Projective Geometry/Classic/Projective Transformations/Transformations of the projective plane

Two-dimensional projective transformations are a type of automorphism of the projective plane onto itself.

Planar transformations can be defined synthetically as follows: point X on a "subjective" plane must be transformed to a point T also on the subjective plane. The transformations uses these tools: a pair of "observation points" P and Q, and an "objective" plane. The subjective and objective planes and the two points all lie in three-dimensional space, and the two planes can intersect at some line.

Draw line l1 through points P and X. Line l1 intersects the objective plane at point R. Draw line l2 through points Q and R. Line l2 intersects the projective plane at point T. Then T is the projective transform of X.

Analysis

Let the xy-plane be the "subjective" plane and let plane m be the "objective" plane. Let plane m be described by

${\displaystyle z=f(x,y)=mx+ny+b}$

where the constants m and n are partial slopes and b is the z-intercept.

Let there be a pair of "observation" points P and Q,

${\displaystyle P:(P_{x},P_{y},P_{z}),}$
${\displaystyle Q:(Q_{x},Q_{y},Q_{z}).}$

Let point X lie on the "subjective" plane:

${\displaystyle X:(x,y,0).}$

Point X must be transformed to a point T,

${\displaystyle T:(T_{x},T_{y},0)}$

also on the "subjective" plane.

The analytical results are a pair of equations, one for abscissa Tx and one for ordinate Ty:

${\displaystyle T_{x}={x(-mQ_{x}P_{z}-nQ_{z}P_{y}+Q_{z}(P_{z}-b))+(ny+b)(Q_{z}P_{x}-Q_{x}P_{z}) \over (mx+ny)(Q_{z}-P_{z})-(mP_{x}+nP_{y})Q_{z}+(Q_{z}-b)P_{z}},\qquad \qquad (12)}$
${\displaystyle T_{y}={y(-nQ_{y}P_{z}-mQ_{z}P_{x}+Q_{z}(P_{z}-b))+(mx+b)(Q_{z}P_{y}-Q_{y}P_{z}) \over (ny+mx)(Q_{z}-P_{z})-(nP_{y}+mP_{x})Q_{z}+(Q_{z}-b)P_{z}}.\qquad \qquad (13)}$

There are (at most) nine degrees of freedom for defining a 2D transformation: Px, Py, Pz, Qx, Qy, Qz, m, n, b. Notice that equations (12) and (13) have the same denominators, and that Ty can be obtained from Tx by exchanging m with n, and x with y (including subscripts of P and Q).

Trilinear fractional transformations

Let

${\displaystyle \alpha =-mQ_{x}P_{z}-nQ_{z}P_{y}+Q_{z}(P_{z}-b),}$
${\displaystyle \beta =n(Q_{z}P_{x}-Q_{x}P_{z}),}$
${\displaystyle \gamma =b(Q_{z}P_{x}-Q_{x}P_{z}),}$
${\displaystyle \delta =m(Q_{z}-P_{z}),}$
${\displaystyle \epsilon =n(Q_{z}-P_{z}),}$
${\displaystyle \zeta =-(mP_{x}+nP_{y})Q_{z}+(Q_{z}-b)P_{z},}$

so that

${\displaystyle T_{x}={\alpha x+\beta y+\gamma \over \delta x+\epsilon y+\zeta }.\qquad \qquad (14)}$

Also let

${\displaystyle \eta =m(Q_{z}P_{y}-Q_{y}P_{z}),}$
${\displaystyle \theta =-mQ_{z}P_{x}-nQ_{y}P_{z}+Q_{z}(P_{z}-b),}$
${\displaystyle \kappa =b(Q_{z}P_{y}-Q_{y}P_{z}),}$

so that

${\displaystyle T_{y}={\eta x+\theta y+\kappa \over \delta x+\epsilon y+\zeta }.\qquad \qquad (15)}$

Equations (14) and (15) together describe the trilinear fractional transformation.

Composition of trilinear transformations

If a transformation is given by equations (14) and (15), then such transformation is characterized by nine coefficients which can be arranged into a coefficient matrix

${\displaystyle M_{T}={\begin{bmatrix}\alpha &\beta &\gamma \\\eta &\theta &\kappa \\\delta &\epsilon &\zeta \end{bmatrix}}.}$

If there are a pair T1 and T2 of planar transformations whose coefficient matrices are ${\displaystyle M_{T_{1}}}$ and ${\displaystyle M_{T_{2}}}$, then the composition of these transformations is another planar transformation T3,

${\displaystyle T_{3}=T_{2}\circ T_{1},}$

such that

${\displaystyle T_{3}(x,y)=T_{2}(T_{1}(x,y)).}$

The coefficient matrix of T3 can be obtained by multiplying the coefficient matrices of T2 and T1:

${\displaystyle M_{T_{3}}=M_{T_{2}}\,M_{T_{1}}.}$

Proof

Given T1 defined by

${\displaystyle T_{1x}={\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}},}$
${\displaystyle T_{1y}={\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}},}$

and given T2 defined by

${\displaystyle T_{2x}={\alpha _{2}x+\beta _{2}y+\gamma _{2} \over \delta _{2}x+\epsilon _{2}y+\zeta _{2}},}$
${\displaystyle T_{2y}={\eta _{2}x+\theta _{2}y+\kappa _{2} \over \delta _{2}x+\epsilon _{2}y+\zeta _{2}},}$

then T3 can be calculated by substituting T1 into T2,

${\displaystyle T_{3x}=T_{2x}(T_{1x},T_{1y})={\alpha _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\beta _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\gamma _{2} \over \delta _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\epsilon _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\zeta _{2}}.}$

Multiply numerator and denominator by the same trinomial,

${\displaystyle T_{3x}={\alpha _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\beta _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\gamma _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1}) \over \delta _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\epsilon _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\zeta _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1})}.}$

Group the coefficients of x, y, and 1:

${\displaystyle T_{3x}={x(\alpha _{2}\alpha _{1}+\beta _{2}\eta _{1}+\gamma _{2}\delta _{1})+y(\alpha _{2}\beta _{1}+\beta _{2}\theta _{1}+\gamma _{2}\epsilon _{1})+(\alpha _{2}\gamma _{1}+\beta _{2}\kappa _{1}+\gamma _{2}\zeta _{1}) \over x(\delta _{2}\alpha _{1}+\epsilon _{2}\eta _{1}+\zeta _{2}\delta _{1})+y(\delta _{2}\beta _{1}+\epsilon _{2}\theta _{1}+\zeta _{2}\epsilon _{1})+(\delta _{2}\gamma _{1}+\epsilon _{2}\kappa _{1}+\zeta _{2}\zeta _{1})}={\alpha _{3}x+\beta _{3}y+\gamma _{3} \over \delta _{3}x+\epsilon _{3}y+\zeta _{3}}.}$

These six coefficients of T3 are the same as those obtained through the product

${\displaystyle {\begin{bmatrix}\alpha _{2}&\beta _{2}&\gamma _{2}\\\eta _{2}&\theta _{2}&\kappa _{2}\\\delta _{2}&\epsilon _{2}&\zeta _{2}\end{bmatrix}}{\begin{bmatrix}\alpha _{1}&\beta _{1}&\gamma _{1}\\\eta _{1}&\theta _{1}&\kappa _{1}\\\delta _{1}&\epsilon _{1}&\zeta _{1}\end{bmatrix}}={\begin{bmatrix}\alpha _{3}&\beta _{3}&\gamma _{3}\\\eta _{3}&\theta _{3}&\kappa _{3}\\\delta _{3}&\epsilon _{3}&\zeta _{3}\end{bmatrix}}.\qquad \qquad (16)}$

The remaining three coefficients can be verified thus

${\displaystyle T_{3y}=T_{2y}(T_{1x},T_{1y})={\eta _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\theta _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\kappa _{2} \over \delta _{2}\left({\alpha _{1}x+\beta _{1}y+\gamma _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\epsilon _{2}\left({\eta _{1}x+\theta _{1}y+\kappa _{1} \over \delta _{1}x+\epsilon _{1}y+\zeta _{1}}\right)+\zeta _{2}}.}$

Multiply numerator and denominator by the same trinomial,

${\displaystyle T_{3y}={\eta _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\theta _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\kappa _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1}) \over \delta _{2}(\alpha _{1}x+\beta _{1}y+\gamma _{1})+\epsilon _{2}(\eta _{1}x+\theta _{1}y+\kappa _{1})+\zeta _{2}(\delta _{1}x+\epsilon _{1}y+\zeta _{1})}.}$

Group the coefficients of x, y, and 1:

${\displaystyle T_{3x}={x(\eta _{2}\alpha _{1}+\theta _{2}\eta _{1}+\kappa _{2}\delta _{1})+y(\eta _{2}\beta _{1}+\theta _{2}\theta _{1}+\kappa _{2}\epsilon _{1})+(\eta _{2}\gamma _{1}+\theta _{2}\kappa _{1}+\kappa _{2}\zeta _{1}) \over x(\delta _{2}\alpha _{1}+\epsilon _{2}\eta _{1}+\zeta _{2}\delta _{1})+y(\delta _{2}\beta _{1}+\epsilon _{2}\theta _{1}+\zeta _{2}\epsilon _{1})+(\delta _{2}\gamma _{1}+\epsilon _{2}\kappa _{1}+\zeta _{2}\zeta _{1})}={\eta _{3}x+\theta _{3}y+\kappa _{3} \over \delta _{3}x+\epsilon _{3}y+\zeta _{3}}.}$

The three remaining coefficients just obtained are the same as those obtained through equation (16). Q.E.D.

Planar transformations of lines

The trilinear transformation given be equations (14) and (15) transforms a straight line

${\displaystyle y=mx+b}$

into another straight line

${\displaystyle T_{y}=nT_{x}+c}$

where n and c are constants and equal to

${\displaystyle n={m(\epsilon \kappa -\zeta \theta )+b(\delta \theta -\epsilon \eta )+(\delta \kappa -\zeta \eta ) \over m(\epsilon \gamma -\zeta \beta )+b(\delta \beta -\epsilon \alpha )+(\delta \gamma -\zeta \alpha )}}$

and

${\displaystyle c={m(\beta \kappa -\gamma \theta )+b(\alpha \theta -\beta \eta )+(\alpha \kappa -\gamma \eta ) \over m(\beta \zeta -\gamma \epsilon )+b(\alpha \epsilon -\beta \delta )+(\alpha \zeta -\gamma \delta )}.}$

Proof

Given y = m x + b, then plugging this into equations (14) and (15) yields

${\displaystyle T_{x}={\alpha x+\beta (mx+b)+\gamma \over \delta x+\epsilon (mx+b)+\zeta }={(\alpha +\beta m)x+(\beta b+\gamma ) \over (\delta +\epsilon m)x+(\epsilon b+\zeta )},}$

and

${\displaystyle T_{y}={(\eta +\theta m)x+(\theta b+\kappa ) \over (\delta +\epsilon m)x+(\epsilon b+\zeta )}.}$

If Ty = n Tx + c and n and c are constants, then

${\displaystyle {\partial T_{y} \over \partial x}=n{\partial T_{x} \over \partial x}}$

so that

${\displaystyle n={\partial T_{y}/\partial x \over \partial T_{x}/\partial y}.}$

Calculation shows that

${\displaystyle {\partial T_{x} \over \partial x}={(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m) \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )]^{2}}}$

and

${\displaystyle {\partial T_{y} \over \partial x}={(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m) \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )]^{2}}}$

therefore

${\displaystyle n={\partial T_{y}/\partial x \over \partial T_{x}/\partial y}={(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m) \over (\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)}.}$

We should now obtain c to be

${\displaystyle c=T_{y}-nT_{x}}$
${\displaystyle ={(\eta +\theta m)x+(\theta b+\kappa )-\left[{(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m) \over (\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)}\right]\cdot [(\alpha +\beta m)x+(\beta b+\gamma )] \over (\delta +\epsilon m)x+(\epsilon b+\zeta )}.}$

Add the two fractions in the numerator:

${\displaystyle c={\left\{[(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)][(\eta +\theta m)x+(\theta b+\kappa )]-[(\epsilon b+\zeta )(\eta +\theta m)-(\theta b+\kappa )(\delta +\epsilon m)][(\alpha +\beta m)x+(\beta b+\gamma )]\right\} \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )][(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)]}.}$

Distribute binomials in parentheses in the numerator, then cancel out equal and opposite terms:

${\displaystyle c={-(\beta b+\gamma )(\delta +\epsilon m)(\eta +\theta m)x+(\epsilon b+\zeta )(\alpha +\beta m)(\theta b+\kappa )+(\theta b+\kappa )(\delta +\epsilon m)(\alpha +\beta m)x-(\epsilon b+\zeta )(\eta +\theta m)(\beta b+\gamma ) \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )][(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)]}.}$

Factor the numerator into a pair of terms, only one of them having the numerus cossicus (x). There is another numerus cossicus in the denominator. The objective now is to get both of these to cancel out.

${\displaystyle c={\left\{[(\theta b+\kappa )(\alpha +\beta m)-(\beta b+\gamma )(\eta +\theta m)](\delta +\epsilon m)x+[(\alpha +\beta m)(\theta b+\kappa )-(\eta +\theta m)(\beta b+\gamma )](\epsilon b+\zeta )\right\} \over [(\delta +\epsilon m)x+(\epsilon b+\zeta )][(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)]}.}$

Factor the numerator,

${\displaystyle c={[(\theta b+\kappa )(\alpha +\beta m)-(\beta b+\gamma )(\eta +\theta m)][(\delta +\epsilon m)x+(\epsilon b+\zeta )] \over [(\epsilon b+\zeta )(\alpha +\beta m)-(\beta b+\gamma )(\delta +\epsilon m)][(\delta +\epsilon m)x+(\epsilon b+\zeta )]}.}$

The terms with the numeri cossici cancel out, therefore

${\displaystyle c={(\alpha +\beta m)(\theta b+\kappa )-(\beta b+\gamma )(\eta +\theta m) \over (\alpha +\beta m)(\epsilon b+\zeta )-(\beta b+\gamma )(\delta +\epsilon m)}}$

is a constant. Q.E.D.

Comparing c with n, notice that their denominators are the same. Also, n is obtained from c by exchanging the following coefficients:

${\displaystyle \alpha \leftrightarrow \delta ,\ \beta \leftrightarrow \epsilon ,\ \gamma \leftrightarrow \zeta .}$

There is also the following exchange symmetry between the numerator and denominator of n:

${\displaystyle \alpha \leftrightarrow \eta ,\ \beta \leftrightarrow \theta ,\ \gamma \leftrightarrow \kappa .}$

The numerator and denominator of c also have exchange symmetry: ${\displaystyle \{\eta \leftrightarrow \delta ,\ \theta \leftrightarrow \epsilon ,\ \kappa \leftrightarrow \zeta \}.}$

The exchange symmetry between n and c can be chunked into binomials:

${\displaystyle n\leftrightarrow c\equiv \{(\alpha +m\beta )\leftrightarrow (\delta +m\epsilon ),\ (\gamma +b\beta )\leftrightarrow (\zeta +b\epsilon )\}.}$

All of these exchange symmetries amount to exchanging pairs of rows in the coefficient matrix.

Planar transformations of conic sections

A trilinear transformation such as T given by equations (14) and (15) will convert a conic section

${\displaystyle Ax^{2}+By^{2}+Cx+Dy+Exy+F=0\qquad \qquad (17)}$

into another conic section

${\displaystyle A'T_{x}^{2}+B'T_{y}^{2}+C'T_{x}+D'T_{y}+E'T_{x}T_{y}+F'=0.\qquad \qquad (18)}$

Proof

Let there be given a conic section described by equation (17) and a planar transformation T described by equations (15) and (16) which converts points (x,y) into points (Tx,Ty).

It is possible to find an inverse transformation T′ which converts back points (Tx,Ty) to points (x,y). This inverse transformation has a coefficient matrix

${\displaystyle M_{T'}={\begin{bmatrix}\alpha '&\beta '&\gamma '\\\eta '&\theta '&\kappa '\\\delta '&\epsilon '&\zeta '\end{bmatrix}}.}$

Equation (17) can be expressed in terms of the inverse transformation:

${\displaystyle A\left({\alpha 'T_{x}+\beta 'T_{y}+\gamma ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)^{2}+B\left({\eta 'T_{x}+\theta 'T_{y}+\kappa ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)^{2}+C\left({\alpha 'T_{x}+\beta 'T_{y}+\gamma ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)+D\left({\eta 'T_{x}+\theta 'T_{y}+\kappa ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)+E\left({\alpha 'T_{x}+\beta 'T_{y}+\gamma ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)\left({\eta 'T_{x}+\theta 'T_{y}+\kappa ' \over \delta 'T_{x}+\epsilon 'T_{y}+\zeta '}\right)+F=0.}$

The denominators can be "dissolved" by multiplying both sides of the equation by the square of a trinomial:

${\displaystyle A(\alpha 'T_{x}+\beta 'T_{y}+\gamma ')^{2}+B(\eta 'T_{x}+\theta 'T_{y}+\kappa ')^{2}+C(\alpha 'T_{x}+\beta 'T_{y}+\gamma ')(\delta 'T_{x}+\epsilon 'T_{y}+\zeta ')+D(\eta 'T_{x}+\theta 'T_{y}+\kappa ')(\delta 'T_{x}+\epsilon 'T_{y}+\zeta ')+E(\alpha 'T_{x}+\beta 'T_{y}+\gamma ')(\eta 'T_{x}+\theta 'T_{y}+\kappa ')+F(\delta 'T_{x}+\epsilon 'T_{y}+\zeta ')^{2}=0.}$

Expand the products of trinomials and collect common powers of Tx and Ty:

${\displaystyle {\begin{matrix}(A\alpha '^{2}+B\eta '^{2}+C\alpha '\delta '+D\eta '\delta '+E\alpha '\eta '+F\delta '^{2})T_{x}^{2}\\+(A\beta '^{2}+B\theta '^{2}+C\beta '\epsilon '+D\theta '\epsilon '+E\beta '\theta '+F\epsilon '^{2})T_{y}^{2}\\+(2A\alpha '\gamma '+2B\eta '\kappa '+C(\alpha '\zeta '+\gamma '\delta ')+D(\eta '\zeta '+\kappa '\delta ')+E(\alpha '\kappa '+\gamma '\eta ')+2F\delta '\zeta ')T_{x}\\+(2A\beta '\gamma '+2B\theta '\kappa '+C(\beta '\zeta '+\gamma '\epsilon ')+D(\theta '\zeta '+\kappa '\epsilon ')+E(\beta '\kappa '+\gamma '\theta ')+2F\epsilon '\zeta ')T_{y}\\+(2A\alpha '\beta '+2B\eta '\theta '+C(\alpha '\epsilon '+\beta '\delta ')+D(\eta '\epsilon '+\theta '\delta ')+E(\alpha '\theta '+\beta '\eta ')+2F\delta '\epsilon ')T_{x}T_{y}\\+(A\gamma '^{2}+B\kappa '^{2}+C\gamma '\zeta '+D\kappa '\zeta '+E\gamma '\kappa '+F\zeta '^{2})=0.\end{matrix}}\qquad \qquad (19)}$

Equation (19) has the same form as equation (18).

What remains to do is to express the primed coefficients in terms of the unprimed coefficients. To do this, apply Cramer's rule to the coefficient matrix MT to obtain the primed matrix of the inverse transformation:

${\displaystyle M_{T'}={1 \over \Delta }{\begin{bmatrix}\left|{\begin{matrix}\theta &\kappa \\\epsilon &\zeta \end{matrix}}\right|&\left|{\begin{matrix}\epsilon &\zeta \\\beta &\gamma \end{matrix}}\right|&\left|{\begin{matrix}\beta &\gamma \\\theta &\kappa \end{matrix}}\right|\\\quad &\quad &\quad \\\left|{\begin{matrix}\kappa &\eta \\\zeta &\delta \end{matrix}}\right|&\left|{\begin{matrix}\zeta &\delta \\\gamma &\alpha \end{matrix}}\right|&\left|{\begin{matrix}\gamma &\alpha \\\kappa &\eta \end{matrix}}\right|\\\quad &\quad &\quad \\\left|{\begin{matrix}\eta &\theta \\\delta &\epsilon \end{matrix}}\right|&\left|{\begin{matrix}\delta &\epsilon \\\alpha &\beta \end{matrix}}\right|&\left|{\begin{matrix}\alpha &\beta \\\eta &\theta \end{matrix}}\right|\end{bmatrix}}\qquad \qquad (20)}$

where Δ is the determinant of the unprimed coefficient matrix.

Equation (20) allows primed coefficients to be expressed in terms of unprimed coefficients. But performing these substitutions on the primed coefficients of equation (19) it can be noticed that the determinant Δ cancels itself out, so that it can be ignored altogether. Therefore

${\displaystyle A'=A(\theta \zeta -\kappa \epsilon )^{2}+B(\kappa \delta -\eta \zeta )^{2}+C(\theta \zeta -\kappa \epsilon )(\eta \epsilon -\theta \delta )+D(\kappa \delta -\eta \zeta )(\eta \epsilon -\theta \delta )+E(\theta \zeta -\kappa \epsilon )(\kappa \delta -\eta \zeta )+F(\eta \epsilon -\theta \delta )^{2}}$
${\displaystyle B'=A(\epsilon \gamma -\zeta \beta )^{2}+B(\zeta \alpha -\delta \gamma )^{2}+C(\epsilon \gamma -\zeta \beta )(\delta \beta -\epsilon \alpha )+D(\zeta \alpha -\delta \gamma )(\delta \beta -\epsilon \alpha )+E(\epsilon \gamma -\zeta \beta )(\zeta \alpha -\delta \gamma )+F(\delta \beta -\epsilon \alpha )^{2}}$
${\displaystyle C'=2A(\theta \zeta -\kappa \epsilon )(\beta \kappa -\gamma \theta )+2B(\kappa \delta -\eta \zeta )(\gamma \eta -\alpha \kappa )+C[(\theta \zeta -\kappa \epsilon )(\alpha \theta -\beta \eta )+(\beta \kappa -\gamma \theta )(\eta \epsilon -\theta \delta )]+D[(\kappa \delta -\eta \zeta )(\alpha \theta -\beta \eta )+(\gamma \eta -\alpha \kappa )(\eta \epsilon -\theta \delta )]+E[(\theta \zeta -\kappa \epsilon )(\gamma \eta -\alpha \kappa )+(\beta \kappa -\gamma \theta )(\kappa \delta -\eta \zeta )]+2F(\eta \epsilon -\theta \delta )(\alpha \theta -\beta \eta )}$
${\displaystyle D'=2A(\epsilon \gamma -\zeta \beta )(\beta \kappa -\gamma \theta )+2B(\zeta \alpha -\delta \gamma )(\gamma \eta -\alpha \kappa )+C[(\epsilon \gamma -\zeta \beta )(\alpha \theta -\beta \eta )+(\beta \kappa -\gamma \theta )(\delta \beta -\epsilon \alpha )]+D[(\zeta \alpha -\delta \gamma )(\alpha \theta -\beta \eta )+(\gamma \eta -\alpha \kappa )(\delta \beta -\epsilon \alpha )]+E[(\epsilon \gamma -\zeta \beta )(\gamma \eta -\alpha \kappa )+(\beta \kappa -\gamma \theta )(\zeta \alpha -\delta \gamma )]+2F(\delta \beta -\epsilon \alpha )(\alpha \theta -\beta \eta )}$
${\displaystyle E'=2A(\theta \zeta -\kappa \epsilon )(\epsilon \gamma -\zeta \beta )+2B(\kappa \delta -\eta \zeta )(\zeta \alpha -\delta \gamma )+C[(\theta \zeta -\kappa \epsilon )(\delta \beta -\epsilon \alpha )+(\epsilon \gamma -\zeta \beta )(\eta \epsilon -\theta \delta )]+D[(\kappa \delta -\eta \zeta )(\delta \beta -\epsilon \alpha )+(\zeta \alpha -\delta \gamma )(\eta \epsilon -\theta \delta )]+E[(\theta \zeta -\kappa \epsilon )(\zeta \alpha -\delta \gamma )+(\epsilon \gamma -\zeta \beta )(\kappa \delta -\eta \zeta )]+2F(\eta \epsilon -\theta \delta )(\delta \beta -\epsilon \alpha )}$
${\displaystyle F'=A(\beta \kappa -\gamma \theta )^{2}+B(\gamma \eta -\alpha \kappa )^{2}+C(\beta \kappa -\gamma \theta )(\alpha \theta -\beta \eta )+D(\gamma \eta -\alpha \kappa )(\alpha \theta -\beta \eta )+E(\beta \kappa -\gamma \theta )(\gamma \eta -\alpha \kappa )+F(\alpha \theta -\beta \eta )^{2}}$

The coefficients of the transformed conic have been expressed in terms of the coefficients of the original conic and the coefficients of the planar transformation T. Q.E.D.

Planar projectivities and cross-ratio

Let four points A, B, C, D be collinear. Let there be a planar projectivity T which transforms these points into points A′, B′, C′, and D′. It was already shown that lines are transformed into lines, so that the transformed points A′ through D′ will also be collinear. Then it will turn out that the cross-ratio of the original four points is the same as the cross-ratio of their transforms:

${\displaystyle [A\ B\ C\ D]=[A'\ B'\ C'\ D'].}$

Proof

If the two-dimensional coordinates of four points are known, and if the four points are collinear, then their cross-ratio can be found from their abscissas alone. It is possible to project the points onto a horizontal line by means of a pencil of vertical lines issuing from a point on the line at infinity:

${\displaystyle [A\ B\ C\ D]=[A_{x}\ B_{x}\ C_{x}\ D_{x}].}$

The same is true for the ordinates of the points. The reason is that any mere rescaling of the coordinates of the points does not change the cross-ratio.

Let

${\displaystyle A:(x_{1},mx_{1}+b),}$
${\displaystyle B:(x_{2},mx_{2}+b),}$
${\displaystyle C:(x_{3},mx_{3}+b),}$
${\displaystyle D:(x_{4},mx_{4}+b).}$

Clearly these four points are collinear. Let

${\displaystyle T_{x}(x,y)={\alpha x+\beta y+\gamma \over \delta x+\epsilon y+\zeta }}$

be the first half of a trilinear transformation. Then

${\displaystyle T_{x}(A)={\alpha x_{1}+\beta (mx_{1}+b)+\gamma \over \delta x_{1}+\epsilon (mx_{1}+b)+\zeta }={(\alpha +\beta m)x_{1}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{1}+(\epsilon b+\zeta )},}$
${\displaystyle T_{x}(B)={\alpha x_{2}+\beta (mx_{2}+b)+\gamma \over \delta x_{2}+\epsilon (mx_{2}+b)+\zeta }={(\alpha +\beta m)x_{2}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{2}+(\epsilon b+\zeta )},}$
${\displaystyle T_{x}(C)={\alpha x_{3}+\beta (mx_{3}+b)+\gamma \over \delta x_{3}+\epsilon (mx_{3}+b)+\zeta }={(\alpha +\beta m)x_{3}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{3}+(\epsilon b+\zeta )},}$
${\displaystyle T_{x}(D)={\alpha x_{4}+\beta (mx_{4}+b)+\gamma \over \delta x_{4}+\epsilon (mx_{4}+b)+\zeta }={(\alpha +\beta m)x_{4}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{4}+(\epsilon b+\zeta )}.}$

The original cross-ratio is

${\displaystyle [x_{1}\ x_{2}\ x_{3}\ x_{4}]={x_{1}-x_{3} \over x_{1}-x_{4}}\cdot {x_{2}-x_{4} \over x_{2}-x_{3}}.}$

It is not necessary to calculate the transformed cross-ratio. Just let

${\displaystyle S(x)={(\alpha +\beta m)x+(\beta b+\gamma ) \over (\delta +\epsilon m)x+(\epsilon b+\zeta )}}$

be a bilinear transformation. Then S(x) is a one-dimensional projective transformation. But Tx(A)=S(A), Tx(B)=S(B), Tx(C)=S(C), and Tx(D)=S(D). Therefore

${\displaystyle [T_{x}(A)\ T_{x}(B)\ T_{x}(C)\ T_{x}(D)]=[S(A)\ S(B)\ S(C)\ S(D)]}$

but it has already been shown that bilinear fractional transformations preserve cross-ratio. Q.E.D.

Example

The following is a rather simple example of a planar projectivity:

${\displaystyle T_{x}={1 \over x},\qquad T_{y}={y \over x}.}$

The coefficient matrix of this projectivity T is

${\displaystyle M_{T}={\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}}}$

and it is easy to verify that MT is its own inverse.

The locus of points described parametrically as ${\displaystyle (\cos \theta ,\,\sin \theta )}$ describe a circle, due to the trigonometric identity

${\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1}$

which has the same form as the canonical equation of a circle. Applying the projectivity T yields the locus of points described parametrically by ${\displaystyle (\sec \theta ,\,\tan \theta )}$ which describe a hyperbola, due to the trigonometric identity

${\displaystyle \sec ^{2}\theta -\tan ^{2}\theta =1}$

which has the same form as the canonical equation of a hyperbola. Notice that points ${\displaystyle (~-1,0)}$ and ${\displaystyle (1,0)}$ are fixed points.

Indeed, this projectivity transforms any circle, of any radius, into a hyperbola centered at the origin with both of its foci lying on the x-axis, and vice versa. This projectivity also transforms the y-axis into the line at infinity, and vice versa:

${\displaystyle T:(0,y)\rightarrow \left({1 \over 0},{y \over 0}\right)=(\pm \infty ,\pm \infty ),}$
${\displaystyle T:(\pm \infty ,\pm \infty )\rightarrow \left({1 \over \pm \infty },{\pm \infty \over \pm \infty }\right)=(0,y).}$

The ratio of infinity over infinity is indeterminate which means that it can be set to any value y desired.

This example emphasizes that in the real projective plane, RP², a hyperbola is a closed curve which passes twice through the line at infinity. But what does the transformation do to a parabola?

Let the locus of points ${\displaystyle (x,x^{2})}$ describe a parabola. Its transformation is

${\displaystyle T:(x,x^{2})\rightarrow \left({1 \over x},{x^{2} \over x}\right)=(x',1/x')}$

which is a hyperbola whose asymptotes are the x-axis and the y-axis and whose wings lie in the first quadrant and the third quadrant. Likewise, the hyperbola

${\displaystyle y={1 \over x}}$

is transformed by T into the parabola

${\displaystyle y=x^{2}\quad }$.

On the other hand, the parabola described by the locus of points ${\displaystyle (x,\pm {\sqrt {x}})}$ is transformed by T into itself: this demonstrates that a parabola intersects the line at infinity at a single point.