Alternate derivation: algebraic

To describe the relationship between R, $\ensuremath{{\bf t}}$, A₁, and A₂ more exactly, and to connect the above equations with those found in [6], we offer the following algebraic derivation.

Recall that a point $\ensuremath{{\bf M}}$ produces an image $\ensuremath{{\bf m}}$ through the equation $\ensuremath{{\bf m}} = \ensuremath{{\tilde P}}\ensuremath{{\bf M}}$. Without loss of generality, we can assume that $\ensuremath{{\bf M}}$ is given with respect to the first camera's coordinate frame to yield the following two imaging equations:

$$\begin{eqnarray*}\lambda_1 \ensuremath{{\bf m}} _1 & = & A_1 \left[\matrix{\ensu... ...matrix{R & \ensuremath{{\bf t}} } \right] \ensuremath{{\bf M}} , \end{eqnarray*}$$

where $\lambda_1$ and $\lambda_2$ are scale factors, $\ensuremath{{\bf I}}$ is the $3 \times 3$ identity matrix and $\ensuremath{{\bf0}}$ is the $3 \times 1$ null vector. By letting $\ensuremath{{\bf M}} = \left[\matrix{\ensuremath{{\bf\hat M}} ^T & 1}\right]^T$ ( $\ensuremath{{\bf\hat M}}$ is $3 \times 1$), we achieve the following relation:

    $$\begin{eqnarray*}\lambda_2 \ensuremath{{\bf m}} _2 & = & A_2 \left[\matrix{R & \... ...da_1 R A_1^{-1} \ensuremath{{\bf m}} _1 + \ensuremath{{\bf t}} . \end{eqnarray*}$$(8) (9)

Geometrically, this equation says that the vector on the left is a linear combination of the two vectors on the right. Therefore, they are all coplanar, and the vector $\ensuremath{{\bf v}} = \ensuremath{{\bf t}}\times R A_1^{-1} \ensuremath{{\bf m}} _1$is perpendicular to that plane:

$$\begin{eqnarray*}\lambda_2 (A_2^{-1} \ensuremath{{\bf m}} _2)^T \ensuremath{{\bf... ...math{{\bf t}}\times R) A_1^{-1})\ensuremath{{\bf m}} _1 & = & 0, \end{eqnarray*}$$

which is identical to (6).

Similarly, the vector $\ensuremath{{\bf w}} = A_2 \ensuremath{{\bf t}}\times A_2 R A_1^{-1} \ensuremath{{\bf m}} _1$is perpendicular to the vectors in (8):

$$\begin{eqnarray*}\lambda_2 \ensuremath{{\bf m}} _2^T \ensuremath{{\bf w}} & = & ... ...h{{\bf t}}\times A_2 R A_1^{-1})\ensuremath{{\bf m}} _1 & = & 0. \end{eqnarray*}$$

This is a surprising result because it gives us a new and equivalent expression for F:

$$F =[A_2 \ensuremath{{\bf t}} ]_x A_2 R A_1^{-1},$$ (10)

which shows that F can be written as the product of an anti-symmetric matrix $[A_2 \ensuremath{{\bf t}} ]_x$ and an invertible matrix A₂ R A₁^-1 [4].