This is an old revision of the document!
Just for Working Sessions
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 $a_{11}$ and $a_{12}$ 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