the pipeline summary is at the end
Note that the last coordinates of x is z, which means “division” in homogenous coordinates
The “Division” matrix of homogenous coordinates including in W:
Denote $e = [0, 0, 1]$ , we can write the above leftmost matrix as P, where E is the identity matrix:
$$ P= \begin{bmatrix} & & & 0\\ & E & & 0\\ & & & 0\\ & e & &0 \end{bmatrix} $$
So we can write
$$ W = P \begin{bmatrix}
R & t\\ 0 &1 \end{bmatrix} = \begin{bmatrix} R & t \\ eR & et \end{bmatrix} $$
where R t are rigid transformation parts.
Denote $H = \begin{bmatrix} \Sigma & p\\ 0 & 1 \end{bmatrix}$, $\Sigma$ is the RS in the paper, the upper left of H. p is the center vector
We have
$$ WH = \begin{bmatrix} R\Sigma & Rp+t \\ eR\Sigma & eRp+et \end{bmatrix} $$
Denote $R = \begin{bmatrix} r_1\\ r_2\\ r_3 \end{bmatrix}, \; \Sigma = [s_1, s_2, 0]$ (as the paper, s1, s2 are vectors, the third column of the matrix is all 0)
( r are rows, s are columns)
We can expand WH as
$$ WH = \begin{bmatrix} r_1s_1 & r_1s_2 & 0 & r_1p+t_1\\ r_2s_1 & r_2s_2 & 0 & r_2p+t_2\\ r_3s_1 & r_3s_2 & 0 & r_3p+t_3\\ r_3s_1 & r_3s_2 & 0 & r_3p+t_3 \end{bmatrix} = [h_1, h_2, 0, h_4] $$