Estimate for ,

where **=** is equality up to scale.
** Minimum number of correspondences:**
4 points (non-collinear) or 4 lines (non-concurrent).

** correspondences**

**n = 4** will be covered as a special case.
Each point correspondence generates two linear equations for
the elements of (dividing by the third component
to remove the unknown scale factor)

and multiplying out

Then points generates **2 n** linear equations,
which are sufficient to solve for .
The problem with projective transformations is that
the matrix is only recovered up to scale.

** Homogeneous Solutions**
There are two methods of dealing with the unknown scale
factor in a homogeneous matrix

- Choose one of the matrix elements to have a certain value. For example, .
- Solve for the matrix up to scale.

** Method I**

If , then the above equations can be written:

for 4 points. A linear solution is then obtained,
in the usual manner, by solving the set of
linear simultaneous equations. Similarly, for **n > 4** points, a solution
can be obtained using a pseudo-inverse.

The problem with this approach is that if the true solution is , then this cannot be reached. Consequently, a poor quality estimate of will be obtained.

** Method II**

The equations,

can be rearranged as

where is the matrix written as a vector. For 4 points,

which has the form , with a matrix. The solution is the (one dimensional) kernel of .

For **n > 4** real point correspondences,
is a matrix,
and there will not be a solution to
. In this case, a sensible procedure
is to again minimise the residuals. It can be shown that
the vector that minimises subject
to ,
is the eigenvector with least eigenvector of .
In the case of **n = 4** the least eigenvalue is zero, and the
eigenvector is the kernel of .

Wed Apr 16 00:58:54 BST 1997