Short answer: Because epipoles themselves lie on the epipolar lines.

Here is the intuition:
A fundamental matrix is given by the equation [math] x_r^T F x_l = 0 [/math]. Now consider an epipolar line [math] l' = F x_l [/math]. The right epipole [math] e_r [/math] lies on this line, so [math] e_r^T l' = 0 [/math] or [math] e_r^T F x_l = 0 [/math] for all [math] x_l [/math].

This implies that [math] e_r^T F = 0 [/math]. Similary, one can prove that [math] F e_l = 0 [/math]. Hence [math] F [/math] has a nullspace which is not just the zero vector. So [math] F [/math] is rank deficient.

The proof that [math] F [/math] has rank 2 actually comes from the fact that [math] F [/math] is constructed from the essential matrix [math] E [/math]:

[math] F = (K_r^{-1})^T E K_l^{-1} [/math] where the [math] K [/math]'s are intrinsic matrices of the two cameras. Now, [math] E = [T_{\times}]R [/math] where [math] R [/math] is the rotation matrix relating the two camera co-ordinate systems and [math] [T_{\times}] = \begin{pmatrix} 0 & -t_z & t_y\\ t_z & 0 & -t_x\\ -t_y & t_x & 0 \end{pmatrix} [/math]. A little bit of manipulation will show that one column of [math] [T_{\times}] [/math] is a linear combination of the other two columns. So [math] [T_{\times}] [/math] has rank 2.

Hence any matrix that you construct by multiplying other matrices with [math] [T_{\times}] [/math] (such as [math] E [/math] and [math] F [/math]) will also have rank 2.