On the coincidence of "diameter points". Definition. Diameter point of an independent set of vectors $\{x_1,..., x_n\}$ in $\mathbb R^m$ ($n\leq m$) is the point $y$ in the linear span of $\{x_1,..., x_n\}$ such that
$$\langle y-x_i, x_i \rangle =0, \forall i=1,..., n.$$
In other words, $y$ is the point such that $y/2$ is the center of the sphere containing $\{x_1,...,x_n\}\cup\{0, y\}$.
Problem. Let $\{ a_1,..., a_n\}$ be a set of independent points/vectors in $\mathbb R^m$ ($n\leq m$). Let $v_i$ be the diameter point of $\{a_j\}_{j\neq i}$ for all $i=1,..., n$. Prove that the diameter point of $\{a_i\}_{i=1,...,n}$ and diameter point of $\{v_i\}_{i=1,...,n}$ coincide.
Comment. The above problem is a problem that I observed intuitively. The problem seems to be very obvious if you think about the rectangular prism where $\{x_1, x_2, x_3\}$ being the unit basis vectors. I checked the correctness of the problem using programming. Now, I try to solve it algebraically, but stuck on the last step as discussed below. Can anyone help? Thanks!
My attempt.
First, let's dive deeper into the definition.
Let $X = [x_1, ..., x_n]\in \mathbb R^{m\times n}$. Then the diameter point $y$ of $\{x_i\}_{i=1,...,n}$ can be computed explicitly as follows,
$$y= X (X^TX)^{-1} ||X||^2,$$
where $||X||^2 = [||x_1||^2,..., ||x_n||^2]^T$.
The proof is easy. Assume that $y = \sum_{i=1}^n z_i x_i$ for $z_i\in \mathbb R$.
Then we can write $$y = X z,$$
where $z=[z_1,..., z_n]^T$.
Then all the conditions $\langle y-x_i, x_i \rangle =0$ is equivalent to the following system of equations,
$$ X^TX z = ||X||^2.$$
Since $X$ is made up by independent columns, $X^TX$ is invertible.
Therefore, $$z=(X^TX)^{-1}||X||^2.$$ Notice that $y=Xz$, we obtain the desired formula of $y$.
Back to our problem.
Let $A= [a_{1},..., a_{i}, ..., a_{n}]\in \mathbb R^{m\times n}$ and $A_{-i}$ is matrix $A$ without column $i$.
We have the diameter point of $\{a_i\}_{i=1,...,n}$ is,
$$ y_1 = A (A^T A)^{-1} ||A||^2 $$
and diameter point of $\{v_i\}_{i=1,...,n}$ is
$$ y_2 = V (V^T V)^{-1} ||V||^2 $$
where $V=[v_1,...,v_n]$ and $v_i= A_{-i} (A^T_{-i} A_{-i})^{-1} ||A_{-i}||^2$ for $i=1,...,n$.
My problem now is to prove that $$y_1=y_2,$$ but I have no ideas how to simplify such a "complicated" equation.
 A: Your claim is only half true. What do I mean by that ?
Let $a$ be the diameter of $a_1,\ldots,a_n$ ($a$ is called "$y_1$" in the OP).
For $a$ to also be the diameter of $v_1,\ldots,v_n$, we would need (i) $\langle a-v_k,v_k \rangle = 0$ for $1\leq k \leq n$ and (ii) $a$ is spanned by $v_1,\ldots,v_n$. Now, what happens here is that (i) is always true, but (ii) is not always true (it will be "generically" true however, as in all except a few exceptional cases the $v_1,\ldots,v_n$ will be linearly independent and hence span
the same subspace as $a_1,\ldots,a_n$, which contains $a$).
Let us start by showing (i). Because of the symmetry in the indices, it suffices to show (i) for $k=n$.
Let ${\cal E}=(e_1,\ldots,e_n)$ be the orthonormal family of vectors obtained by applying the Gram-Schmidt process to ${\cal A}=(a_1,\ldots,a_n)$. Then the matrix $A=(a_{ij})$ of ${\cal A}$ in ${\cal E}$ is upper triangular and invertible (the diagonal coefficients are nonzero).
Write $a=\sum_{k=1}^{n}\alpha_k e_k$ ; by hypothesis $\langle a,a_j \rangle = \langle a_j,a_j \rangle$ for $1\leq j \leq n$. Consider $w=a-\alpha_n e_n$. Since
$e_n$ is orthogonal to every $a_j$ with $j\lt n$, we have $\langle w,a_j \rangle= \langle a_j,a_j \rangle$ for those $j$. On the other hand, $w$ is in
the subspace spanned by $e_1,\ldots,e_{n-1}$ which coincides with the subspace spanned by $a_1,\ldots,a_{n-1}$. Those two conditions together imply that $w=v_n$.
Then $\langle v_n,v_n \rangle = \langle a-\alpha_ne_n,v_n \rangle= \langle a,v_n \rangle$ which shows (i) for $k=n$ as wished.
Next, we provide a counterexample to (ii). Let $f(t)=t^3-2t^2+2t-2$. Since $f(1)=-1$ and $f(2)=2$, we see that $f$ has a root $\zeta\in [1,2]$. Then, consider
$$
A=\left(\begin{matrix}1 & \zeta & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)
$$
It is easy to see that in this case,
$$
\begin{array}{c}
v_1=a_2+a_3=\left(\begin{matrix}\zeta \\ 1 \\ 1\end{matrix}\right),
v_2=a_1+a_3=\left(\begin{matrix}1 \\ 0 \\ 1\end{matrix}\right), \\
v_3=(\zeta^2-\zeta+1)(a_2-a_1)=\left(\begin{matrix}1 \\ \zeta^2-\zeta+1 \\ 0\end{matrix}\right)
\end{array}
$$
while
$$
a=\left(\begin{matrix}1 \\ \zeta^2-\zeta+1 \\ 1\end{matrix}\right)
$$
Note that all the $v_i$'s are in the plane defined by the equation $z-x=(1-\zeta)y$, but $a$ is not. So $a$ is not spanned by the $v_i$'s. This finishes the proof.
