Hierarchy of Transformation

1. Affine and projectve transformations

AffineProjective

For two arbitrary ideal points which are of the form $(x, y, 0)^t$ , these points are always on the line $l_{\infty} = (0, 0, 1)^t$ since $(x_1, y_1, 0)^t x (x_2, y_2, 0)^t = l_{\infty}$ .
A point on $l_{\infty}$ is transformed to another point on $l_{\infty}$ by affine transformation. But it is transformed to a point on the vanishing line by projective transformation.

Projective transformation causes the distortion such that parallel lines in affine space are not parallel any more because a vanishing line comes about.
Let the vanishing line be $l = (l_1, l_2, l_3)^t$ , then $l$ is mapped to $l_{\infty}$ by a matrix $H$ as follows:

\begin{equation} H = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ l_1 & l_2 & l_3 \end{array}\right) \end{equation}

While a point is transformed by $H$ , a normal vector is transformed by $H^{-t}$ . So $H^{-t}l = l_{\infty}$ .

\begin{equation} H^{-t}l = \left(\begin{array}{ccc} 1 & 0 & -\frac{l_1}{l_3} \\ 0 & 1 & -\frac{l_2}{l_3} \\ 0 & 0 & \frac{1}{l_3} \end{array}\right) \left(\begin{array}{c} l_1 \\ l_2 \\ l_3 \end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) = l_{\infty} \end{equation}

Therefore, $H^{-t}$ transforms any line $l$ into $l_{\infty}$ , which means that it restores the projective distortion. But it still remains the affine distortion.

TwoLines

\begin{aligned} cos\theta &= \frac{l_1 m_1 + l_2 m_2}{\sqrt{(l^2_1 +ㅣ^2_2)(m^2_1 +m^2_2)}} = \frac{l^t C^{*}_{\infty}m}{\sqrt{(l^t C^{*}_{\infty}l)(m^t C^{*}_{\infty}m)}} \\ \\ C^{*}_{\infty} &= \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) : \text{dual degenerate conic} \end{aligned}

Even though the projective distortion is removed, there is still the affine distortion, which a square shape turns into a parallelogram shape. In other words, the angle between lines is not preserved.
Assume that two lines $l$ , $m$ , and dual degenerate conic $C^{*}_{\infty}$ in the original space are transformed to $l'$ , $m'$ and $C'^{*}_{\infty}$ by affine transformation. Then $l' = H^{-t}l$ , $m' = H^{-t}m$ , and $C'^{*}_{\infty} = HC^{*}_{\infty}H^t$ .
For the angle $\theta$ between $l$ and $m$ , the numerator of $cos\theta$ is $l^t C^{*}_{\infty} m = (l'^t H)(H^{-1}C'^{*}_{\infty} H^{-t})(H^t m') = l'^t C'^{*}_{\infty} m'$ .
When $l$ and $m$ are perpendicular, $l'^t C'^{*}_{\infty} m' = 0$ . It yields as follows:

\begin{aligned} l'^t C'^{*}_{\infty} m' = l'^t H C^{*}_{\infty} H^t m' = (l'_1, l'_2, l'_3) \left(\begin{array}{cc} A & \overrightarrow{t} \\ \overrightarrow{0} & 1 \end{array}\right) \left(\begin{array}{cc} I & \overrightarrow{0} \\ \overrightarrow{0} & 0 \end{array}\right) \left(\begin{array}{cc} A & \overrightarrow{0} \\ \overrightarrow{t} & 1 \end{array}\right) \left(\begin{array}{c} m'_1 \\ m'_2 \\ m'_3 \end{array}\right) = 0 \\ \\ (l'_1, l'_2, l'_3) \left(\begin{array}{cc} AA^t & \overrightarrow{0} \\ \overrightarrow{0} & 0 \end{array}\right) \left(\begin{array}{c} m'_1 \\ m'_2 \\ m'_3 \end{array}\right) = (l'_1, l'_2) \left(\begin{array}{cc} s_{11} & s_{12} \\ s_{21} & s_{22} \end{array}\right) \left(\begin{array}{c} m'_1 \\ m'_2 \end{array}\right) = 0 \end{aligned}

where $\left(\begin{array}{cc} s_{11} & s_{12} \\ s_{21} & s_{22} \end{array}\right) = S = AA^t$ .

$S$ is symmetric, $s_{12} = s_{21}$ , and $s_{22}$ can be set to $1$ since $S$ is defined up to scale.
$s_{11} l'_1 m'_1 + s_{12} (l'_1 m'_2 +l'_2 m'_1) = 0$ .
If we get two pairs of $l$ and $m$ which are perpendicular, then $S$ can be calculated.
Having calculated $S$ , $H$ can be esimated assuming $A$ is positive-definite. From $S = AA^t$ , the eigenvalues of $A$ are the positive square roots of teh eigenvalues of $S$ .
After removing the affine distortion, there still remains the similarity distortion.

As a point on $l_{\infty}$ is transformed to another point on $l_{\infty}$ by affine transformation, a point on the plane $\Pi_{\infty} = (0, 0, 0, 1)^t$ is also transformed to another point on $\Pi_{\infty}$ .
As a point on $l_{\infty}$ is transformed to a point on the vanishing line by projective transformation, a point on $\Pi_{\infty} = (0, 0, 0, 1)^t$ is transformed to a point on the specific plane.

ProjectiveDistortion

When a cube is transformed in projective space, each a pair of parallel lines is not parallel any more and intersects at a point which is not at infinity such as $v_1$ , $v_2$ , and $v_3$ .
These points $v_1$ , $v_2$ , and $v_3$ consist of a plane $\Pi$ which does correspond to $\Pi_{\infty}$ in affine space.
If we know this plane $\Pi$ , it can be transformed to $\Pi_{\infty}$ by some matrix $H$ , so the projective distortion can be removed.
Let $\Pi = (\pi_1, \pi_2, \pi_3, \pi_4)^t = (\overline{\pi}^t, \pi_4)^t$ , then the following $H$ restores $\Pi$ to $\Pi_{\infty}$ :

\begin{equation} H = \left(\begin{array}{cc} A & \overrightarrow{0} \\ \overline{\pi}^t & \pi_4 \end{array}\right), H^{-1} = \left(\begin{array}{cc} A^{-1} & \overrightarrow{0} \\ -\frac{1}{\pi_4} \overline{\pi}^t A^{-1} & \frac{1}{\pi_4} \end{array}\right), H^{-t} = \left(\begin{array}{cc} A^{-t} & -\frac{1}{\pi_4} A^{-t} \overline{\pi} \\ \overrightarrow{0} & \frac{1}{\pi_4} \end{array}\right) \end{equation}

\begin{equation} H^{-t} \Pi = \left(\begin{array}{cc} A^{-t} & -\frac{1}{\pi_4} A^{-t} \overline{\pi} \\ \overrightarrow{0} & \frac{1}{\pi_4} \end{array}\right) \left(\begin{array}{c} \overline{\pi} \\ \pi_4 \end{array}\right) = \left(\begin{array}{c} \overrightarrow{0} \\ 1 \end{array}\right) = \Pi_{\infty} \end{equation}

$\overline{\pi}$ is the orientation of $\Pi$ and $\Pi$ is a homogeneous representation, so $\Pi \sim \frac{1}{\pi_4} \Pi$ .

Transforms

[1] Hartley, R. and Zisserman, A. (2003) Multiple View Geometry in Computer Vision. 2nd Edition, Cambridge University Press, Cambridge.