Another commonly used normalization is to set *F=1*. If we use the same
notations as in the last subsection, the problem becomes to minimize the
following function:

where is the sixth element of vector , i.e., .

Indeed, we seek for a least-squares solution to under the constraint . The equation can be rewritten as

where is the matrix formed by the first *(n-1)* columns of
, is the last column of and is the
vector . The problem can now be solved by the technique
described in sect:Ax=b.

In the following, we present another technique for solving this kind of problems, i.e.,

based on eigen analysis, where we consider a general formulation, that is
is a matrix, is a *m*-vector, and is the
last element of vector . The function to minimize is

As in the last subsection, the symmetric matrix can be decomposed as in (5), i.e., . Now if we normalize each eigenvalue and eigenvector by the last element of the eigenvector, i.e.,

where is the last element of the eigenvector , then the last element of the new eigenvector is equal to one. We now have

where and . The original problem (7) becomes:

Find such that is minimized with

subject to.

After some simple algebra, we have

The problem now becomes to minimize the following unconstrained function:

where is the Lagrange multiplier. Setting the derivatives of with respect to through and yields:

The unique solution to the above equations is given by

where . The solution to the problem (7) is given by

Note that this normalization (*F=1*) has singularities for all conics going
through the origin. That is, this method cannot fit such conics because they
require to set *F=0*. This might suggest that the other normalizations are
superior to the *F=1* normalization with respect to singularities. However, as
shown in [20], the singularity problem can be overcome by shifting the
data so that they are centered on the origin, and better results by setting
*F=1* has been obtained than by setting *A+C=1*.

Thu Feb 8 11:42:20 MET 1996