I have following task: Let $A$ be the adjacency matrix and $L$ the Laplacian matrix of a simple undirected connected graph $G$. Show that for an arbitrary vector of real numbers $x \in \Bbb R^n$ the graph Laplacian associates the following quadratic form with the graph $G$:
$$x^T\!Lx = \frac{1}{2} \sum_{i,j} A_{ij} (x_i - x_j)^2$$
I have studied some literature on the Laplacian and its quadratic form, but the nearest I could get to was $$x^T\!Lx = \sum_{i,j} A_{ij} (x_i - x_j)^2.$$
So please could somebody enlighten me on where the $1/2$ term comes from?
The 1/2 factor comes from the fact that you count every edge twice in the sum (as (i,j) and then as (j,i)). The computation is
$$\sum_{i,j} A_{i,j} (x_i-x_j)^2 = \sum_{i,j} A_{i,j} x_i^2 + \sum_{i,j} A_{i,j} x_j^2 - 2\sum_{i,j} A_{i,j} x_ix_j.$$
The first two terms can be summed using the degree $D_{ii} = \sum_{j} A_{ij}$ to obtain
$$\sum_{i,j} A_{i,j} (x_i-x_j)^2 = \sum_{i} D_{ii}x_i^2 + \sum_{j} D_{jj}x_j^2 - 2\sum_{i,j} A_{i,j} x_ix_j.$$
Now relabel $j$ as $i$ in the second sum to get
$$\sum_{i,j} A_{i,j} (x_i-x_j)^2 = 2\sum_{i} D_{ii}x_i^2 - 2\sum_{i,j} A_{i,j} x_ix_j.$$
The right hand side is exactly $2 x^T(D-A)x^T$, and the graph Laplacian is $L=D-A$ (here $D$ is the diagonal matrix with entries $D_{ii}$).