Direct linear transformation: Difference between revisions

Revision as of 05:50, 15 May 2020

Revision as of 19:20, 5 October 2020

Line 24:

== Example ==

Suppose that <math> k \in \mathbb{N} </math>. Let <math> \mathbf{x}_{k} = (x_{1k}, x_{2k}) \in \mathbb{R}^{2} </math> and <math> \mathbf{y}_{k} = (y_{1k}, y_{2k}, y_{3k}) \in \mathbb{R}^{3} </math> be two sets of known vectors, and thewe problem iswant to find the <math> 2 \times 3 </math> matrix <math> \mathbf{A} </math> such that

: <math> \alpha_{k} \, \mathbf{x}_{k} = \mathbf{A} \, \mathbf{y}_{k} </math> &nbsp; for <math> \, k = 1, \ldots, N </math>

where <math> \alpha_{k} \neq 0 </math> is the unknown scalar factor related to equation ''k''.

Line 36:

and multiply both sides of the equation with <math> \mathbf{x}_{k}^{T} \, \mathbf{H} </math> from the left

:<math> \begin{align}

:<math> \alpha_{k} \, \mathbf{x}_{k}^{T} \, \mathbf{H} \, \mathbf{x}_{k} = \mathbf{x}_{k}^{T} \, \mathbf{H} \, \mathbf{A} \, \mathbf{y}_{k} </math> &nbsp; for <math> \, k = 1, \ldots, N .</math>

(\mathbf{x}_{k}^{T} \, \mathbf{H}) \, \alpha_{k} \, \mathbf{x}_{k} &= (\mathbf{x}_{k}^{T} \, \mathbf{H}) \, \mathbf{A} \, \mathbf{y}_{k} \\

:<math> \alpha_{k} \, \mathbf{x}_{k}^{T} \, \mathbf{H} \, \mathbf{x}_{k} &= \mathbf{x}_{k}^{T} \, \mathbf{H} \, \mathbf{A} \, \mathbf{y}_{k} </math> &nbsp; for <math> \, k = 1, \ldots, N .</math>

\end{align}

</math>

Since <math> \mathbf{x}_{k}^{T} \, \mathbf{H} \, \mathbf{x}_{k} = 0, </math> the following homogeneous equations, which no longer contain the unknown scalars, are at hand

: <math> 0 = \mathbf{x}_{k}^{T} \, \mathbf{H} \, \mathbf{A} \, \mathbf{y}_{k} </math> &nbsp; for <math> \, k = 1, \ldots, N .0</math>

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>: