This is exercise 3.5.5 in Gathman's notes. We proved so far that zero set of degree $2$ homogeneous polynomials in variables $x_0,x_1,x_2$ can be identified as $\mathbb{P}^5$, and conics can be identified as an open subset $U\subset\mathbb{P}^5$. Also, given a point $P\in\mathbb{P}^2$, and let $F: \mathbb{P}^2\mapsto\mathbb{P}^5$ be the Veronese embedding. Then a conic $C\in U$ passing through $P$ if and only if $F(P)\cdot C = 0$. The last part requires to prove that given $P_1,\ldots, P_5\in\mathbb{P}^2$ such that no three of them is on the same line, then there is a unique conic passing through the $5$ points.
What I prove so far is that the matrix $$\begin{bmatrix}F(P_1)\\\vdots\\F(P_5)\end{bmatrix}$$ is a $5\times 6$ matrix, and hence the null set have dimension at least $1$ which corresponds to vanishing of degree $2$ homogeneous polynomials passing though $P_1,\ldots, P_5$. I also showed that if no three of $P_1,\ldots, P_5$ lies on the same line, then that degree $2$ homogeneous polynomial must be a conic (irreducible).
However, to prove the uniqueness, I need to show that $F(P_i)$ are linearly independent, and I got stuck on this. I read answers from another post, and it suggests that linear relation on $F(P_1),\ldots, F(P_5)$ can be pulled back to linear relation on $P_1,\ldots, P_5$. I don't see why this is true.
Best Answer
You will need the condition that $P_i$'s are in general position, which means an open subset (in the Zariski sense) of all the possible configurations $\{(P_1,\ldots,P_5)\}$.
Precisely, you need to show the condition that $\{F(P_i)\}$ linearly independent is an open condition. Indeed, by basic linear algebra, the dependency of $\{F(P_i)\}$ is equivalent to that the determinants $D_i=D_i(P_1, \ldots, P_5)$ $(i=1,\ldots,6)$ of all the $5\times 5$ submatrices of $\begin{bmatrix}F(P_1)\\\vdots\\F(P_5)\end{bmatrix}$ are zero, which is a closed condition.