Nov 27, 2013 and Dec 6, 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} \label{eq org} 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 $\mathbb{R}^d$.

Definition. We call an operator $H$ local, if for each variable $f_{\xi}$ for $\xi=1,…m$ there is a point $x_{j}$ in $\mathbb{R}^d$ such that $f_{\xi}$ under the operation of $H$ is influenced by variables $\varphi_j$ only if they belong to the point $x_{j}$.

Examples. Consider the matrix \begin{equation} A = \left( \begin{array}{cc} a_{11} & a_{12}
a_{21} & a_{22} \end{array} \right) := \left( \begin{array}{cc} 1 & 1
1 & -1 \end{array} \right) \end{equation} where $\varphi_{1}$ and $\varphi_{2}$ do belong to two different points $x_1$ and $x_2$ in physical space $\mathbb{R}^2$. Then $A$ is not a local matrix, since the first component of $A\varphi$ is influenced by both $\varphi_1$ and $\varphi_2$, i.e. by variables located in two different points $x_1$ and $x_2$ in space. The matrix \begin{equation} B = \left( \begin{array}{cc} b_{11} & b_{12}
b_{21} & b_{22} \end{array} \right) := \left( \begin{array}{cc} 0 & 1
3 & 0 \end{array} \right) \end{equation} however is local, since the output $f_1$ is only influenced by $\varphi_2$ which is located at $x_2$ and $f_{2}$ is only influenced by $\varphi_1$ located at $x_1$.

Lemma. An operator is local if and only if by reordering of the variables it can be transformed into a diagonal operator.

Remark. A reordering operation is equivalent to the application of a permutation matrix $P$.

Proof.

$\Box$

Question. Our question is: can we find transformations $T: X \rightarrow X$ of $X$ and $S: Y \rightarrow Y$ of $Y$, such that $\tilde{H} := S H T^{-1}$ is local?

An approach using Singular Value Decomposition. By singular value decomposition SVD we have \begin{equation} \label{svd} H = U \Lambda V^{T}, \end{equation} where $\Lambda$ is a digonal matrix and the matrices $U$ and $V$ consist of orthonormal vectors. When $V^T$ and $U$ are invertible, we can proceed as follows. Now, we define $$ T := V^{T} $$ and $$ S = U^{-1} $$ Then, we have \begin{equation} S H T^{-1} = U^{-1} H (V^{T})^{-1} = \Lambda. \label{work1} \end{equation} We have found the desired transformation by SVD. $\Box$

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

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

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

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