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Old 02-06-2006   #3 (permalink)
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Re: Tricky puzzle

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.


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