In the book "Multiple View Geometry in Computer Vision" of R. Hartley, A. Zisserman there is an Algorithm 4.7, describing how to find affine homography. I give link the to the book later, but I also describe in words now.
We take the affine transformation matrix of the form $A=$ $\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ 0 & 0 & 1 \end{bmatrix}$
and we translate our initial point correspondences to the origin, so that we get rid of $a_{13}$ and $a_{23}$ elements, they become $0$. Then we introduce matrix of points embedding $M_i^T=(x_i,y_i,x_i',y_i')$, where $x_i,y_i$ are coordinates of point correspondences in the first image and $x'_i,y'_i$ -- of the second image. Then we perform SVD on the matrix M, since we are searching solutions for the equation $Mv=0$, where $v=[v_1,v_2]$ are the vectors, which are the basis of the codimension-2 subspace variety in $\mathbb{R}^4$, formed by the correspondences. After we obtain the result of SVD, we take the 2 largest eigenvalues of the matrix $V$, and recover, estimate our submatrix of interest $A' =$ $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix}$ by $A'=CB^{-1}$ where $C$ and $B$ are, if $V_1$ and $V_2$ are the largest eigenvectors: $[V_1,V_2]$=$\begin{bmatrix} B\\ C \end{bmatrix}$. So that means we obtain $V$ matrix of SVD, which is 4x4, take only 2 first columns, get 4x2 matrix, divide it vertically in half, getting $B$ and $C$ both 2x2 and then evaluate $A'=CB^{-1}$.
And my question and problem is exactly to understand, why we find $C$ and $B$ matrixes and how can they provide in the end solution to get $A'$,submatrix of an affine transformation matrix.
(I understand the DLT procedure of homography estimation and that there we also take SVD and the solution is the smallest singular value) but this procedure, and why 2 largest singular values required, I don't understand.
https://books.google.it/books?id=si3R3Pfa98QC&lpg=PP1&hl=it&pg=PA130#v=onepage&q&f=false
Ok, I think I got closer to understanding, why $A'=CB^{-1}$: $\begin{bmatrix} x'_{i}\\ y'_{i} \\ 1 \\ \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & 0 \\ 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} x_{i} \\ y_{i} \\ 1\\ \end{bmatrix}$ from which we get, just by multiplying (omitting the scaling $w$ factor, because it is 1): \begin{equation*} \begin{cases} x'_{i}=a_{11}x_1+a_{12}y_1\\ y'_{i}=a_{21}x_1+a_{22}y_1\\ \end{cases} \end{equation*} then we get \begin{equation*} \begin{cases} a_{11}x_1+a_{12}y_1-x'_{i}=0\\ a_{21}x_1+a_{22}y_1-y'_{i}=0\\ \end{cases} \end{equation*} and if we use the embedding $M_i^T$ on the right we get the matrix $(x_i,y_i,x'_i,y'_i) \begin{bmatrix} a_{11} & a_{21}\\ a_{12} & a_{22}\\ -1 & 0 \\ 0 & -1 \\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}_i$ if we rewrite the system of equations once again \begin{equation*} \begin{cases} a_{11}x_1+a_{12}y_1=x'_{i}+0 \cdot y'_{i}\\ a_{21}x_1+a_{22}y_1=y'_{i}+0 \cdot x'_{i}\\ \end{cases} \end{equation*} $ \underbrace{\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{bmatrix}}_B \begin{bmatrix} x_{i}\\ y_{i}\\ \end{bmatrix}= \underbrace{ \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}}_C \begin{bmatrix} x'_{i}\\ y'_{i}\\ \end{bmatrix} $
But I still don't understand the moment about largest singular values.