The Fundamental Theorem of Projective Planes

Theorem 1 (The Fundamental Theorem of Projective Planes): Let $\mathbf{z_1} = [1, 0, 0]$, $\mathbf{z_2} = [0, 1, 0]$, $\mathbf{z_3} = [0, 0, 1]$, and $\mathbf{z_4} = [1, 1, 1]$. If $\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s}$ are four distinct points in the projective plane $\mathbb{P}^2(F)$ such that no three of them are collinear, then there exists a unique (up to a scalar) invertible $3 x 3$ matrix $M$ whose entries are from $F$ such that the collineation $\phi_M : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ defined for all $\mathbf{x} \in \mathbb{P}^2(F)$ by $\phi_M(\mathbf{x}) = \mathbf{x}M$ is such that $\phi_M(\mathbf{z_1}) = \mathbf{p}$, $\phi_M(\mathbf{z_2}) = \mathbf{q}$, $\phi_M(\mathbf{z_3}) = \mathbf{r}$, and $\phi_M(\mathbf{z_4}) = \mathbf{s}$.

• Proof: Let $\mathbf{p} = [p_1, p_2, p_3], \mathbf{q} = [q_1, q_2, q_3], \mathbf{r} = [r_1, r_2, r_3], \mathbf{s} = [s_1, s_2, s_3] \in \mathbb{P}^2(F)$. Consider the $3 \times 3$ augmented matrix:

(1)

\begin{align} \quad M^{*} = \begin{bmatrix} \mathbf{p}\\ \mathbf{q}\\ \mathbf{r} \end{bmatrix} = \begin{bmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3\\ r_1 & r_2 & r_3 \end{bmatrix} \end{align}

• Since no three pairs of points of $\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s}$ are collinear, we have that $\det (M^{*}) \neq 0$ and so $M^{*}$ is invertible. Let $\phi_{M^{*}} : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ be the collineation associated to $M^{*}$ so that for all $\mathbf{x} \in \mathbb{P}^2(F)$ we have that $\phi_{M^{*}}(\mathbf{x}) = \mathbf{x}M^{*}$. Then for the points $\mathbf{z_1} = [1, 0, 0], \mathbf{z_2} = [0, 1, 0], \mathbf{z_3} = [0, 0, 1] \in \mathbb{P}^2(F)$ we have that:

(2)

\begin{align} \quad \phi_{M^{*}} (\mathbf{z_1}) = \mathbf{z_1}M^{*} = \mathbf{p} \\ \quad \phi_{M^{*}} (\mathbf{z_2}) = \mathbf{z_2}M^{*} = \mathbf{q} \\ \quad \phi_{M^{*}} (\mathbf{z_3}) = \mathbf{z_3}M^{*} = \mathbf{r} \end{align}

• We now only need to have $M^{*}$ be such that $\phi_{M^{*}} (\mathbf{z_4}) = \mathbf{s}$. Let $\alpha, \beta, \gamma \in F$ and consider the matrix $M$ below:

(3)

\begin{align} \quad M =\begin{bmatrix} \alpha \mathbf{p}\\ \beta \mathbf{q}\\ \gamma \mathbf{r} \end{bmatrix} = \begin{bmatrix} \alpha p_1 & \alpha p_2 & \alpha p_3\\ \beta q_1 & \beta q_2 & \beta q_3\\ \gamma r_1 & \gamma r_2 & \gamma r_3 \end{bmatrix} \end{align}

• Provided that $\alpha, \beta\ , \gamma \neq 0$ we will have that $\det (M) \neq 0$ so that then $M$ is invertible, and the collineation $\phi_M : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ defined for all $\mathbf{x} \in \mathbb{P}^2(F)$ by $\phi_M(\mathbf{x}) = \mathbf{x}M$ also satisfies $\phi_M(\mathbf{z_1}) = \mathbf{p}$, $\phi_M(\mathbf{z_2}) = \mathbf{q}$, and $\phi_M(\mathbf{z_3}) = \mathbf{r}$ since:

(4)

\begin{align} \quad \phi_M(\mathbf{z_1}) = \mathbf{z_1}M = [1, 0, 0] M = [\alpha, 0, 0] M^{*} = [1, 0, 0] M^{*} = \mathbf{p} \\ \quad \phi_M(\mathbf{z_2}) = \mathbf{z_2}M = [0, 1, 0] M = [0, \beta, 0] M^{*} = [0, 1, 0]M^{*} = \mathbf{q} \\ \quad \phi_M(\mathbf{z_3}) = \mathbf{z_3}M = [0, 0, 1] M = [0, 0, \gamma] M^{*} = [0, 0, 1]M^{*} = \mathbf{r} \end{align}

• We want to find $\alpha, \beta, \gamma \in F$ such that $\phi_M(\mathbf{z_4}) = \mathbf{s}$. We note that:

(5)

\begin{align} \quad \phi_M(\mathbf{z_4}) &= \mathbf{z_4}M \\ \quad \phi_M(\mathbf{z_4}) &= \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha p_1 & \alpha p_2 & \alpha p_3\\ \beta q_1 & \beta q_2 & \beta q_3\\ \gamma r_1 & \gamma r_2 & \gamma r_3 \end{bmatrix} \\ \quad \phi_M(\mathbf{z_4}) &= \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 & 0\\ 0 & \beta & 0\\ 0 & 0 & \gamma \end{bmatrix} \begin{bmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3\\ r_1 & r_2 & r_3 \end{bmatrix} \\ \quad \phi_M(\mathbf{z_4}) &= \begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} M^{*} \end{align}

• So we want:

(6)

\begin{align} \quad \phi_M(\mathbf{z_4}) &= \mathbf{s} \quad (*) \\ \quad \begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} M^{*} &= \mathbf{s} \\ \quad \begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} &= \mathbf{s} M^{*-1} \end{align}

• Notice that $\alpha, \beta, \gamma \neq 0$ otherwise the existence of $M$ is not guaranteed. However, from $(*)$, we can find explicit formulas for $\alpha, \beta, \gamma$ by applying Cramer's rule:

(7)

\begin{align} \alpha = \frac{\begin{vmatrix} s_1 & s_2 & s_3\\ q_1 & q_2 & q_3\\ r_1 & r_2 & r_3 \end{vmatrix}} {\begin{vmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3\\ r_1 & r_2 & r_3 \end{vmatrix}} \quad , \quad \beta = \frac{\begin{vmatrix} p_1 & p_2 & p_3\\ s_1 & s_2 & s_3\\ r_1 & r_2 & r_3 \end{vmatrix}} {\begin{vmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3\\ r_1 & r_2 & r_3 \end{vmatrix}} \quad , \quad \gamma = \frac{\begin{vmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3\\ s_1 & s_2 & s_3 \end{vmatrix}} {\begin{vmatrix} p_1 & p_2 & p_3\\ q_1 & q_2 & q_3\\ r_1 & r_2 & r_3 \end{vmatrix}} \end{align}

• Since no three of the points $\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s}$ are collinear, we see that therefore $\alpha, \beta, \gamma \neq 0$ as desired.

• All that is left to prove is that the matrix $M$ is unique. Suppose that two such matrices $M$ and $N$ satisfy the conditions above. Since both $\phi_N : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ is a bijection, we have that $\phi_N^{-1} : \mathbb{P}^2(F) \to \mathbb{P}^2(F)$ exists and more over, so $\phi_M \circ \phi_{N^-1} = \phi_{MN^{-1}}$ is another collineation. Since $\phi_M(\mathbf{z_1}) = z_1$, $\phi_M(\mathbf{z_2}) = \mathbf{z_2}$, and $\phi_M(\mathbf{z_3}) = \mathbf{z_3}$, we must have that for some scalars $\mu_1, \mu_2, \mu_3 \in F$ that:

(8)

\begin{align} \quad MN^{-1} = \begin{bmatrix} \mu_1 & 0 & 0\\ 0 & \mu_2 & 0\\ 0 & 0 & \mu_3 \end{bmatrix} \end{align}

• But we also have that $\mathbf{z_4} = \phi_{MN^{-1}}(\mathbf{z_4})$ and $\mathbf = \phi_{MN^{-1}} \mathbf{z_4}MN^{-1} = [\mu_1, \mu_2, \mu_3]$. Therefore $\mu = \mu_1 = \mu_2 = \mu_3$ and so $MN^{-1} = \mu I$ and $M = \mu N$, so $M$ is unique up to a scalar. $\blacksquare$