For any $x, y, z\in\mathbb R^n(n>2)$, there exists an isometry mapping all but the first two coordinates of $x, y$ and $z$ to zero.
How to prove the statement above? I know that translation and rotation are isometries, but I’m just wondering if we can formulate a strict proof.
I tried in the following way but got stuck.
Let $L_{*\star}$ be the distance between $*$ and $\star$ and let $(x_1’,x_2’,0,\ldots,0), (y_1’,y_2’,0,\ldots,0), (z_1’,z_2’,0,\ldots,0)$ be the corresponding points. Since it is a distance-preserving mapping, we have
$$\left\{\begin{array}{l} L_{xy}=\sqrt{(x_1’-y_1’)^2+(x_2’-y_2’)^2}\\ L_{xz}=\sqrt{(x_1’-z_1’)^2+(x_2’-z_2’)^2}\\ L_{yz}=\sqrt{(y_1’-z_1’)^2+(y_2’-z_2’)^2} \end{array}\right.$$
I got stuck when trying to prove that system of equations has at least one solution. I suspect I’m doing it in the wrong way. Could you give me some help? Thank you!
Given points $x,y,z\in \mathbb{R}^n$, we can define an isometry $M:\mathbb{R}^n\rightarrow \mathbb{R}^n$ by $M(w) = T(w-z)$ where $T$ is an orthogonal matrix.
General idea
Consider the $n\times 2$ matrix with column vectors $x-z$, $y-z$. Apply rotators or reflectors as in QR decomposition to construct the matrix $T$.
Example
Let $x=[2,3,3,1]^\top$, $y=[2,3,5,2]^\top$, $z=[1,1,1,1]^\top$
Initially we form the matrix
\begin{equation} [x-z \ y-z] = \left[ \begin{matrix} 1 & 1\\ 2 & 2\\ 2 & 4\\ 0 & 1 \end{matrix} \right] \end{equation}
The QR decomposition of this matrix is (numerically with floating point precision)
\begin{equation} \left[ \begin{matrix} -0.3333333 & -0.24759378& -0.89364759& -0.17023569\\ -0.66666667 & -0.49518757 & 0.29595756 & 0.47196817\\ -0.66666667& 0.61898446 & 0.15086624& -0.38685032\\ -0. & 0.55708601& -0.30173248 & 0.77370064 \end{matrix} \right] \left[ \begin{matrix} -3. & -4.33333333\\ 0. & 1.79505494 \\ 0. & 0. \\ 0. & 0. \end{matrix} \right] \end{equation}
Let $T$ be the transpose of the matrix $Q$ which is orthogonal. Then $M$ will map $x$ to the first column of $R$, $y$ to the second column of $R$, and $z$ to 0.