Adjacency matrix defines a distance metric

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Let $A$ be adjacency matrix of a graph (perhaps weighted). Prove that \begin{equation} \sum_i \sum_j A_{ij} (f_i- f_j)^2 = \mathbf{f}^T L \mathbf{f} \end{equation}

where $\mathbf{f}$ holds values of a scalar function defined on vertices of the matrix, $f_i$ is the value of function on $i^{th}$ vertex, and $L$ is the graph Laplacian corresponding to adjacency matrix $A$. How do you approach this?

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By definition of the multiplication of matrices, the RHS is $\sum\limits_i\sum\limits_jL_{ij}f_if_j$ hence $L_{ij}$ is the coefficient of $f_if_j$ in the quadratic form $\mathbf f\mapsto\mathbf f^TL\mathbf f$. The LHS is $\sum\limits_i\sum\limits_jA_{ij}(f_i^2-2f_if_j+f_j^2)\mathbf 1_{i\ne j}$, from which the coefficient of $f_if_j$ in the quadratic form can be extracted. A solution for $L$ is $L_{ij}=-A_{ij}$ for every $i\ne j$ and $L_{ii}=\sum\limits_{j\ne i}(A_{ij}+A_{ji})$ for every $i$. Thus, if $A$ is the matrix $(A_{ij})_{ij}$ with $A_{ii}=0$ for every $i$, then $L=D-A$ where $D_{ii}=L_{ii}$ for every $i$.

In the unweighted oriented case when $A_{ij}=1$ if $ij$ is an edge and $A_{ij}=0$ otherwise, $L_{ii}$ is the sum of the out- and in-degrees of $i$, and $(-L_{ij})$ is the indicator function of the fact that $ij$ is an edge, for every $i\ne j$.