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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 ATMA = M, det(A) = 1.
Proof
Suppose that A is a matrix such that, for [x] on C, [Ax] is also on C.
Take x on C.
Then (Ax)TM(Ax) = 0, so (x)TATMAx,
so x lies on the conic defined by ATMA.
But x is any point of C, so M and ATMA define the same conic, C.
By the Lemma,
ATMA = kM for some non-zero k.
Taking determinants, since det(AT) = det(A), (det(A))2det(M) = k3det(M).
As M is non-singular, (det(A))2 = k3, so k > 0. We can replace A by (1/λ)A,
where λ is chosen so that λ2 = k3 and λ, det(A) have the same sign.
Clearly, the matrix (1/λ)A has determinant 1, and t([x]) = [(1/λ)Ax]
maps C to itself.
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