What if in the linear system the unknowns are not but the matrix ? Of course, if we have a sufficient number of such systems , , … with the same matrix and independent second hands, we can stack them, and solve

…

for the unknowns . A similar problem arises when looking for a planar homography:

(1)

given a sufficient number of couples of homogeneous points . It is to be noted that the problem is not identical as we have here meaning proportionality, not equality; this because, with homogeneous coordinates, and , with , do represent the same point, and as a consequence the homography matrix , while being full rank for regular homographies, is also defined up to a scale factor.

The correct method to solve this problem (eg Hartley&Zisserman, *Multiple View Geometry in Computer Vision*, §4.1) is to transform the system by applying a cross product with to both hands, so that the second hand vanishes and we are left with the set of homogeneous linear equations , containing for each sample three equations in the unknown entries of . Only two of such equations are linearly independent, so one can be discarded.

This method is the so called DLT, short for *Direct Linear Transformation*. It works fine, albeit the cross product smells of *deus ex machina* math trick. One can get the same result in a way that, in my opinion, is more elegant if one recognizes that the real problem is not the homogeneity of the involved quantities in itself, but the related fact that *are not observable*: the observables are the ratios instead (provided that we have no point at infinity). So one should manipulate the equations (1) to make these ratios emerge. Start with

whence:

Last, divide by :

and you are left with the *two* DLT equations.

### Like this:

Like Loading...

*Related*

The question is: do the two DLT equations we are left with ensure a single H?

Two equations are not enough; as said, you need four or more samples (couples of points) for a total of eight or more equations. Eight are sufficient even if H has nine entries, because H is defined up to a scale factor.

I add here that no three of the four points must be aligned.

Pingback: Reconstruction of 3D points from images in a calibrated binocular system | Tramp's Trifles