Let $M = (m_{ij})_{(i,j) \in \llbracket 1,n\rrbracket^2} \in \mathcal{S}_n(\mathbb{R})$ such that for every pair $(i,j) \in \llbracket 1,n\rrbracket^2$, $m_{ij} \geqslant 0$ and $m_{ii} = 0$. Conversely, we assume that the eigenvalues of $\Phi(M)$ are all non-positive and we set $\Psi(M) = -\frac{1}{2}\Phi(M)$ and $r = \operatorname{rg}(\Psi(M))$.
1) Show that there exists a matrix $U \in \mathcal{M}_{r,n}(\mathbb{R})$ such that ${}^t UU = \Psi(M)$.
2) We denote $U_1, U_2, \cdots, U_n$ the columns of the matrix $U$. We seek to show that for every $(i,j) \in \llbracket 1,n\rrbracket^2$, $m_{ij} = \|U_i - U_j\|^2$.
a) Show that the $(U_i)$ are centered, that is, $\sum_{i=1}^n U_i = 0$.
b) Show that the matrix $N = (n_{ij})$ defined by:
$$\forall (i,j) \in \llbracket 1,n\rrbracket^2, \quad n_{ij} = \|U_i - U_j\|^2$$
satisfies $\Psi(N) = \Psi(M)$.
c) Show that $M = N$ and conclude.