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

$z=f(x,y)=mx+ny+b$

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,

$P:(P_{x},P_{y},P_{z}),$

$Q:(Q_{x},Q_{y},Q_{z}).$

Let point X lie on the "subjective" plane:

$X:(x,y,0).$

Point X must be transformed to a point T,

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

also on the "subjective" plane.

The analytical results are a pair of equations, one for abscissa T_x and one for ordinate T_y:

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

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

There are (at most) nine degrees of freedom for defining a 2D transformation: P_x, P_y, P_z, Q_x, Q_y, Q_z, m, n, b. Notice that equations (12) and (13) have the same denominators, and that T_y can be obtained from T_x by exchanging m with n, and x with y (including subscripts of P and Q).

Trilinear fractional transformations

Let

$\alpha =-mQ_{x}P_{z}-nQ_{z}P_{y}+Q_{z}(P_{z}-b),$

$\beta =n(Q_{z}P_{x}-Q_{x}P_{z}),$

$\gamma =b(Q_{z}P_{x}-Q_{x}P_{z}),$

$\delta =m(Q_{z}-P_{z}),$

$\epsilon =n(Q_{z}-P_{z}),$

$\zeta =-(mP_{x}+nP_{y})Q_{z}+(Q_{z}-b)P_{z},$

so that

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

Also let

$\eta =m(Q_{z}P_{y}-Q_{y}P_{z}),$

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

$\kappa =b(Q_{z}P_{y}-Q_{y}P_{z}),$

so that

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

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

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

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

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

such that

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

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

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

Proof

Given T₁ defined by

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

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

and given T₂ defined by

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

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

then T₃ can be calculated by substituting T₁ into T₂,

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

Multiply numerator and denominator by the same trinomial,

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

Group the coefficients of x, y, and 1:

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

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

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

The remaining three coefficients can be verified thus

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

Multiply numerator and denominator by the same trinomial,

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

Group the coefficients of x, y, and 1:

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

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

$y=mx+b$

into another straight line

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

where n and c are constants and equal to

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

and

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

Proof

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

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

and

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

If T_y = n T_x + c and n and c are constants, then

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

so that

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

Calculation shows that

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

and

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

therefore

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

We should now obtain c to be

$c=T_{y}-nT_{x}$

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

Add the two fractions in the numerator:

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

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

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

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.

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

Factor the numerator,

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

The terms with the numeri cossici cancel out, therefore

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

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:

$\alpha \leftrightarrow \delta ,\ \beta \leftrightarrow \epsilon ,\ \gamma \leftrightarrow \zeta .$

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

$\alpha \leftrightarrow \eta ,\ \beta \leftrightarrow \theta ,\ \gamma \leftrightarrow \kappa .$

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

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

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

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

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

into another conic section

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

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 (T_x,T_y).

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

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

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

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

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

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

Expand the products of trinomials and collect common powers of T_x and T_y:

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

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 M_T to obtain the primed matrix of the inverse transformation:

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

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

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

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

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

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

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

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

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:

$[A\ B\ C\ D]=[A'\ B'\ C'\ D'].$

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:

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

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

$A:(x_{1},mx_{1}+b),$

$B:(x_{2},mx_{2}+b),$

$C:(x_{3},mx_{3}+b),$

$D:(x_{4},mx_{4}+b).$

Clearly these four points are collinear. Let

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

be the first half of a trilinear transformation. Then

$T_{x}(A)={\alpha x_{1}+\beta (mx_{1}+b)+\gamma \over \delta x_{1}+\epsilon (mx_{1}+b)+\zeta }={(\alpha +\beta m)x_{1}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{1}+(\epsilon b+\zeta )},$

$T_{x}(B)={\alpha x_{2}+\beta (mx_{2}+b)+\gamma \over \delta x_{2}+\epsilon (mx_{2}+b)+\zeta }={(\alpha +\beta m)x_{2}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{2}+(\epsilon b+\zeta )},$

$T_{x}(C)={\alpha x_{3}+\beta (mx_{3}+b)+\gamma \over \delta x_{3}+\epsilon (mx_{3}+b)+\zeta }={(\alpha +\beta m)x_{3}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{3}+(\epsilon b+\zeta )},$

$T_{x}(D)={\alpha x_{4}+\beta (mx_{4}+b)+\gamma \over \delta x_{4}+\epsilon (mx_{4}+b)+\zeta }={(\alpha +\beta m)x_{4}+(\beta b+\gamma ) \over (\delta +\epsilon m)x_{4}+(\epsilon b+\zeta )}.$

The original cross-ratio is

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

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

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

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

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

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:

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

The coefficient matrix of this projectivity T is

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

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

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

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

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

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

which has the same form as the canonical equation of a hyperbola. Notice that points $(~-1,0)$ and $(1,0)$ 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:

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

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

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 $(x,x^{2})$ describe a parabola. Its transformation is

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

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

$y={1 \over x}$

is transformed by T into the parabola

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

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