I came arcross the following problem while studying projective geometry:
Let $f \in \mathbb{R}[x, y, z]$ a homogeneous polynomial of degree 2 and let $V(f) = \{(x: y: z) \in \mathbb{R}\mathbb{P}^2 | f(x, y, z) = 0\}$ be the curve defined by it. Show that a sufficient $\mathbb{R}\mathbb{P}^2 \rightarrow \mathbb{R}\mathbb{P}^2$ projective transformation maps $V(f)$ into one of the following sets:
- empty set
- $(0:0:1)$ point
- $x = 0$ line
- $xy = 0$ pair of lines
- $x^2 + y^2 - z^2 = 0$ circle
I could come up with specific examples of polynomials and projective transformations that map the curve to for example the $x = 0$ line but I was not able to prove it in general.
Any help is appreciated.
Start by writing your quadric polynomial as
$$f=p^T\cdot A\cdot p$$
for some symmetric matrix $A\in\mathbb R^3$ with $p=(x,y,z)^T$ the column vector of a generic point. This gives you a connection to the matrices used to express projective transformations. Applying a projective transformation $M$ to your quadric is the same as conjugating the matrix with the inverse transformation, i.e. the transformed quadric will have matrix $\left(M^{-1}\right)^T\cdot A\cdot M^{-1}$. That's because $p$ lies in $A$ iff $M\cdot p$ lies on this transformed version of $A$.
Observe that any real symmetric matrix is diagonalizable. So you can use a projective transformation to make that matrix diagonal, which is an important first simplification. Additional projective transformations can help you reorder and rescale the diagonal entries, but won't be able to change the signs of them. So in the end you're left with case distinctions based on the signs of the entries of a diagonal matrix, or if you prefer the signs of the eigenvalues of the original matrix.
The above is based on chapter 9 of Perspectives on Projective Geometry by J. Richter-Gebert.