Simultaneous diagonalization of two transverse conics ($P^{T}AP,P^{T}BP\text{ diagonal}$)

Two complex plane conics meeting in four distinct points can be diagonalized by the same invertible change of homogeneous coordinates. Their intersection points form a projective frame: three on a line would force that line to be a common component by Bézout theorem, contradicting a finite intersection. Move the four points to the coordinate points and $[1:1:1]$. Each equation becomes $ayz+bzx+cxy=0$ with $a+b+c=0$. Substituting $x=v+w,y=w+u,z=u+v$ gives $a{u}^2+b{v}^2+c{w}^2$; the substitution matrix has determinant two. Composing this with the frame change yields the displayed simultaneous congruence of the symmetric coefficient matrices. This is congruence of quadratic forms, not a claim of simultaneous similarity of arbitrary matrices.