Define a quadratic curve by $\{(x,y) \in \mathbb{R}^2 \; | \; Ax^2+Bxy+Cy^2+Dx+Ey+F = 0\}$. I want to show that given $5$ points in the plane there exists a quadratic curve that passes through each of them. The goal is to use only linear algebra and I want to be sure this approach doesn't miss anything.
Assume we have a collection of five points in the plane say $\{(x_i,y_i)\}_{1 \leq i \leq 5}$. Let $$V = \{Ax^2+Bxy+Cy^2+Dx+Ey+F \; | \; A,...,F \in \mathbb{R} \}$$ so it's set of all polynomials which define quadratic curves. If we give this set the same pointwise operations that we do for $\mathbb{F}[x]$ we have a vector space, and moreover $\dim(V) = 6$. The reason it has 6 dimensions is because it's basically just $\mathbb{F}[x] \cup \mathbb{F}[y]$ but we don't count the constant term twice and we have to add the mixed $xy$ term.
So next define a map $T:V \rightarrow \mathbb{R}^5$ by $f \mapsto (f(x_1,y_1),f(x_2,y_2),...,f(x_5,y_5))$. We can check this transformation is linear because for $f,g \in V$ and $\alpha \in \mathbb{R}$ we have that $$ T(f+\alpha g) = ((f+\alpha g)(x_i,y_i)) = (f(x_i,y_i)+\alpha g(x_i,y_i)) = T(f)+\alpha T(g) $$ Then by the rank-nullity theorem we know that $\dim(\ker(T)) \geq \dim(V) - \dim(W)$ and so here $\dim(\ker(T)) \geq 6-5 = 1$. This means that there exists some non-zero quadratic curve $f \in V$ such that $f(x_i,y_i) = 0$ for $1\leq i \leq 5$. Hence we have a quadratic curve that passes through whatever the five points were, and moreover because this is a subspace of $V$ it's technically a family of quadratic curves because we can scale them.
Are there any mistakes in the proof idea? Are there any possible cases I've neglected? How might I refine the proof to make it more concise? Thanks in advance for the help!