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--- misc2 [2013/12/06 16:10] – potthast
+++ misc2 [2023/03/28 09:14] (current) – external edit 127.0.0.1
@@ Line 1: / Line 1: @@
 ====== Just for Working Sessions ======
-Nov 27, 2013.
+Nov 27, 2013 and Dec 6, 2013.
 We consider some observation operator $H: X\rightarrow Y$ defined on the space $X$. Our goal
@@ Line 26: / Line 26: @@
 **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
+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 $x_{j}$.
@@ Line 32: / Line 32: @@
 Consider the matrix
 \begin{equation}
-A = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)
+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
+where $\varphi_{1}$ and $\varphi_{2}$ do belong to two different points $x_1$ and $x_2$ in physical
-space $\R^2$. Then $A$ is not a local matrix, since the first component of $A\varphi$ is
+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.
+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.** An operator is //local// if and only if by reordering of the variables
+**Lemma.**
-it can be transformed into a diagonal operator.
+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 $P$.
+matrix $P$, i.e. a matrix which has exactly one element 1 in each row and column, with
+all other elements zero.
-//Proof.//
-$\Box$ \\
+//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 $\Box$ \\