I started reading David Eisenbud’s Commutative Algebra but got stuck at a part of Exercise 1.15. One is supposed to show that in $\mathbb{P}^2(\mathbb{C})$ there exist “only three types of loci represented by quadratic equations”.
I understand intuitively that quadratic forms should have $n+1$ types of vanishing loci in $\mathbb{P}^n(\mathbb{C})$, namely the vanishing loci of $x_0^2 + \dotsb + x_i^2$ for $i = 0, \dotsc, n$. (Since every such quadratic form comes from a symmetric bilinear form on $\mathbb{C}^{n+1}$, and every non-degenerate symmetric bilinear form on $\mathbb{C}^{n+1}$ admits an orthonormal basis since $\mathbb{C}$ has characteristic $\neq 2$ and is algebraically closed).
But I don’t see how to distinguish these $n+1$ vanishing loci from each other. Eisenbud also doesn’t specify up to what kind of transformation we want to classify these vanishing loci; I asumme it’s up to the action of $\operatorname{GL}_{n+1}(\mathbb{C})$ on $\mathbb{P}^n(\mathbb{C})$.
Question: How can we show that the vanishing loci of $x_0^2 + \dotsb + x_i^2$ in $\mathbb{P}^n(\mathbb{C})$ for $i = 0, \dotsc, n$ are non-equivalent under the action of $\operatorname{GL}_{n+1}(\mathbb{C})$ on $\mathbb{P}^n(\mathbb{C})$?
My guess was to look at the corresponding vanishing loci in $\mathbb{C}^{n+1}$ and show that they are non-equivalent under the action of $\operatorname{GL}_{n+1}(\mathbb{C})$. This is easy for $n = 1$, where $x_0^2 = 0$ defines a line and $x_0^2 + x_1^2 = 0$ the union of two non-parallel lines. But already for $n = 2$ I’m not sure how the case $x_0^2 + x_1^2 + x_2^2 = 0$ looks like, and how it can be distinguished from $x_0^2 = 0$ (which is a plane) and $x_0^2 + x_1^2 = 0$ (which is the union of two non-parallel planes).
Every quadric has the equation $xAx^t=0$ where $x=(x_0,\ldots,x_n)$ and $A$ is a symmetric matrix. It's equivalent to $xBx^t=0$ iff $B=PAP^t$ for some non-singular matrix $P$. This entails $A$ and $B$ having the same matrix rank. Then $x_0^2+\cdots+x_k^2=0$ corresponds to a matrix of rank $k+1$.