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I am working with about 300 disconnected networks of different sizes. I calculate different graph-level centralization measures for these networks using the STATNET and iGraph packages in R.

However, I find that the nodes in subgraphs of N=2 get assigned the highest value of 1 for the Eigenvector centrality measure with iGraph. As a result, networks with a lot of isolated dyads get very high graph-level Eigenvector centralization scores.

In my networks this is not a valid result, because these networks are poorly connected and thus should, theoretically, have a low centralization score.

Does anyone know how these measures handle disconnected graphs? And are there ways to deal with this? Also, are there other ways to assess the structure of these networks?

Any help is welcome. Thank you!

wake_wake
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1 Answers1

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Eigenvector centrality is not well-defined for disconnected graphs since the centrality scores of the individual components are independent of each other; one can inflate the centrality scores of one component by multiplying the with a large constant (say, 10000), then normalize the centrality scores again to sum up to 1; the resulting vector would still satisfy the eigenvector centrality equation. Therefore, you should calculate eigenvector centralities for connected graphs only. If you have multiple components in your graph, break it down to connected components first and then calculate and compare eigenvector centralities for the individual components only.

Tamás
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  • Thanks! I wonder how this theoretically will work when I compare the following types of graphs? Type A: no large component, mainly dyads and some triads. Type B: very big largest component, some dyads and triads. Which network will have the highest graph-level Eigenvector centrality score, if any at all? (I might misunderstood your earlier comment) – wake_wake Nov 06 '14 at 21:25
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    My point is that there is no such thing as a "unique" eigenvector centrality vector for a graph because if vector v satisfies the eigenvector centrality, then k times v (where k is an arbitrary constant) also satisfies it. That's why people usually normalize v such that its coordinates sum up to 1. When your graph is disconnected, it becomes even more complicated because it is enough to multiply the coordinates of v that correspond to the same connected component (and leave the rest at their original values) to get another solution for the eigenvector centrality equation. – Tamás Nov 07 '14 at 22:04
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    So, in your particular case, I could make type A have the highest graph-level eigenvector centrality score if I wanted to (by picking a connected component of it, multiplying the eigenvector centrality scores by a sufficiently large number, and then renormalizing the entire eigenvector centrality vector) - but I could do the same with any connected component of B as well if I wanted to make type B have the largest eigenvector centrality. The point is that you should only ever compare the absolute values of eigenvector centralities for *connected* networks *of the same size*. – Tamás Nov 07 '14 at 22:06
  • This makes very much sense. Thank you! – wake_wake Nov 07 '14 at 23:01