# why images are related by an affine transformation in following specific case?(background in computer vision required)

In two view case, the epipolar geometry can be constructed as follows:

According to Multiple View Geometry in computer vision from page 155 to page 157: $$x=PX=K[I,0]X,x'=P'X=K'[R,t]X$$ where $K,K'$ are internal parameters matrices of two cameras, and $R$ is the rotation matrix of camera centered at $C'$ relative to camera centered at $C$(given that the world frame coincides with the frame of camera centered at $C$). Now consider following case:

When two images are taken with only camera translation and rotation aligned with the image plane (no perspective skew), then they can be related by an affine matrix, denote this affine matrix as $A$. So we have $$x'=Ax=\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ 0 & 0 & 1\end{bmatrix}x$$ how to draw above conclusion?

Analogous derivation can be found in the section "2.feature matching" of paper Recongnising Panoramas

Where does it say in the book that "When two images are taken with only camera translation and rotation aligned with the image plane (no perspective skew), then they can be related by an affine matrix" ?

I tried to prove it, but the following seems to prove the opposite, that those points are not related by any homography at all.

Because the motion is planar, and let it be in XY plane, with Z axis perpendicular to the image plane, the rotation is:

$$R=\begin{pmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{pmatrix}$$

The inhomogeneous translation vector is:

$$C=\begin{pmatrix}C_1 \\ C_2 \\ 0 \end{pmatrix}$$

Then the derivation is as follows:

$$x=K[I | 0]\begin{pmatrix}X_1 \\ X_2 \\ X_3 \\ 1 \end{pmatrix}$$

$$K^{-1}x=\begin{pmatrix}X_1 \\ X_2 \\ X_3\end{pmatrix}$$

$$x'=K'R\bigg[I \big| -\begin{pmatrix}C_1 \\ C_2 \\ 0 \end{pmatrix}\bigg]\begin{pmatrix}X_1 \\ X_2 \\ X_3 \\ 1 \end{pmatrix} = K'R\Bigg(\begin{pmatrix}X_1 \\ X_2 \\ X_3\end{pmatrix} - \begin{pmatrix}C_1 \\ C_2 \\ 0 \end{pmatrix}\Bigg)= K'R\Bigg(K^{-1}x - \begin{pmatrix}C_1 \\ C_2 \\ 0 \end{pmatrix}\Bigg)$$

This is not a linear transformation, therefore is not homography. The same thing follows of course if a motion is simply a translation. If there is a rotation only, or change in camera parameters K, or both, then points will be related under homography. But if a camera center changes, it is no longer true.