Let $X$ be a set of $d=2n$ points in general position in $\mathbb{P}^n$. Then $I(X)$ is generated by homogeneous polynomials of degree $2$.
I have proved that $h_X(2)=d$ (where $h_X$ is the Hilbert function of $X$), but i don't know how to deduce the statement.
There is probably also a way with Hilbert polynomial but let me write an elementary solution (I read it from the book by Harris : Algebraic Geometry, a first course ). The proof is a bit tricky but I find it pretty nice.
Let $q \in \mathbb P^n$ such that every quadratic polynomial which vanishes on $X$ also vanishes on $q$. $(\star)$
We will show $q \in X$. Write $X = X_1 \cup X_2$ with $|X_i| = n$, by hypothesis there is exactly one hyperplane $H_i$ passing by $X_i$, given by the linear equation $f_i = 0$. So $f_1 \cdot f_2$ vanishes on $X$, i.e $q \in H_1 \cup H_2$ by $(\star)$. Let's assume $q \in H_1$.
We can take $p_1, \dots, p_k$ minimal with the property that $q \in \text{span}\{p_i\} $. By the above, $k \leq n$. Now add $ n - k + 1$ points $q_{j_i} \in X$, and take $\Sigma = \{p_2, \dots, p_k, q_{j_1}, \dots \}$: the hyperplane $\Lambda$ containing $\Sigma$ does not contains $p_1$ by the general position hypothesis. So $\Lambda$ can't contain $q$ since by hypothesis $q \in \text{span}\{p_i\} $. Using $(\star)$ again $q$ is in the other hyperplane spanned by the points in $X \backslash \Sigma$. But since there are at least $n$ choices for $\Sigma$ we deduce that $q$ is in the intersection of $n$ hyperplanes in general position which also contains $p_1$; so their intersection is $p_1$ and in particular $q = p_1$.
\\in text: here, because it does not work, in LaTeX becase it is essentially always wrong. – Mariano Suárez-Álvarez Apr 28 '17 at 19:07