is eigenvector centrality defined intra-component?

[IgnacioOQ]

I'm doing some exploration and found that if you have three nodes (a) (b)---(c), where (a) is isolated and the other two are connected by an undirected link, then when applying nw:eigenvector-centrality all turtles/nodes receive a value of 1.
The manual (http://ccl.northwestern.edu/netlogo/5.0/docs/nw.html#eigenvector-centrality) states that this should report "false" since the network is not connected. My suspicion is that the implementation is measuring the eigenvector of each turtle relative to its component.
Am I right?
Many thanks!

[qiemem]

Your suspicion is correct. However, this was a changed with NetLogo 6 (iirc); I believe you are looking at old documentation there. The current documentation is here: https://ccl.northwestern.edu/netlogo/6.0/docs/nw.html#nw:eigenvector-centrality

The current documentation, however, only states that our implementation should agree with Gephi's, but does not explicitly state that it is normalized relative to each component. It probably should.

[IgnacioOQ]

Many thanks!
Just to follow up, the normalization secures that "the highest eigenvector centrality a node can have is 1", but the example before also shows that it is not normalizing so that for each component the eigenvector adds up to 1 [i.e. In (b)---(c) both have a value of 1 rather than 1/2]
Is this what you meant by "normalized relative to each component"?

[qiemem]

Ah, good call. That is ambiguous. For each component, the centralities are normalized so that the largest centrality will be 1; i.e. all centralities are divided by the maximum centrality on their component.

[IgnacioOQ]

Thanks again!
What about defining a function(v) that uses nw:weak-component-clusters, applies nw:eigenvector-centrality to each of the elements in the list, adds the values up to get the new normalizing parameter k, and divides the given v by k?
Or digging into the source code, normalize by suck a k rather than the maximum centrality on their component.
In any case, problem solved.
Thanks for the quick response!