Abstract

Let F be a field, G = (V,E) be an undirected graph on n vertices, and let S(F,G) be the set of all symmetric n × n matrices whose nonzero off-diagonal entries occur in exactly the positions corresponding to the edges of G. Let mr(F,G)be the minimum rank over all matrices in S(F,G). We give a field independent proof of a well-known result that for a tree the sum of its path cover number and minimal rank is equal to the number of vertices in the tree. The rank-spread of a vertex v of G is the difference between the minimal ranks of G and G - v, the graph obtained by deleting v and all its incident edges from G. The rank-spread of an edge is defined similarly. We derive a formula that expresses the minimal rank of a tree as the difference of sums of rank-spreads, the first being the sum of the rank-spreads of all the vertices and the second the sum of the rank-spreads of all the edges. We show that this is a special case of a more general inequality for all graphs. In proving the above results we explore how rank-spreads change as graphs are vertex-summed.

Degree

College and Department

Physical and Mathematical Sciences; Mathematics

Rights

http://lib.byu.edu/about/copyright/