Showing that a matrix is positive (semi-)definite Let $G = (V,E)$ be a connected graph and $T$ one of its spanning trees. Let $w \in[0,1]^{|V|-1}$ be a weight for the spanning tree, i.e. we assign to each of the spanning tree's edges a number in $[0,1]$. For $(i,j) \in V \times V$ let $x_{ij}^T(w)$ be the minimum of the weights on the unique path from $i$ to $j$ in $T$. For $i=j$ we assign $x_{ij}^T(w) = 1$.

Theorem: The $|V| \times |V|$ real symmetric matrix $x_{ij}^T(w)$ is positive semidefinite. If we assume $w \in [0,1[^{|V|-1}$ it is even positive definite.

This positivity property is necessary for the absolute convergence of the reshuffled sum using the Loop Vertex Expansion. It is the central property of the forest formula, however I am wondering: Is there a simple proof not related to the forest formula at all?
//edit: The proof I already know is in Rivasseau's and Wang's "How To Resum Feynman Graphs".
 A: Fun problem. Let the weights of the edges be $w_1 \leq w_2 \leq \cdots \leq w_{n-1}$. For convenience, define $w_0=0$ and $w_1=1$. Let $e_1$, $e_2$, ..., $e_{n-1}$ be the edges, in the same order as the weights. Let $D$ be the matrix in the problem.
For $1 \leq k \leq n$, let $D_k$ be the matrix where $(D_k)_{ij}=1$ if $D_{ij} \geq w_k$ and $(D_k)_{ij}=0$ otherwise. Then
$$D = \sum_{k=1}^n (w_k-w_{k-1}) D_k.$$
So it is enough to show that each $D_k$ is positive semidefinite. Moreover, it is enough to check this in the case that $w_k > w_{k-1}$, as otherwise the coefficient of $D_k$ is $0$.
So, assume $w_{k-1} < w_k$. We have $(D_k)_{ij}=1$ if and only if the path from $i$ to $j$ does NOT pass through edges $e_1$, $e_2$, ..., $e_{k-1}$. In other words, let $F_k$ be the forest obtained by deleting $e_1$, $e_2$, ..., $e_{k-1}$ from $T$; then $D_k$ is the block diagonal matrix whose blocks correspond to the connected components of $F_k$ and where each block is a square matrix entirely made up of $1$'s. This matrix is definitely positive semi-definite.
Moreover, if $w_{n-1} <1$, then $D_n$ is the identity matrix, so the sum is positive definite.
After writing this up, I checked the proof in Rivasseau and Wang (Theorem 2.2) and this is basically the same thing, so I don't know if you will like it any better.
