Past exam of the geometry course of the University of Oxford / 2008 / B / 5

For a projective frame, choose representative vectors $v_1,\dots ,v_{n+1}$ forming a basis. Express the final point by $\sum c_i{v}_i$. Every $c_i$ is nonzero, since otherwise some $n+1$ of the given representatives would be dependent. Rescale the first representatives to $c_i{v}_i$, so the last point is represented by their sum. Do the same for the target frame, with basis $w_i$. The linear map taking each rescaled source basis vector to the corresponding target vector induces the required projective transformation. For uniqueness, a transformation fixing the first $n+1$ coordinate points is represented by a diagonal matrix. Fixing their sum forces every diagonal entry to agree. It is therefore scalar, giving a unique projective transformation.

If three intersection points of the conics lay on a line, Bézout's theorem would force that line to be a component of either nonsingular conic, impossible. Thus the four points form a projective frame and can be moved to the three coordinate points and $[1:1:1]$. Vanishing at the coordinate points removes the square terms from each conic equation; vanishing at the last point gives $a+b+c=0$.

Substitution gives, by direct expansion,The substitution matrixhas determinant two, hence is invertible over $\mathbb{C}$. If the initial frame transformation is represented in the direction old coordinates equal $Q$ times new coordinates, take $P=QS$. Both quadratic forms then have no mixed terms, proving $P^{T}AP$ and $P^{T}BP$ are simultaneously diagonal. The same argument applies when an individual conic is reducible but the intersection consists of exactly four distinct points: a line containing three would be a common component, contradicting finiteness of the intersection.