Prove that, for all $v\in\left(\mathbb E^3\right)^n$, $\langle Lv,v\rangle=\frac{1}{2}\sum_{i,j}^n a_{ij} \|v_i-v_j\|^2.$ Given a nonnegative, symmetric, $n\times n$ matrix $A$ the Laplacian $L$ of $A$ is defined to be $L=D-A$, where $D=\operatorname{diag}(d_1,\dots,d_n)$ and $d_l=\sum_{j=1}^n a_{lj}$. The Laplacian as just defined has its origins in graph theory where the matrix $A$ is the adjacency matrix of a graph $G$ and many of the properties of $G$ can be read out from $L$.
Denote 3-dimensional Euclidean space by $\mathbb E^3$ and let $\left(\mathbb E^3\right)^n$ be its n-fold product endowed with the induced inner product structure. Prove that, for all $v\in\left(\mathbb E^3\right)^n$,
$$\langle Lv,v\rangle=\frac{1}{2}\sum_{i,j}^n a_{ij} \|v_i-v_j\|^2.$$
Any suggestions please?
 A: I'll first demonstrate the result for one dimensional euclidean space and then show how it implies the result for $\mathbb{R}^3$.  Let $v \in \mathbb{R}^n$ where $n$ is the number of vertices in the graph, i.e. $v$ is an assignment of one .  As above, the graph Laplacian is defined as $$L = D - A$$
We can express the action of $L$ on a vector $v$ as:
$$(Lv)_i = \sum_j a_{ij} (v_i - v_j)$$
Now we can compute
$$\langle Lv, v\rangle = \sum_i (Lv)_i v_i = \sum_i \left(\sum_j a_{ij} (v_i - v_j)\right)v_i = \sum_{i,j} a_{ij} \frac{1}{2}(v_i - v_j)^2$$
The second equality follows from the following:
\begin{eqnarray*}
\sum_{i,j} a_{ij} \frac{1}{2}(v_i - v_j)^2 &=& \frac{1}{2} \left(\sum_{i,j} a_{ij} \frac{1}{2}(v_i - v_j)^2\right) \\
&=& \frac{1}{2} \left(\sum_{i,j} a_{ij} (v_i^2 - v_iv_j + (v_j^2 -v_iv_j^2)\right) \\
&=& \frac{1}{2} \left(\sum_{i,j} a_{ij} (v_i - v_j)v_i + \sum_{i,j} a_{ij}(v_j -v_i)v_j\right) \\
&=& \frac{1}{2} \left(\sum_{i,j} a_{ij} (v_i - v_j)v_i + \sum_{j,i} a_{ji}(v_i -v_j)v_i\right) \\
&=& \frac{1}{2} \left(\sum_{i,j} a_{ij} (v_i - v_j)v_i + \sum_{j,i} a_{ij}(v_i -v_j)v_i\right) \\
&=& \sum_{i,j} a_{ij} (v_i - v_j)v_i\\
\end{eqnarray*}
where we have used the symmetry of the matrix $a_{ij} = a_{ji}$.
Now we need to consider assignments of three reals instead of one to each vertex of the graph.  This operator $L : (\mathbb{R}^3)^n \to (\mathbb{R}^3)^n$ can be expressed in terms of independent copies of the $L$ defined for one dimension.  Let the superscript $i$ in $v^i$ denote the $i$th component of the value assigned to each vertex.  Then
$$Lv = L_1v^1 + L_2v^2 + L_3v^3$$
where $L_i$ acts on the ith component. Hence,
\begin{eqnarray*}
\langle Lv, v \rangle &=& \langle \sum_k L_kv^k, \sum_j v^j \rangle \\
&=& \sum_k \langle L_kv^k, v^k\rangle \\
&=& \sum_k \sum_{i,j} a_{ij} \frac{1}{2}(v_i^k - v_j^k)^2 \\
&=& \sum_{i,j} a_{ij} \frac{1}{2}\sum_k (v_i^k - v_j^k)^2 \\
&=& \frac{1}{2}\sum_{i,j}^n a_{ij} \|v_i-v_j\|^2\\
\end{eqnarray*} 
