Assumption about the points of the real projective plane

Question

Before I ask I just want to say this is a homework problem so I'm looking for a hint not a full solution.
The question asks to show that for any 4 points $A, B, C, D$, such that no 3 are collinear, there exists an automorphism $T$ of the real projective plane where $T(A)=[1, 0, 0], T(B)=[0, 1, 0], T(C)=[0, 0, 1], T(D)=[1, 1, 1]$ Where the points are given in homogeneous coordinates.
Now I know from my own studies that the automorphism group of the real projective plane is the projective general linear group, but I don't know what to do with this.
At first I was thinking one could start by looking at the coordinates just as being usual in $\mathbb{R}^3$ and taking rotations to map each of $A, B, C$ onto the appropriate axis, and then $D$ would follow since it's a fourth vector in a 3-D space, so it's dependent on the first ones. Afterwards we have the equivalence of points up to scalar multiplication.
However, simple rotations don't seem to be able to get and keep all these points where I want them to be.
My other thought was that we could work with affine transformations of $\mathbb{R}^2$, and then extend these to projective automorphisms, but that seems like a slog through many cases.

Any hints?

EDIT: I also know we can describe relations between points and lines by taking dot and cross products of homogeneous coordinates, but I don't see how to make use of this to construct valid automorphisms.

Answer 1

You’re on the right track. You can take advantage of the fact that the columns of a transformation matrix are the images of the basis vectors, so a matrix representation of the transformation $T$ such that $T(A)=\mathbf e_1$, $T(B)=\mathbf e_2$ and $T(C)=\mathbf e_3$ is $$M=\begin{bmatrix}A&B&C\\1&1&1\end{bmatrix}^{-1}.$$ We know that this matrix is nonsingular because the three points are not colinear. The problem you’re no doubt running into is that we also need $T(D)=\mathbf e_1+\mathbf e_2+\mathbf e_3$, which means we must have $M^{-1}[1,1,1]^T=A+B+C=D$. This of course doesn’t hold for an arbitrary set of four coordinate tuples, but you’re working with homogeneous coordinates: for $\lambda\ne0$, the tuples $[x:y:z]$ and $[\lambda x:\lambda y:\lambda z]$ represent the same point. So, try adjusting the columns of $M^{-1}$ by scaling each one so that $M^{-1}[1,1,1]^T=D$.

Answer 2

I figured I'd post here the solution I handed in to my prof, for anyone else who comes across this.

Let $\vec{A}, \vec{B}, \vec{C}, \vec{D}$ be 4 distinct lines through the origin of $\mathbb{R}^3$ determined by the scalar multiples of the points $A, B, C, D$ respectively.
Suppose that no 3 of these lines are colinear in $\Pi_\mathbb{R}$.
As $\{A, B, C, D\}$ is a set of 4 distinct vectors in $\mathbb{R}^3$, they must be linearly dependent. Thus $\exists a, b, c\in\mathbb{R}$ such that $aA+bB+cC=D$. \newline Assume towards a contradiction that $\{A, B, C\}$ was a linearly dependednt set of vectors in $\mathbb{R}^3$.
Thus $\exists x, y\in\mathbb{R}$ such that $xA+yB=C$.
Now $\forall\lambda\in\mathbb{R}, (x\lambda)A+(y\lambda)B=\lambda C$, so $\vec{C}$ lies in the two-dimensional subspace spanned the scalar multiples of $A$ and $B$, and so $\vec{A}, \vec{B}, \vec{C}$ are collinear in $\Pi_\mathbb{R}$, a contradiction. Therefore $A, B, C$ forms a basis for $\mathbb{R}^3$.
Since any scalar mutliples of the standard basis $e_1, e_2, e_3$ also form a basis for $\mathbb{R}^3$, by the results of elementary linear algebra there exists a change of basis matrix $T$ such that $TA=fe_1, TB=ge_2, TC=he_3$, for any $f, g, h\in\mathbb{R}$.
Thus we can choose $f, g, h$ so that $af=bg=ch=1$. \newline Now $T(D)=T(aA+bB+cC)=aTA+bT+cTC=afe_1+bge_2+che_3$\newline $=e_1+e_2+e_3=\begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}$. \newline Suppose $\vec{x}, \vec{y}$ define distinct lines through the origin of $\mathbb{R}^3$.
Thus \newline $\vec{x}-\vec{y}\neq \vec{0}\Rightarrow T(\vec{x}-\vec{y})\neq\vec{0}\Rightarrow T\vec{x}-T\vec{y}\neq\vec{0}\Rightarrow T(\vec{x})\neq T(\vec{y})$. Thus $T$ is injective on lines through the origin of $\mathbb{R}^3$. \newline Take $Q\in\mathbb{R}^3, \lambda\in\mathbb{R}$. Clearly change of basis matrices are invertible, and thus we left/right inverses, and therefore these matrices are bijective on points of $\mathbb{R}$.
Thus \newline $\exists P\in\mathbb{R}^3$ such that $T(P)=Q$, so for the line $\lambda P, T(\lambda P)=\lambda TP=\lambda Q$.
Therefore $T$ is surjective on the lines through the origin in $\mathbb{R}^3$. \newline We conclude that $T$ is bijective on lines through the origin of $\mathbb{R}^3$, and thus on points of $\Pi_\mathbb{R}$.
Suppose the points $\vec{P}, \vec{Q}$ are collinear in $\Pi_\mathbb{R}$.
Thus they both lie on a plane spanned by some $\vec{x}, \vec{y}$, so $\exists p_1, p_2, q_1, q_2\in\mathbb{R}$ such that $\vec{P}=p_1\vec{x}+p_2\vec{y}, \vec{Q}=q_1\vec{x}+q_2\vec{y}$.
We have then $TP=p_1(T\vec{x})+p_2(T\vec{y}), TQ=q_1(T\vec{x})+q_2(T\vec{y})$ so $TP, TQ$ both lie on the plane spanned by $T\vec{x}, T\vec{y}$, and thus are collinear in $\Pi_\mathbb{R}$.
Therefore $T$ is a collineation preserving bijection, and thus an autmorphism of the projective plane such that for any 4 points of $\Pi_\mathbb{R}$, where no 3 are collinear, $T$ maps the 4 points to $e_1, e_2, e_3, \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}$\newline Let both $S$ and $T$ be automorphisms sending $\vec{A}, \vec{B}, \vec{C}, \vec{D}$ to the prescribed points above.
Since $\{\vec{A}, \vec{B}, \vec{C}\}$ forms a basis of $\mathbb{R}^3$, for $\vec{x}\in\mathbb{R}^3, \exists\mu, \nu, \lambda\in\mathbb{R}$ such that \newline $\mu \vec{A}+\nu \vec{B}+\lambda \vec{C}=\vec{x}$. Thus, \newline $T(\vec{x})=\mu(T\vec{A})+\nu(T\vec{B})\lambda(T\vec{C})=\mu e_1+\nu e_2+\lambda e_3=\mu(S\vec{A})+\nu(S\vec{B})+\lambda(S\vec{C})=S(\vec{X})$. Since $S$ is invertible, $(S^{-1}T)\vec{x}=\vec{x}$. By the arbitrary choice of $\vec{x}$ this holds for all vectors, therefore $S^{-1}T=I$ the $3\times 3$ identity matrix. Thus $T=S$, and so the transformation must be unique, up to the choice of $\vec{A}, \vec{B}, \vec{C}, \vec{D}$ in $\Pi_\mathbb{R}$. \newline $\blacksquare$