I just found this definition of the projective space over a vector space: "Given a vector space V of dimension $n+1$, we will denote by $\mathbb{P}^n= \mathbb{P}(V)$ the projective space of all hyperplanes of V"
I tried to prove that this definition is equivalent to the "classic" one i.e. the one where I define $\mathbb{P}^n$ by the set of all the lines of the vector space V, but I couldn't do it. Can you please help me?
On
There is a bijection between the hyperplanes and the projective points. But defining in terms of hyperplanes is strange (as mentioned in another answer, it is the dual; where did you find this definition, by the way?).
Projective geometry is sometimes defined, not in terms of the one-dimenional spaces, but in terms of ALL the subspaces of $V$. This gives all objects of the projective space (projective points, lines, planes, $\ldots$, hyperplanes). We can then speak of the projective dimension of these objects (a little confusing because it is one less than the vector space dimension, i.e. a one-dimensional vector subspace becomes a projective point with projective dimension $0$).
Basically, a main property of projective spaces is the notion of duality, which is a method to obtain a new projective space (the dual) with a (bijective) map that reverses inclusion, that is, points are mapped to hyperplanes in the dual, and a line through two points is mapped to the intersection of the two corresponding hyperplanes. The dual space is isomorphic to the original in most circumstances (except for some examples of nonclassical projective planes).
There is a bijection between the set of lines and hyperplanes in any vector space, but it is not canonical (it depends on a choice).
Namely, choose a non-degenerate bilinear form (this you can do for instance by choosing a basis and taking the standard non-degenerate bilinear form with respect to it). Then the orthogonal complement of a line is a hyperplane and this gives a bijective correspondence.