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misc2 [2023/03/28 09:14] (current) – external edit 127.0.0.1
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 +====== 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 at most one 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 $x_1$. The matrix
 +\begin{equation}
 +\label{C example}
 +C = \left( \begin{array}{cc} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array} \right)
 +:= \left( \begin{array}{cc} 1 & 0 \\ 3 & 0 \end{array} \right)
 +\end{equation}
 +is local as well, since both output variables $f_{1}$ and $f_{2}$ are influenced by $\varphi_1$
 +only. 
 +
 +**Lemma.** 
 +If we have a local operator for which each measusment is influenced by a different point
 +$x_{j} \in \mathbb{R}^d$, then by reordering of the variables 
 +it can be transformed into a diagonal operator. In general, when a point influences two or more
 +output variables, diagonalization by reordering is not possible.  
 +
 +//Remark.// A reordering operation is equivalent to the application of a permutation 
 +matrix $P$, i.e. a matrix which has exactly one element 1 in each row and column, with 
 +all other elements zero. 
 +
 +//Proof.// We first assume that in the state space $X = \mathbb{R}^n$ each element belong
 +to one and only one point $x_{j}$, $j=1,...,n$ with $x_{j}\in \mathbb{R}^d$, where all $x_{j}$
 +are different. Then, each column of the matrix is multiplied with a variable which 
 +belongs to a different point $x_j \in \mathbb{R}^d$. The output variable $f_{\ell}$, $\ell=1,...,m$ 
 +is influenced by the entries in the $\ell$-th row of the matrix $H$. Thus, this is local
 +if and only if at most one of the entries is non-zero. This applies to every row, i.e. in 
 +each row there is at most one non-zero element. By the assumption that different measurements
 +are influenced by different points, this means that there can be at most one nonzero entry
 +in each column as well. But that means that the operator $H$ looks like a scaled version of 
 +a permutation matrix $P$, with scaling $0$ allowed. Clearly, by reordering we can make this
 +into a diagonal matrix. In general, we take (\ref{C example}) as counter example, 
 +and the proof is complete $\Box$ \\
 +
 +
 +**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
 +
 +
 +
 +