix is invertible and symmetric.
It is probably trivial, probably not but I really don't know how to prove it.
They define a non degenerate scalar product, a scalar product verifying :
$\forall w ~ \langle v |w\rangle =0 \Rightarrow v=0 $
The fact that the matrix is symmetric follows by the symmetry of the scalar product; e.g. consider some basis vectors $\{e_i\}$, then
$$ M_{ij} = \langle e_i | e_j \rangle $$
such that $M_{ij} = M_{ji}$.
The fact that it's invertible follows since $M_{ij}$ has trivial kernel, which is immediate from your definition of nondegeneracy. Allow there to exist some $v = \sum_i v^i e_i \ne 0$ such that $v\in \ker M$, that is
$$ M_{ij}v^j = 0, \quad\forall i $$
but, for any $w = \sum_j w^je_j$ we have $M_{ij}v^jw^i = 0 = \langle w|v\rangle = \langle v|w\rangle$, yet $v\ne 0$ by assumption, which immediately implies the above since $w$ was arbitrary. By contrapositive, then, $M$ has a trivial kernel and is therefore invertible.
