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misc2 [2013/12/06 16:41]
potthast
misc2 [2015/10/17 18:48] (current)
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 located at $x_2$ and $f_{2}$ is only influenced by $\varphi_1$ located at $x_1$. The matrix located at $x_2$ and $f_{2}$ is only influenced by $\varphi_1$ located at $x_1$. The matrix
 \begin{equation} \begin{equation}
 +\label{C example}
 C = \left( \begin{array}{cc} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array} \right) 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) := \left( \begin{array}{cc} 1 & 0 \\ 3 & 0 \end{array} \right)
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 If we have a local operator for which each measusment is influenced by a different point 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 ​ $x_{j} \in \mathbb{R}^d$,​ then by reordering of the variables ​
-it can be transformed into a diagonal operator. ​+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 ​ //Remark.// A reordering operation is equivalent to the application of a permutation ​
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 in each column as well. But that means that the operator $H$ looks like a scaled version of  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 a permutation matrix $P$, with scaling $0$ allowed. Clearly, by reordering we can make this
-into a diagonal matrix, and the proof is complete $\Box$ \\+into a diagonal matrix. In generalwe take (\ref{C example}) as counter example,  
 +and the proof is complete $\Box$ \\
  
  
misc2.txt · Last modified: 2015/10/17 18:48 (external edit)