Matthew Knepley <[email protected]> writes: >> >> A graph Laplacian certainly does transform under coordinate >> >> transformation and indeed, we use that property to design effective >> >> coarsening strategies. That one basis strikes you as intrinsically >> >> "more canonical" does not mean it isn't a linear operator. >> >> >> > >> > That is one operator. This is argument by anecdote. An arbitrary graph >> > is not a linear operator, but an arbitrary matrix definitely is (the >> > coordinate representation of one). >> >> Dude, we solve linear systems and eigenproblems for arbitrary graphs. >> It isn't an anecdote. >> >> Barry (rightly) objects to a 2D array representing a function on a grid >> being considered a Matrix. We don't "apply" it as a linear operator. >> There is no "vector" on which it operates. >> >> But we absolutely do with a graph. Our vectors are functions at the >> vertices of the graph. Applying the graph Laplacian tells us about >> local compatibility of the field over the vertices. It is entirely >> analogous to fields over a grid. You don't need a concept of "grid >> refinement" to have matrices. >> > > I don't think makes sense. You are saying, because a linear operator (the > graph Laplacian) can be defined using the graph, then the graph is identical > with this operator.
It's a bijection; given an operator (with zero row sums - Laplacian, or ignoring the diagonal, or allowing a "vertex weight"), you can draw the graph. > I do not agree. Whereas the matrix means nothing else but the linear > operator which it represents. I don't agree, but this is mindless pedantry.
