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How to compute eigenvectors and eigenvalues?

Dear Arnau,

In this and a subsequent message, you seem to incorrectly infer that the two
equal eigenvalues of the matrix imply that it's singular. The rank of the
matrix is equal to the number of *nonzero* eigenvalues, here 3, and so the
matrix is nonsingular. That two eigenvalues are equal simply implies that
the corresponding eigenvectors span a subspace of dimension 2; the 2
corresponding eigenvectors reported by R are (within rounding error) a basis
for this subspace.

I hope this helps,
 John

--------------------------------
John Fox
Senator William McMaster
  Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox