You should be able to use the result from an article entitled "Inverse Sum of Matrix Sums" by Kenneth S. Miller. He gives this relation (assuming that H has rank one and that both G and G + H are invertible, obviously) as a lemma at the beginning of the article:
inverse(G+H) = inverse(G) - 1/(1+g) inverse(G) * H * inverse(G)where g=trace(inverse(G) * H).
If you have pre-calculated inverse(G)before the loop, the extra calculations inside the loop should be quite limited. Note, however, that I did not implement this to see if this really gives significant acceleration.
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