Let C be the standard projective conic with equation xy + yz + zx = 0.
This passes through the standard ppoints X[1,0,0], Y[0,1,0], Z[0,0,1].
To simplify the algebra, we scale the equation to 2xy + 2yz + 2zx = 0.
Then the equation can be written as x^{T}Kx = 0, where K is the matrix
K = 
 
0 
1 
1 
 
 
1 
0 
1 
 
 
1 
1 
0 
 

Now suppose that t is the projective transformation t([x]) = [Ax], where
A is a nonsingular 3x3 matrix.
Here, we will determine explicitly the matrices A for which t(C) = C.
Note that t(X) = [a_{1}], where a_{1} is the first column of A.
Similarly, we have
t(Y) = [a_{2}] and t(Z) = [a_{3}], where
a_{2}, a_{3} are the remaining columns of A.
If t maps C to C, then, as X,Y,Z lie on C,
the columns of A must correspond
to three ppoints X',Y',Z' of C. Since A is nonsingular, these will be distinct.
Suppose that, if X' = [u_{1}], Y' = [a_{2}], Z' = [a_{3}].
To say that the column a_{1}
corresponds to X' means
that a_{1} = λ_{1}u_{1} for some nonzero λ_{1},
and similarly
a_{2} = λ_{2}u_{2} and
a_{3} = λ_{3}u_{3}.
Suppose that X',Y',Z' are distinct ppoints on C. The three point theorem
shows that there is a unique t mapping C to C, and (X,Y,Z) to (X',Y',Z'). The
uniqueness part implies that the multipliers are determined up to an overall
multiplier. We now proceed to show how these multipliers can be found.
From earlier work, we know that t(C) = C if and only if
A^{T}KA = κK with κ ≠ 0.
There, we showed that we can scale A to achieve κ = 1. Here, it is
as well to
postpone this scaling.
If A has the columns as above, then A^{T} has rows
λ_{1}u_{1}^{T},λ_{2}u_{2}^{T},
λ_{3}u_{3}^{T}, so that
the (i,j) entry of A^{T}KA is λ_{i}λ_{j}
(u_{i}^{T}Ku_{j}).
Provided that X',Y',Z' are on C, u_{i}^{T}Ku_{i} = 0
for i = 1,2,3. Thus the diagonal
entries of A^{T}KA are zero, so agree with those of κA for any κ.
As A^{T}KA is symmetric (since K is symmetric), we need only choose the λ_{i}
such that the elements of A^{T}KA above the diagonal agree with those of κA,
i.e. such that
(*)
λ_{1}λ_{2}(u_{1}^{T}Ku_{2}) =
λ_{1}λ_{3}
(u_{1}^{T}Ku_{3}) =
λ_{2}λ_{3}
(u_{2}^{T}Ku_{3}) = κ (≠ 0).
We must check that, for i ≠ j, u_{i}^{T}Ku_{j} ≠ 0.
By earlier work, the equation of
the polar of V = [v] is v^{T}Kx = 0. For V on C,
the polar is the tangent at V.
This meets C only in V. Thus, if [w] is another ppoint on C,
v^{T}Kw ≠ 0.
Provided that the [u_{i}] are distinct ppoints on C, we have the result.
Now we can see that the in the equations (*), we can choose any nonzero
value for λ_{1}. Then the values of λ_{2} and λ_{3}
and that of κ are determined.
Putting
λ_{1} = α.u_{2}^{T}Ku_{3}, we get
λ_{2} = α.u_{1}^{T}Ku_{3} and
λ_{3} = α.u_{1}^{T}Ku_{2},
and finally
κ = α^{2}.u_{2}^{T}Ku_{3}.
u_{2}^{T}Ku_{3}.
u_{2}^{T}Ku_{3}.
Changing α simply scales the matrix A and
so produces the same projective transformation t.
(**) Note that, as A^{T}KA = κK, det(A)^{2} = κ^{3}.
The above calculation shows how to find the matrix A from any three distinct
ppoints on C. To allow us to construct explicitly the elements of S(C,P(2)), we
need to find all triples of ppoints on C.
This is achieved by the parametrization theorem. Each ppoint on C other than
Z[0,0,1] has the form [r] = [r,1r,r(r1)], with r real. Note that, if r ≠ s, then
these produce different ppoints since we cannot have s = kr and 1s = k(1r)
unless k = 1 and r = s.
We thus have two kinds of triple, those including Z, and those not including Z.
The calculation of the u_{i}^{T}Ku_{j}
is fairly straightforward  we leave the details
as an exercise.
Suppose that z = (0,0,1), r = (r,1r,r(r1)), s = (s,1s,s(s1)).
(***) r^{T}Kz = 1 and r^{T}Ks = (rs)^{2}.
If we choose the ppoints [r], [s],[z] (in order) then we obtain a matrix A.
By (*),(**),(***), det(A)^{2} = (rs)^{6}. Again by direct calculation, we find that
(1/(rs))A has determinant 1. It has columns
(r/(rs),(1r)/(rs),r(r1)/(rs)),
(s/(rs),(1s)/(rs),s(s1)/(rs)) and (0,0,rs).
Other cases involving Z are similar  the columns are permuted, and rs may
have to be replaced by sr to get determinant +1.
If u = (u,1u,u(u1)), the ppoints [r],[s],[u] lead to a matrix A which has
det(A)^{2} = {(rs)(su)(ur)}^{6}. If we scale by 1/{(rs)(su)(ur)}, then we can
check that the scaled matrix has determinant 1.
Thus, we can write down the elements of S(C,P(2)) explicitly, choosing in each
case the matrix scaled to have determinant 1.
The terminally curious can see typical elements.

