Tricky puzzle

[Qfwfq]

Consider a graph, not directed, the nodes of which we'll call n_i and the edges r_i. To each r_i associate a value a_i, complex if you like. Now construct a matrix A having elements as follows:

Each diagonal element a_ii is the sum of any a_u such that r_u connects n_i to some other node, zero otherwise. Note: no edge may connect a node to itself.

Each non-diagonal element a_ij is -a_u if an edge r_u connects n_i and n_j (the sum, if the graph isn't simple), zero otherwise.

Obvious note, A will be symmetric. Chose one node n_p and add a non-zero value a to a_pp. Now invert this matrix; using ^ to indicate inversion, we could call it B = (A + P)^ where P has the one element of value a and all others zero.

I say: B_pp = 1/a. Easy to prove?

[CraigD]

Originally Posted by Qfwfq

Consider a graph...

Can you provide a short example graph and matrixes A, P, and B?

[Qfwfq]

http://mathworld.wolfram.com/topics/GraphTheory.html

http://mathworld.wolfram.com/Graph.html

A trivial graph with one node and no edges would give the one by one matrix with element 0. An edge between two nodes would give a two by two symmetric matrix with elements differing only by sign. In any case the sum for each row, or column, is 0.
Admittedly I wrote the post a bit hurriedly, I could have been clearer about constructing the matrix. I could have said that a_ij and a_ji are both equal to minus the summed a values of any edges between the n_i and n_j, zero if there are no edges between those two nodes. Of course, you might as well have at most one edge between each pair of nodes.

I was interested to know if there is a proof simpler and more straightforward than the indirect reason I had for stating the thesis, but yesterday I found enough time to find one. A necessary condition for the invertibility of A + P is that the graph be a single connected component and the value of a not zero, I'm still trying to glean a bit more about sufficient conditions. It would be comfortable since I might need to use this in the job I'm doing. Actually it could be stated in terms of the matrix elements except that I'm not 100% sure whether or not the graph might help about sufficient conditions for invertibility.