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misc2 [2013/12/06 16:37] potthastmisc2 [2013/12/06 16:41] potthast
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 //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.// We first assume that in the state space $X = \mathbb{R}^n$ each element belong //Proof.// We first assume that in the state space $X = \mathbb{R}^n$ each element belong
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 is influenced by the entries in the $\ell$-th row of the matrix $H$. Thus, this is local 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  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.  +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 
-$\Box$ \\+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, and the proof is complete $\Box$ \\
  
  
misc2.txt · Last modified: 2023/03/28 09:14 by 127.0.0.1