Homography matrix problem

1. How is homography matrix calculated?
2. How do you use homography matrix?
3. Why does homography matrix have 8 degrees of freedom?
4. Why do you need 4 points for homography?

How is homography matrix calculated?

What is the homography matrix? The homography matrix is a 3x3 matrix but with 8 DoF (degrees of freedom) as it is estimated up to a scale. It is generally normalized (see also 1) with h_33 = 1 or h_11^2 + h_12^2 + h_13^2 + h_21^2 + h_22^2 + h_23^2 + h_31^2 + h_32^2 + h_33^2 = 1 .

How do you use homography matrix?

This spatial relationship is represented by a transformation known as a homography, H, where H is a 3 x 3 matrix. To apply homography H to a point p, simply compute p' = Hp, where p and p' are (3-dimensional) homogeneous coordinates. p' is then the transformed point.

Why does homography matrix have 8 degrees of freedom?

Also, homography is defined upto a scale (c in above equation) i.e. it can be changed by a non zero constant without any affect on projective transformation. Thus, homography has 8 degree of freedom even though it contains 9 elements (3x3 matrix) i.e. the number of unknowns that need to be solved for is 8.

Why do you need 4 points for homography?

Given that 1 point-to-point correspondence represents 2 constraints, then 4 point-to-point correspondences corresponds to 8 constraints. Given this and given that homographies have 8 degrees of freedom, at least 4 point-to-point correspondences are necessary to estimate a homography.