misc2
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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 $\mathbb{R}^n$, | ||
+ | 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, | ||
+ | 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 $\varphi_2$, | ||
+ | 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 | ||
+ | $x_{j} \in \mathbb{R}^d$, | ||
+ | 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 $x_{j}\in \mathbb{R}^d$, | ||
+ | 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 $\Box$ \\ | ||
+ | |||
+ | |||
+ | **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 | ||
+ | |||
+ | |||
+ | |||
+ |