Lets look at the following laplacian: $L = I - \frac{1}{d}A$ and the graph $H = (U, V)$.
I am trying to derive the known quadratic form of this laplacian $x^T Lx = \frac{1}{d}\sum_{u, v \in H}(x_u - x_v)^2$ but i cant manage to do it.
This is what i came up with so far:
\begin{align} x^T Lx &= x^T Ix - x^T\frac{1}{d}Ax \\ &= x^Tx - \frac{1}{d}\sum_{u, v \in H}x_ux_v \\ &= \sum_{i}x^2 - \frac{1}{d}\sum_{u, v \in H}x_ux_v \end{align}
I cant see how i should get the $-2x_ux_v$ term and the $x_v^2$ terms needed out of the present terms.
Does anyone see the solution to this?
