Using linear algebra to show that there exists a projective transformation so that four points on a projective line switch places of pairs of points.

Question

"Let A,B,C,D be four points on a projective line. Show that there's a projective transformation which switches places of A and C, and also B and D."

Lets take this in this in the real three dimensional space. A projective line then is just a plane going through the origin. And the points are now vectors on the plane. Now because this is a plane we have that two of these vectors can be used as a base, and we can take the other two as another base. We know that these vectors are all linearly independent because they are unique homogenous coordinates. So my thinking of how to show this is use that we can use an invertible matrix (A) to take any base of the plane to any other. Now this would prove the the thing if I could also prove that $A^2*(x,y,z) = k(x,y,z)$ for all A, where k is a scalar.

Is this true, and if so is it relatively easy to prove, or should I try find another way of proving the statement?

To clarify terms: projective lines are lines of this form: ax+by+cz=0. The points here have homogenous coordinates, which means that for a point P, P = kP for any non zero k. Projective transformations are just invertible linear transformations, and therefore have a corresponding invertible matrix.

Answer 1

The fundamental theorem of projective geometry says that there's a unique projective transformation of the projective line taking any three distinct points to $0, 1, \infty$. Alternatively, if $A,B,C$ and $A',B',C'$ are triples of distinct points, there's a unique projective transformation taking $A$ to $A'$, etc.

If we let $A' = C, C' = A, B' = D$, then that unique projective transformation swaps $A$ and $C$, and sends $B$ to $D$, and it must send $D$ somewhere, say $E$. The only question is whether $E = B$.

Now projective transformations preserve cross-ratios. If you apply that idea to what you know about the transformation above, you find that $\frac{DB}{CB} = \frac{DE}{CE}$ which, after some fiddling, shows that $E = B$.

Perhaps a better way to state the claim of the theorem is that if a projective transformation swaps $A$ and $C$ and sends $B$ to $D$, then it also sends $D$ to $B$.

Answer 2

Your reasoning is correct. Any three linearly independent vectors in $\mathbb{R}^3$ can be mapped to any other three. It is enough to show this from $\mathbf{i},\mathbf{j},\mathbf{k}$ to $\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3$, via $U:=[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]$ ($u_i$ as column vectors); for then taking two such bases, the matrix $VU^{-1}$ converts from one to the other.

So in the case of four vectors $v_1,v_2,v_3,v_4$ spanning a plane, with $u$ perpendicular to the plane, then let $U:=[u,v_1,v_2]$, $V:=[u,v_3,v_4]$. The matrix $VU^{-1}$ maps the basis $u,v_1,v_2$ to $i,j,k$ and then to $u,v_3,v_4$, so $v_1,v_2$ are mapped to $v_3,v_4$.

There is no guarantee that $v_3,v_4$ map back to $v_1,v_2$ with this matrix. However, there is still leeway in choosing $U$ and $V$, because of projectivity. In fact, $U$ can be modified to be $\widetilde{U}=[\alpha_0 u,\alpha_1v_1,\alpha_2v_2]$, $\widetilde{V}=[\alpha_5 u,\alpha_3v_3,\alpha_4v_4]$; then $\widetilde{V}\widetilde{U}^{-1}$ maps $v_1\mapsto j/\alpha_1\mapsto(\alpha_3/\alpha_1)v_3$, but projectively this still represents point $C$. The extra requirements $Av_3=v_1$, $Av_4=v_2$ lead to six equations that can be used to find these six unknowns $\alpha_i$.