Projective equivalence between conics Consider the two projective conics in $\mathbb{P}^2_{\mathbb{R}}$:


*

*$C: x_0^2+x_1^2-x_2^2=0$

*$C': x_0^2-x_1^2-x_2^2=0$
Why are they projectively equivalent? Clearly the two associated matrices $A=\begin{pmatrix}
1 &0 & 0\\ 
 0&  1& 0\\ 
 0&  0& -1
\end{pmatrix},A'=\begin{pmatrix}
1 &0 & 0\\ 
 0&  -1& 0\\ 
 0&  0& -1
\end{pmatrix}$ are not congruent as they have different signature, but then the only option is that they're congruent modulo a unity of $\mathbb{R}$, i.e. there exist an invertible $3\times 3$ real matrix $M$ and $\alpha \in \mathbb{R} \setminus \{0\}$ such that $A'=\alpha M^tAM$.
However I couldn't find such matrix and number, is there anything wrong in my reasoning, do you have any hint?
 A: Remember that you’re dealing with homogeneous equations and matrices here, i.e., equivalence classes of matrices that are nonzero scalar multiples of each other. In this framework, the signatures $++-$ and $--+$ are equivalent. Multiply both sides of the second equation by $-1$ and it becomes obvious that it’s the same as the first one, but with $x_0$ and $x_2$ swapped. Equivalently, if you multiply $A'$ by $-1$, the result is obviously a permutation of $A$, so the transformation matrix $$M=\pmatrix{0&0&1\\0&1&0\\1&0&0}$$ will do the trick.  
Nondegenerate conics can be classified by their intersections with the line at infinity. If we consider $0:0:1$ to be this line, then the original conic doesn’t intersect it at all, so is an ellipse. Hyperbolas, on the other hand, intersect the line at infinity at two distinct real points, as does the second conic. So, if we choose a transformation that maps some line that intersects the ellipse at two points to the line at infinity, we can transform the ellipse into a hyperbola. The matrix $M$ does just that: it exchanges the lines $1:0:0$ and $0:0:1$.  
I will note that here $M$ turns out to be symmetric and its own inverse, so we can be fairly cavalier about how points, lines and quadratic forms transform, but in general we need to be careful about this. Under the point transformation $\mathbf p'=M\mathbf p$, lines transform as $\mathbf l'=M^{-T}\mathbf l$ and quadratic forms as $Q'=M^{-T}QM^{-1}$.
