I was trying to figure out why the laplacian matrix of a graph $$L=D-A$$ is named so. For this, I draw a 2D grid and tried to find the laplacian at some point $f_{x,y}$
I can write down the following: $$ \Delta f(x, y) \approx \frac{f(x-h, y)+f(x+h, y)+f(x, y-h)+f(x, y+h)-4 f(x, y)}{h^{2}} $$ Then I find the Laplacian matrix $$ L=\left[\begin{array}{ccccc} {1} & {0} & {0} & {-1} & {0} \\ {0} & {1} & {0} & {-1} & {0} \\ {0} & {0} & {1} & {-1} & {0} \\ {-1} & {-1} & {-1} & {4} & {-1} \\ {0} & {0} & {0} & {-1} & {1}\end{array}\right] $$
multiplying by the function $$ L=\left[\begin{array}{c} {f(x-h,y)} \\ {f(x,y+h)} \\ {f(x+h,y)} \\ {f(x,y)} \\ {f(x,y-h)}\end{array}\right] $$
I get $$ L f = -f(x-h, y)-f(x+h, y)-f(x, y-h)-f(x, y+h)+4 f(x, y) $$ I can't figure out what to do with the sign and the $h^2$ term. Or is it just that this is not the reason for the naming?
In terms of "sign": there's a long debate between geometers and analysts about whether the Laplacian should be the trace of the Hessian or its negative. From the functional analytic point of view (which will be useful also for the graph Laplacian), defining the Laplacian as the negative trace of the Hessian has the advantage in giving a positive operator.
The $h^2$ term is unimportant: you are working on a graph so you are just multiplying by an overall factor. Unless you are thinking of the graph as some discretization of the plane and are actually interested in the limit of taking the step-size $h\to 0$, whether you keep track of the $h^2$ is not something to worry about.