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Ion,

An eigenvector is only defined up to a constant, so if I have an equation 
Av = rv (where A is a matrix, r is an eigenvalue, and v is its associated 
eigenvector), then -v solves this equation if and only if v does.

The components of the eigenvector give you the eigenvector centralities of 
the nodes.

To see what's going on with your example, note that the network is small 
enough to do the calculation by hand.  That will be the easiest way to 
see how it works. You can construct the adjacency matrix of the star and 
calculate the eigenvector centrality vector directly from it.

Also, in addition to the book below that you already own, another source 
is this listserv's archives. (The question about getting negative values 
for eigenvector centrality has come up before.)

-----
Mason


>
> I am using ORA Software (see McCulloh, Armstrong and Johnson's book =
> "Social
> Network Analysis With Applications", or
> http://www.casos.cs.cmu.edu/projects/ora/index.php).=20
>
> =20
>
> I input a 6-node, star network.=20
>
> I ask for eigenvector centrality scores for all vertices.
>
> Given the general description of what eigenvector centrality does, I =
> expect
> the peripheral vertices to have high scores, and the central vertex to =
> have
> a lower score.
>
> =20
>
> The results are as follows:
>
> Central vertex: -1.0=20
>
> All other vertices: -0.4472136
>
> =20
>
> My questions:
>
> Why are the results negative?
>
> Why does the central vertex have a higher absolute value (should the
> absolute values even be considered)?
>
> Are the results at all interpretable in terms of eigenvector centrality;
> and, if so, how?
>
> =20
>
> I am grateful for any assistance in resolving this conundrum.
>
> =20
>
> Ion Georgiou
>
> =20
>

----
Mason

--------------------------------

  Mason A. Porter
  Professor of Nonlinear and Complex Systems
  Oxford Centre for Industrial and Applied Mathematics
  Mathematical Institute, University of Oxford

  Homepage: http://people.maths.ox.ac.uk/porterm
  Blog: http://masonporter.blogspot.com
  Twitter: @masonporter
  Skype: tepid451
  ----------------------------------------------------------------------------
  "I will be the lion." (Me)
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