A homography has only 8 degrees of freedom (since it is scale invariant). As a result, we need 8 constraints to solve for the homography matrix H. Every point correspondence gives exactly 2 constraints, and we therefore only need 4 correspondences to provide enough constraints to get H.
Going back to this slide, our matrix A would be $8 \times 9$ in this case; 8 rows representing the 8 constraints. We would then compute the smallest right singular vector to find our solution to the homogeneous least squares problem Ax = 0.
why is it 4 point correspondences?
A homography has only 8 degrees of freedom (since it is scale invariant). As a result, we need 8 constraints to solve for the homography matrix H. Every point correspondence gives exactly 2 constraints, and we therefore only need 4 correspondences to provide enough constraints to get H.
Going back to this slide, our matrix A would be $8 \times 9$ in this case; 8 rows representing the 8 constraints. We would then compute the smallest right singular vector to find our solution to the homogeneous least squares problem Ax = 0.