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Dominant eigenvector displayed as third (Marco Visser)

Yes, Spencer, your observation is correct, because the characeristic equation det(A - \lambda*I) is a sixth degree polynomial: \lambda^6 - 5 = 0.  So the eigenvalues are the complex numbers (generally) that are located at equal angles on the circle of radius 5^(1/6), at angles 2*pi*k/6, where k runs from 0 to 5.  Thus, the roots are:

z_k = 5^(1/6) * exp(i * 2*pi*k/6), k= 0, 1, ..., 5.

where i = sqrt(-1).

Ravi.

----- Original Message -----
From: Spencer Graves <spencer.graves at pdf.com>
Date: Friday, June 29, 2007 6:51 pm
Subject: Re: [R] Dominant eigenvector displayed as third (Marco Visser)
To: Marco Visser <visser_md at yahoo.com>
Cc: r-help at stat.math.ethz.ch