Direct linear transformation: Difference between revisions

Revision as of 05:42, 16 October 2012

Revision as of 10:24, 17 October 2012

Line 43:

In order to solve <math> \mathbf{A} </math> from this set of equations, consider the elements of the vectors <math> \mathbf{x}_{k} </math> and <math> \mathbf{y}_{k} </math> and matrix <math> \mathbf{A} </math>:

: <math> \mathbf{x}_{k} = \begin{pmatrix} x_{1k} \\ x_{2k} \end{pmatrix} </math>, &nbsp; <math> \mathbf{y}_{k} = \begin{pmatrix} y_{1k} \\ y_{2k} \\ y_{3k} \end{pmatrix} </math>, &nbsp; and &nbsp; <math> \mathbf{A} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} </math>

and the above homogeneous equation becomes

Line 61:

:<math> \mathbf{0} = \mathbf{B} \, \mathbf{a} </math>

where <math> \mathbf{B} </math> is a <math> N \times 6 </math> matrix which holds the vectors <math> \mathbf{b}_{k} </math> in its rows. This means that <math> \mathbf{a} </math> lies in the [[null space]] of <math> \mathbf{B} </math> and can be determined, for example, by a [[singular value decomposition]] of <math> \mathbf{B} </math>; <math> \mathbf{a} </math> is a right singular vector of <math> \mathbf{B} </math> corresponding to a singular value that equals zero. Once <math> \mathbf{a} </math> has been determined, the elements of <math> \mathbf{A} </math> can be found by a simple rearrangement from a 6-dimensional vector to a <math> 2 \times 3 </math> matrix. Notice that the scaling of <math> \mathbf{a} </math> or <math> \mathbf{A} </math> is not important (except that it must be non-zero) since the defining equations already allow for unknown scaling.In

In practice the vectors <math> \mathbf{x}_{k} </math> and <math> \mathbf{y}_{k} </math> may contain noise which means that the similarity equations are only approximately valid. As a consequence, there may not be a vector <math> \mathbf{a} </math> which solves the homogeneous equation <math> \mathbf{0} = \mathbf{B} \, \mathbf{a} </math> exactly. In these cases, a [[total least squares]] solution can be used by choosing <math> \mathbf{a} </math> as a right singular vector corresponding to the smallest singular value of <math> \mathbf{B}. </math>