Nov 27, 2013.

We consider some observation operator $H: X\rightarrow Y$ defined on the space $X$. Our goal is to solve an operator equation of the type \begin{equation} H \varphi = f \end{equation} with $f \in Y$ given. Here, we think of $X$ as a finite dimensional space. Then, it is isomorphic to $\mathbb{R}^n$, with $n \in \mathbb{N}$. In the same way, the finite dimensional space $Y$ will be isomorphic to $\mathbb{R}^m$ for some $m \in \mathbb{N}$. In general, a linear operator $H$ will consist of sums of multiples of the elements of vectors $\varphi \in X$. If we consider each element $\varphi_j$ of $$ \varphi = \left( \begin{array}{cc} \varphi_1
\vdots
\varphi_n \end{array} \right) $$ for $j=1,…n$ to belong to a particular point $x_j$ in physical space $\mathbb{R}^d$ of dimension $d \in \{1,2,3\}$, then in general the operator $H$ is not local in the sense that its outcome belongs to individual points in space and only depends on the input in these points. In general, the space $Y$ will not be local in the sense that each of its variables belongs to one and only one point in the physical space $\mathbb{R}^d$.

We call an operator $H$ local, if each variable $f_{\xi}$ for $\xi=1,…m$ is influenced by variables $x_{}$

Our question is: can we find some transformation $T: X \rightarrow X$, such that $\tilde{H} := H T^{-1}$ is local?

By singular value decomposition SVD we have $$ H = U \Lambda V^{T}, $$ where $\Lambda$ is a digonal matrix and the matrices $U$ and $V$ consist of orthonormal vectors. Now, we define $$ T := V^{T} $$ and $$ \tilde{f} = U \overline f

$$ \tilde{\varphi} = T\varphi$$

$$ H\varphi = HT^{-1}T\varphi $$

$$ = H^{\sim}T\varphi = H^{\sim}\varphi^{\sim} $$

such that $H^{\sim}$ is local