The optical center

• The projective camera $\tilde{\bf P}$ has a rank 3 whereas it has 4 columns. Clearly, it has a on dimensional right null space. Suppose the null space is generated by the 4-vector ${\bf C}$ , that is
  $$\tilde{\bf P}{\bf C} = {\bf0}$$

• Claim: ${\bf C}$ is the optical center of the camera $\tilde{\bf P}$ .
  Proof. Consider the line containing ${\bf C}$ and any other point ${\bf A}$ in 3-space. Points on this line can be represented as

  $${\bf X}(\lambda) = \lambda {\bf A} + (1 - \lambda){\bf C}$$

  Under, the mapping ${\bf x}= \tilde{\bf P}{\bf X}$ , points on this line are projected to

  $${\bf x}= \tilde{\bf P}{\bf X}(\lambda) = \lambda \tilde{\bf P}{\bf A} + (1 - \lambda)\tilde{\bf P}{\bf C} = \lambda \tilde{\bf P}{\bf A}$$

  since $\tilde{\bf P}{\bf C} = {\bf0}$ . Since every point on the line are mapped on to the same image point, the line must be a ray through the camera center. It follows that ${\bf C}$ is the camera center because for all choices of ${\bf A}$ the line passes through the optical center. $\qedsymbol$

• Writing
  $$\tilde{\bf P} = \left[ {\bf P} \mid -{\bf P}{\bf t} \right]$$
  where ${\bf P}$ is $3 \times 3$ non-singular we have that ${\bf t}$ is the optical center.
  $$\left[ {\bf P} \mid -{\bf P}{\bf t} \right] \left[ \begin{array}{c} {\bf t} \\ 1 \end{array} \right] = {\bf0}$$

• An image point ${\bf x}$ defines a line $\lambda {\bf P}^{-1}{\bf x} + {\bf t}$ in 3-space:
  $${\bf X} = \left[ \begin{array}{c} X \\ Y\\ Z\\ 1 \end{array} \ri... ...ray}\right] + \left[ \begin{array}{c} t_x \\ t_y\\ t_z\\ 1 \end{array} \right]$$
  Clearly all points ${\bf X}$ project to ${\bf x}$ under ${\bf x}= \tilde{\bf P}{\bf X}$ .