Transform $m-$simplex in $\mathbb{R}^n$ so that $n-m$ components are $0$ Given are $m+1$ points in $\mathbb{R}^n$ with $m< n$. These $m+1$ points define a non-degenerated $m-$simplex. How to rotate (and shift) the points in $\mathbb{R}^n$ so that we can write the simplex with $m$ instead of $n$ coordinates, i.e. the last $n-m$ components of all coordinates are zero? The Euclidean distances between the points shall not be changed. Reflections are allowed.
Could Principal Component Analysis be used or is this method reserved for statistical data? There should be a simpler way of finding an orthogonal $n \times n$ matrix that rotates the points. Is additional shifting needed or rotation sufficient?
Example
Given are points $\bf{x}$,$\bf{y}$,$\bf{z}$ that form a $2$-simplex (triangle) in $\mathbb{R}^4$. We have $m=2,n=4$, and because $n-m=2$ we can set two components to $0$.
How to rotate (and shift) the points $\bf{x}$=$(x_1,x_2,x_3,x_4)$, $\bf{y}$=$(y_1,y_2,y_3,y_4)$, $\bf{z}$=$(z_1,z_2,z_3,z_4)$ so that they can be written like $\bf{x'}$=$(x'_1,x'_2,0,0)$, $\bf{y'}$=$(y'_1,y'_2,0,0)$, $\bf{z'}$=$(z'_1,z'_2,0,0)$?
 A: You have $m+1$ affinely independent points
$\big\{\mathbf v_0, \mathbf v_1, \dots,\mathbf v_{m}\big\}$ and in particular are interested in taking convex combinations of these, a subset of affine combinations.
First apply a translation:
Define $\mathbf a_k:=\mathbf v_k-\mathbf v_0$
We are now working in a vector subspace of $\mathbb R^n$ generated by
$\big\{\mathbf a_1 \dots,\mathbf a_{m}\big\}$
These are linearly independent vectors.
Now collect these in matrix
$A:=\bigg[\begin{array}{c|c|c|c} \mathbf a_1 & \mathbf a_2 &\cdots & \mathbf a_{m}\end{array}\bigg]$
where $A$ is injective.  Using "full" QR factorization
$A=QR$
$Q\in O_n(\mathbb R)$ and $R$ is "upper triangular" but not square (in fact it is tall and skinny).  This means that $r_{i,i}\neq 0$ in all cases and $r_{i,j}=0$ for $i\gt j$.  Then
$Q^{-1}A=Q^{T}A$  gives the desired result of coordinate vectors that are zero in each component after the $m$th.

To recap: we first apply a translation $\mathbf v_k\mapsto \mathbf v_k -\mathbf v_0$, then a rotation $(\mathbf v_k -\mathbf v_0)\mapsto Q^T(\mathbf v_k -\mathbf v_0)$, both of which are rigid motions.
As an aside, if you let $E$ be the type 3 elementary matrix with a $-1$ in the bottom right corner then
$A=QR=QE^2R=(QE)(ER)=(QE)R$ and checking the determinant, it must be the case that $Q\in \mathbb SO_n(\mathbb R)$ or $(QE)\in \mathbb SO_n(\mathbb R)$. I.e. you can use always assume you used a bona fide rotation if you like.
A: I think the principle component analysis is not what you want. It's for reducing statistical data, you seem to have a different problem.
What you want can be achieved by successively applying rotations in $\mathbb{R}^n$ in the planes spanned by the unit axes, like this one that rotates by $\alpha$ in the $e_3$-$e_4$-plane:
$$
\mathbf{Q}_{34}(\alpha)=\left[\begin{matrix} 
1      & 0      &            0 &            0 & 0 ... \\
0      & 1      &            0 &            0 & 0 ... \\
0      & 0      & \cos{\alpha} & -\sin{\alpha} & 0 ... \\
0      & 0      & \sin{\alpha} & \cos{\alpha} & 0 ... \\
0      & 0      &            0 &            0 & 1 ... \\
\vdots & \vdots & \vdots       &       \vdots & \vdots 
\end{matrix}\right]
$$
Let's assume you have three points in $\mathbb{R}^6$ labeled $\mathbf{P}_{1,2,3}$. Then you can transform these like so
$$
\mathbf{P}_1'  = \mathbf{Q}_{36}(\alpha) \cdot \mathbf{Q}_{46}(\beta) \cdot \mathbf{Q}_{56}(\gamma) \cdot \mathbf{P}_1 \\
\mathbf{P}_2'  = \mathbf{Q}_{36}(\alpha) \cdot \mathbf{Q}_{46}(\beta) \cdot \mathbf{Q}_{56}(\gamma) \cdot \mathbf{P}_2 \\
\mathbf{P}_3'  = \mathbf{Q}_{36}(\alpha) \cdot \mathbf{Q}_{46}(\beta) \cdot \mathbf{Q}_{56}(\gamma) \cdot \mathbf{P}_3
$$
Then you have three angles $\alpha,\beta,\gamma$ to make the sixth' components of $\mathbf{P}_{1,2,3}'$ zero. Then you do
$$
\mathbf{P}_1''  = \mathbf{Q}_{25}(\alpha') \cdot \mathbf{Q}_{35}(\beta') \cdot \mathbf{Q}_{45}(\gamma') \cdot \mathbf{P}_1' \\
\mathbf{P}_2''  = \mathbf{Q}_{25}(\alpha') \cdot \mathbf{Q}_{35}(\beta') \cdot \mathbf{Q}_{45}(\gamma') \cdot \mathbf{P}_2' \\
\mathbf{P}_3''  = \mathbf{Q}_{25}(\alpha') \cdot \mathbf{Q}_{35}(\beta') \cdot \mathbf{Q}_{45}(\gamma') \cdot \mathbf{P}_3'
$$
and use $\alpha',\beta',\gamma'$ to make the fifth components of $\mathbf{P}_{1,2,3}''$ vanish, and so on.
Put into a Mathematica notebook the algorithm looks like this:
Remove["Global`*"]
p1 = {-(1/(2 Sqrt[3])), 1/Sqrt[3], -(1/(2 Sqrt[3])), -(1/2), 0, 1/2};
p2 = {1/2, 0, -(1/2), -(1/(2 Sqrt[3])), 1/Sqrt[3], -(1/(2 Sqrt[3]))};
p3 = {1/(2 Sqrt[3]), -(1/Sqrt[3]), 1/(2 Sqrt[3]), -(1/2), 0, 1/2};
p4 = {-(1/2), 0, 1/2, -(1/(2 Sqrt[3])), 1/Sqrt[3], -(1/(2 Sqrt[3]))};
(* P1 is the new origin, so we have to deal only with three points *)
(* t is the list of location vectors that we want to rotate such that *)
(* the trailing components are zero *)
t = {p2 - p1, p3 - p1, p4 - p1};

n = Length[p1];
m = Length[t]; 
alphas = Array[alpha, m];
For[ss = n, ss >= m + 1, ss--,
    Print[ss];
    Print[MatrixForm[t]];
    (* create rotation matrices *)
    Rs = Table[mat = IdentityMatrix[n];
               mat[[ss - i, ss - i]] = Cos[alphas[[i]]];
               mat[[ss, ss]] = Cos[alphas[[i]]];
               mat[[ss - i, ss]] = Sin[alphas[[i]]];
               mat[[ss, ss - i]] = -Sin[alphas[[i]]];
               mat, {i, 1, m}];
    (* apply rotations *)
    tmp = Rs[[1]] . Transpose[t];
    For[ii = 2, ii <= m, ii++, tmp = Rs[[ii]] . tmp];
    (* solve for the unknown angles with random initial values to break symmetries *)
    erg = FindRoot[tmp[[ss]], Table[{alphas[[ii]], Random[]}, {ii, 1, m}]];
    (* update the list of points*)
    t = Transpose[tmp /. erg];
]
Print[MatrixForm[t]]
Graphics3D[Tetrahedron[{{0, 0, 0}, t[[1, 1 ;; 3]], t[[2, 1 ;; 3]], t[[3, 1 ;; 3]]}]]

