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Let $x = (x_{0}, \ldots, x_{n})$ and $y = (y_{0}, \ldots, y_{n})$ be two vectors, and $d = (d_{0} = x_{0} - y_{0}, \ldots, d_{n} = x_{n} - y_{n})$ their element-wise euclidean distances. My question is, that is it possible to construct (invertible) projections $p, q$ so that the element-wise euclidean distances for $p(x), q(y)$ would be all equal, i.e. $d_{0}=d_{1}=\dots=d_{n}$? I quess making $x$ and $y$ orthogonal would give a good start? |
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If for invertible projection you mean just an ivertible linear map $p$ such that $p^2=p$ then the answer is yes. If $x$ and $y$ are 0 then the identity map works. Assume that $\| x\|\leq\|y\|\neq 0$. Take $p=Id$. Let $D$ be the line $\mathbb{R}(1,\ldots,1)$. Take $y'\in x+D$ such that $\|y'\|=\|y\|$. It is easy to construct an invertible projection $q$ such that $q(y)=y'$. Then $p(x)=x$ and $q(y)=y'\in x+D$ and so their difference lies in $D$. |
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