Alternate derivation: from the epipolar line

Faugeras [2] approaches the problem from a slightly different direction by using the fact that the point $\ensuremath{{\bf m}} _2$ must lie on the epipolar line corresponding to $\ensuremath{{\bf m}} _1$:

$$\ensuremath{{\bf m}} _2^T \ensuremath{{\bf l}} = 0.$$ (11)

That line contains two points, the epipole $\ensuremath{{\bf e}}$ (the projection of the first camera's optical center into the second camera) and the point at infinity associated with $\ensuremath{{\bf m}} _1$:

$$\ensuremath{{\bf l}} = \ensuremath{{\bf e}}\times \ensuremath{{\bf m}} _\infty.$$

In [2, pp. 40-41] it is shown that the epipole is given by

$$\ensuremath{{\bf e}} = \ensuremath{{\tilde P}} _2 \left[ \mat... ...th{{\tilde P}} _2 \left[ \matrix{-P_1^{-1} p_1 \cr 1} \right],$$

and the point at infinity by

$$\ensuremath{{\bf m}} _\infty = P_2 P_1^{-1} \ensuremath{{\bf m}} _1.$$

Therefore, the epipolar line is:

$$\begin{eqnarray*}\ensuremath{{\bf l}} & = & \ensuremath{{\bf e}}\times \ensurema... ...uremath{{\bf t}}\times (A_2 R A_1^{-1}) \ensuremath{{\bf m}} _1, \end{eqnarray*}$$

where we have used the substitutions in (5). Combining with (11), we get the desired result:

$$F =[A_2 \ensuremath{{\bf t}} ]_x A_2 R A_1^{-1}.$$