How is the fundamental matrix (in computer vision) derived? In this paper, page 92, the so called fundamental matrix in computer vision is derived. 
Some notation:
$M = (x,y,z)^T$ is a 3d point and $
\left[
    \begin{array}{cc}
      M\\
      1
    \end{array}
\right] 
$ represents the homogeneous coordinate $(x,y,z,1)^T$
There are two pinhole cameras defined as:
$P_i$ a 3x4 projection matrix for camera $i$:
$$m_i = P_i \left[
    \begin{array}{cc}
      M\\
      1
    \end{array}
\right] 
= s_i \left[ 
    \begin{array}{cc}
      u_i\\
      v_i\\
      1
    \end{array}
\right]$$
The projection can be decomposed into $A_i\left[R_i\ t_i\right]$, where $A_i$ are the intrinsic parameters of the camera (an upper 3x3 triangular matrix), $R_i$ is a 3x3 rotation matrix (rotates the camera relative to the coordinate systems axises), and $t_i$ is a 3x1 translation vector. Note: $\left[R_i\ t_i\right]$ is a 3x4 matrix ($t$ is the last column).
$A_i$ defines the camera intrinsic parameters (focal length, scale factors, etc) I don't think these parameters are relevant to my question ,but the matrix looks like this:
$$
A_i = \begin{bmatrix}a&b&c\\0&d&e\\0&0&1\end{bmatrix}
$$
The first camera ($i=1$) is positioned at origo and its axises aligns with the coordinate axises:
$$P_1 = A_1 \left[I\ \Bbb{0} \right]$$
The second camera is translated by $t$ and rotated by $R$ and has its own intrinsic parameters $A_2$:
$$P_2 = A_2\left[R\ t \right]$$
With these notations, we have the following two equations to project the 3d point $M$ to the plane of each camera:
$$
s_1m_1 = A_1[I\ 0] \left[
    \begin{array}{cc}
      M\\
      1
    \end{array}
\right] 
\tag{1}
$$
$$
s_2m_2 = A_2[R\ t] \left[
    \begin{array}{cc}
      M\\
      1
    \end{array}
\right] 
\tag{2}
$$
Epipolar geometry
Given two images from two different cameras of the same scene, a ray from the camera center to a point $M$ will project as a line in each camera plane. $M$ is projected to $m_1$. There is a line $l_{m_1}$ in the other camera plane in which $m_2$ must be on. This line is called the epipolar line.
An image explains it better:

So if we know $m_1$, and we need to find $m_2$ (the corresponding point to $m_1$), then we could limit the search to the epipolar line $l_2$ (which goes through $e_2$ and $m_2$). In this way, we can search in one dimension instead of two for the corresponding points in the other camera image. Of course we cannot find $l_2$ via $m_2$ since we are looking for $m_2$.
The fundamental matrix $F$ is defined such that: $l_2 = F\ m_1$. And the constraint that $m_2$ will be on this line is: $m_2^T l_2 = 0$. 
So the constraint to find $F$ is: $m_2^T F m_1 = 0$. 
Unfortunately I'm not able to see (neither geometrically nor algebraically) how $F$ is derived / deduced.
Question
From these two equations, in the paper (and I've seen it in other papers as well), the following is deduced (which proofs the existence of $F$):
$$ m_2^{T}A_2^{-T}TRA_1^{-1}m_1= 0 $$
with the note: "by eliminating $M$, $s1$ and $s2$", and: $T$ is an antisymmetric matrix defined by $t$ such that, where $\times$ is the cross product, $Tx = t \times x$ for all 3D vectors.
How exactly can these be eliminated given (1) and (2)?
I thought I could eliminate $M$ this way:
$
A_2^{-1}s_2m_2 = RM+t
$
$RA_1^{-1}s_1m_1 + t = RM+t$
$RA_1^{-1}s_1m_1 + t = A_2^{-1}s_2m_2$
At this point, I suppose T is used since $Tt$ would be $t$ cross $t$ which is zero. 
$TRA_1^{-1}s_1m_1 = TA_2^{-1}s_2m_2$
Thankful for a hint or two.
 A: You were almost there. From equations (1) and (2), we have $s_1A_1^{-1}\tilde{\mathbf m}_1 = M$ and $s_2A_2^{-1}\tilde{\mathbf m}_2 = RM + \mathbf t = s_1RA_1^{-1}\tilde{\mathbf m}_1 + \mathbf t$. The cross product of two vectors is orthogonal to them both, so we have $$\begin{align} 0 = (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge s_2A_2^{-1}\tilde{\mathbf m}_2) &= (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge(s_1RA_1^{-1}\tilde{\mathbf m}_1 + \mathbf t)) \\
&= (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge s_1RA_1^{-1}\tilde{\mathbf m}_1 + \mathbf t\wedge\mathbf t) \\
&= (s_2A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge s_1RA_1^{-1}\tilde{\mathbf m}_1) \\
&= s_1s_2(A_2^{-1}\tilde{\mathbf m}_2)^T(\mathbf t\wedge RA_1^{-1}\tilde{\mathbf m}_1) \\
&= s_1s_2\tilde{\mathbf m}_2^T(A_2^{-1})^T TRA_1^{-1}\tilde{\mathbf m}_1.\end{align}$$ We now drop the irrelevant scale factors and are left with the desired equation.
A: It seems helpful to note that for any pair of vectors $x,y$ on $3$ components, we have
$$
x^T[m_2^{T}A_2^{-T}TRA_1^{-1}m_1]y = \\
(A_2^{-1}m_2 x)^T(TRA^{-1}m_1y) = \\
(A_2^{-1}m_2 x)^T(t \times [RA^{-1}m_1y])
$$
It would be sufficient to prove that the image of $A_2^{-1}m_2$ lies in the plane spanned by $t$ and the image of $RA^{-1} m_1$, assuming that I've interpreted the data types here correctly.
