A projective conic C has a matrix equation x^{T}Mx = 0,
where M is a symmetric 3x3 matrix.
The standard conic C_{0} : x^{2}+y^{2}z^{2} = 0 has matrix
equation x^{T}Dx = 0, where D = diag(1,1,1).
We know that any conic C is projective congruent to C_{0}, so the
symmetry group of C is
conjugate to that of C_{0}.
The Three Point Theorem
If P,Q and R are ppoints on the projective conic C, then there is a unique
projective transformation which maps C to C_{0} and maps the points P,Q,R
to X,Y,Z respectively.
We can apply this to any three ppoints on C_{0} to obtain a corresponding element
of the projective symmetry group of C_{0}. This shows that the group is very large.
It also allows us to describe the elements to some extent. Given three ppoints
P,Q,R on C_{0}, then the inverse p of the map guaranteed by the theorem maps
X to P, Y to Q and Z to R. If we think of the matrix description of t, we see that
the matrix has columns consisting of homogeneous coordinates for P, Q and R.
The theorem states that the map is uniquely defined by the ppoints, the actual
multiples involved are constrained  we may use any verion of P, but then the
multiples of Q and R are determined.
Our main result needs the following fact which is not entirely obvious.
Lemma
If the symmetric matrices M and N define the same conic C, then
N = kM for some nonzero k.
proof
The Projective Symmetry Theorem
If the projective conic C has matrix M, then the projective symmetry group of C,
S(C,P(2)), consists of the maps t([x]) = [Ax], with A^{T}MA = M, det(A) = 1.
Each such A determines a different element of the group.
proof

