Find a linear mapping to map coplanar points to coplanar points I have a set of coplanar points $(X_1, X_2,\ldots,X_k)$ in $4$-dim space. I want to find a linear mapping to map these points to $3$-dim space and the resultant points $(X'_1,X'_2,\ldots,X'_k)$ are also coplanar on the $3$-dim space.
I have tried the following but it does not work:
Pick $4$ coplanar points $X_1$, $X_2$, $X_3$ and $X_4$ which are not colinear, form $V_1=X_2-X_1$, $V_2=X_3-X_1$, and $V_3=X_4-X_1$, then perform Gramm-Schmitt procedure on $V_1$, $V_2$ and $V_3$ to get $3$ orthonormal vectors $e_1$, $e_2$, and $e_3$. Then construct the linear mapping by concatenating the $3$ orthonrmal vectors together: $M = [e_1,e_2,e_3]^T$.
However, I find that the resultant points $(X'_1=MX_1,\ X'_2=MX_2,\ldots, X'_k=MX_k)$ are not coplanar on the $3$-dim space.
Any idea how to construct a proper linear mapping that I want?
 A: The points $X_1,...,X_k$ are on a hyperplane $W$ in a $4$ dimensionnal vector space $V$. Then $W$ is a $3$ dimensionnal vector space and we can can consider any plane in $W$ and project the points $(X_1,...,X_k)$ onto that plane. This gives us points $X_1',...,X_k'$ which are coplanar in $W$ a $3$ dimensionnal vector space. Since projections are linear we're done.
A: The given points $\{ x_i \}$ satisfy $ c^T x = d $ a hyperplane in $\mathbb{R}^4$.
Since the vector $c \in \mathbb{R}^4 $ , then to find it, we need four points $x_1, x_2, x_3, x_4 $, such that we can write
$ (x_i - x_1)^T c = 0 $ for $i= 2, 3, 4 $
Thus forming a $3 \times 4$ homogeneous system in $c$.  Using Gaussian elimination, the above system can be reduced and the direction of vector $c$ can be found.  Now, any vector in this hyperplane can be expressed as
the following linear combinations
$ x = v_0 + V u $
where $c^T v_0 = c^T x_1 = d$ and $V = [v_1, v_2, v_3] \in \mathbb{R}^{4 \times 3} $ satisfies $ c^T V = 0 $ , i.e. , vector $v_1, v_2, v_3 $ are orthogonal to $c$.  Vector $u = [u_1, u_2, u_3]^T $ is a vector of arbitrary real parameters.
Now define the linear transformation
$y = A x = A v_0 + A V u $
where $A $ is a $ 3 \times 4 $ real constant matrix.  Since we want the vectors $y$ to lie in one plane, then $A V$ must have precisely one column equal to the zero vector.  And this can happen if the rows of $A$ are all orthogonal to exactly one of the three vectors $v_1, v_2, v_3$.  Thus to determine $A$, we select one of the vectors, say $v_1$, and solve the linear system
$ v_1^T z = 0 $
Assuming we have run the vectors $ v_1, v_2 , v_3 $ through Gram-Schmidt, and that they are now orthonormal, then the solution to this last equation is
$z = \lambda c + \alpha v_2 + \beta v_3 $
Now $A_i$, the $i$-th row of $A$, can be selected as
$ A_i = z_i^T = \lambda_i c^T + \alpha_{i} v_2^T + \beta_{i} v_3^T $
Post-Multiplying $A$ by $[v_1, v_2, v_3 ] $ yields,
$A V = \begin{bmatrix} 0 && \alpha_{1} && \beta_{1} \\ 0 && \alpha_{2} && \beta_{2} \\ 0 && \alpha_{3} && \beta_{3} \end{bmatrix} $
Any choice of $\alpha_1, \alpha_2, \alpha_3, \beta_1 , \beta_2, \beta_3 $ that makes the vector $[\alpha_1, \alpha_2, \alpha_3]^T$ independent of the vector $[\beta_1 , \beta_2, \beta_3]^T$ would work.
