Background

Moving one dimension up, we consider two planes $\pi_1$ and $\pi_2$ sitting in a three-dimensional projective space $V$ . Then there are two bijections, one from $\pi_1$ to $\pi_2$ , and the other from $\pi_2$ to $\pi_1$ , such that their product is a bijection $\phi$ of, say $\pi_1$ onto itself, such that $\phi$ sends points to points, lines to lines, and fixes a line (the line of intersection of $\pi_1$ and $\pi_2$ ).

Definition

Certainly any linear transformation that is the identity when restricted to a subspace of codimension 1 induces a perspectivity. Conversely, a perspectivity induces a linear transformation, if the underlying vector space $V$ is at least three dimensional.

Proof. Assume $\operatorname{dim}(V)\ge 3$ . By the fundamental theorem of projective geometry, a perspectivity $\phi$ on $PG(V)$ induces a semi-linear transformation $f$ on $V$ . Now, by assumption, there is a two dimensional plane $\Pi$ , such that $\phi(S)=S$ for all $S\le \Pi$ . Let $kv$ and $kw$ be two points that span $\Pi$ (so $v$ and $w$ are linearly independent). So, $\phi(kv)=kv$ and $\phi(kw)=kw$ . This means $f(v)=av$ and $f(w)=bw$ , where $a,b\in k$ . Since $k(v+w)$ is also a point in $\Pi$ , $\phi(k(v+w))=k(v+w)$ and so $f(v+w)=c(v+w)$ . By additivity of $f$ , $a=c=b$ . Next, for an arbitrary non-zero element $d\in k$ , $k(dv+w)$ is a point in $\Pi$ and hence fixed by $\phi$ . Translating this to $f$ , we have $f(dv+w)=r(dv+w)$ . By semi-linearity of $f$ , we get $rdv+rw=d^{\alpha}f(v)+f(w)=d^{\alpha}cv+cw$ . So $rd=d^{\alpha}c$ and $r=c$ . Solving this we get $cd=d^{\alpha}c$ or $d^{\alpha}=cdc^{-1}$ . This shows that $\alpha$ is an inner-automorphism of $k$ . But $k$ is a field, $\alpha$ is the identity on $k$ . As a result, $f$ is a linear transformation.

Remarks.

• In the literature, the maps $\rho$ and $\sigma$ (see Background above) are also known as perspectivies. To make a distinction between this and the formal definition given earlier, we will call, say $\rho$ , a perspectivity between two projective figures $\ell_1$ and $\ell_2$ with center $P$ . A projective figure is just a collection of points (one-dimensional subspaces) in a projective geometry. In this way, we can define a projectivity between two projective figures $X$ and $Y$ as a bijection $\rho:X\to Y$ such that there exists a point $P$ with the property that
  for any $x\in X$ , points $P,x,$ and $\rho(x)$ are collinear.
  Two projective figures are perspective if there is a perspectivity between them.
• In a projective geometry, every pair of points are perpective.
• If $V$ is finite dimensional, then every collineation is a product of a finite number of perspectivities. The number of perspectivities required is at most $\operatorname{dim}(V)$ . In particular, the picture becomes apparant when the collineation can be diagonalized. For if a collineation $p$ can be represented by a diagonal matrix whose diagonal entries are $d_1,\ldots, d_n$ from the top left to the bottom right, then
  $\begin{pmatrix} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_n \end{pmatrix}$ = $\begin{pmatrix} d_1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & d_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} \cdots \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_n \end{pmatrix}$
  where each of the matrices on the right represents a perspectivity. This is only an idealized situation where the linear transformation can be diagonalized. There are perspectivities that induce non-diagonalizable linear transformations; see central collineations for detail.
• (Second Fundamental Theorem of Projective Geometry). From the two assertions above, one can show that

  if $\operatorname{dim}(V)=n < \infty$ , and if $\lbrace v_1,\ldots,v_n \rbrace$ and $\lbrace w_1,\ldots, w_n\rbrace$ are two sets of points in general position, then there is a unique projectivity $p$ such that $p(v_i)=w_i$ .

• Since a projective geometry is a bounded lattice, the concept of two points being perspective with one another can be generalized to elements found in a bounded lattice: two elements $a,b$ in a bounded lattice $L$ are in perspective if they have a common complement $c$ , that is, $a\vee c=b\vee c=1$ and $a\wedge c=b\wedge c=0$ . If we apply this back to the case when $L$ is a projective geometry, we see that $a$ and $b$ are in perspective precisely when there is a perspectivity taking $a$ to $b$ . If $a\ne b$ , then $c$ is the hyperplane that is fixed by the perspectivity.