misc2
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misc2 [2013/12/06 16:20] – potthast | misc2 [2023/03/28 09:14] (current) – external edit 127.0.0.1 | ||
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**Definition.** | **Definition.** | ||
We call an operator $H$ //local//, if for each variable $f_{\xi}$ for $\xi=1, | We call an operator $H$ //local//, if for each variable $f_{\xi}$ for $\xi=1, | ||
- | 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 $x_{j}$. | influenced by variables $\varphi_j$ only if they belong to the point $x_{j}$. | ||
Line 35: | Line 35: | ||
:= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) | := \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) | ||
\end{equation} | \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 $\mathbb{R}^2$. 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 $\varphi_2$, | influenced by both $\varphi_1$ and $\varphi_2$, | ||
- | 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.** | + | **Lemma.** |
- | it can be transformed into a diagonal operator. | + | If we have a local operator |
+ | $x_{j} \in \mathbb{R}^d$, | ||
+ | it can be transformed into a diagonal operator. | ||
+ | output variables, diagonalization by reordering is not possible. | ||
//Remark.// A reordering operation is equivalent to the application of a permutation | //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 $x_{j}\in \mathbb{R}^d$, | ||
+ | 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, | ||
+ | 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 | ||
misc2.1386343218.txt.gz · Last modified: 2023/03/28 09:14 (external edit)