Two different approaches to estimating a homography matrix

Question

I've been using the following reference to compute a homography given set of N>=4 points.

https://cseweb.ucsd.edu/classes/wi07/cse252a/homography_estimation/homography_estimation.pdf

The approach is to construct a homogenous system of equations of the form $Ah = 0$ and solve it using SVD. $h$ is a 9 element vector containing the entries of the homography matrix.

Here's how this article constructs the $A$ matrix. Note the terms $a_x$ and $a_y$

Link to image 1

Now there's another article that attempts to solve for the homography matrix in the same way.

http://www.robots.ox.ac.uk/%7Evgg/presentations/bmvc97/criminispaper/node3.html

However, their $A$ matrix is the negative of the $A$ matrix from the first article.

Link to image 2

Question: Why are there two different formulations of the $A$ matrix? They yield two different homographies.

(Sorry for not embedding images, don't have enough rep)

Thanks!

Answer 1

The resulting systems of equations are equivalent, since $$ Ah = 0 \iff (-A)h=0. $$ Let $A=U_1\Sigma_1 V_1^{\top}$ and $-A=U_2\Sigma_2 V_2^{\top}$ be SVDs of $A$ and $-A$, respectively. The homogeneous least-squares solution $h$ is obtained as the rightmost column in either $V_1$ or $V_2$. Depending on the particular SVD implementation, the results may either be equal or differ by a minus sign. Since the homography matrix is only determined up to a scalar multiple anyway, this possible minus sign does not matter.