Let $K$ be a field and $S_2$ be the linear space of all homogeneous polynomials $F \in K[X,Y,Z]$ of degree $2$. Let $P_1,...,P_n$ be points of the projective space $\mathbb{P}^2$. Define as: $$S_2(P_1,...,P_n) = \{F \in S_d : F(P_i)=0, \forall i=1,...,n\}$$ I want to proof that, if $n \leq 5$, and any four of $P_1,...,P_n$ are not collinear, then we have $\dim_K S_2(P_1,...,P_n) = 6-n$.
My thoughts:
In the first place, for $F = \sum_{i+j+k=2} a_{ijk}X^iY^jZ^k$ and $P_i=(x_i:y_i:z_i)$ , we have that $F(P_i)=0, \forall i =1,...,n \iff (a_{200},...,a_{002}) \in Ker\,T$, being $T:K^6 \to K^n$ the linear transformation given by the matrix:
$$A=\begin{bmatrix}x_1^2 & y_1^2 & z_1^2 & x_1y_1 & y_1z_1 & x_1z_1\\ x_2^2 & y_2^2 & z_2^2 & x_2y_2 & y_2z_2 & x_2z_2 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ x_n^2 & y_n^2 & z_n^2 & x_ny_n & y_nz_n & x_nz_n \end{bmatrix}.$$
We have $\dim_K Im\,T \leq n$ by the co-domain dimension, so we have $dim_K S_2(P_1,...,P_n) = dim_K Ker\,T \geq 6-n$. But I'm not sure how to proof that $dim_K S_2(P_1,...,P_n) \leq 6-n$. How can I guarantee that? Any leads?
NB: are you sure you don't mean "any three points are not collinear"?
Because of the condition that no three points are collinear,
there exists a projective transformation $T$ such that
I'll let you finish the proof if you want. Just to give you some hints:
The only degree-$2$ polynomials that send $T(P_1)$, $T(P_2)$ and $T(P_3)$ all to $0$ are spanned by $xy$, $yz$ and $xz$.
If you have more than three points, think about which linear functionals send $(ab,bc,ac)$ and $(de,ef,df)$ to zero, where $T(P_4)=(a,b,c)$ and $T(P_5)=(d,e,f)$.