Eigenvectors are defined only up to a scalar constant (assuming distinct
eigenvalues). However, your `by hand' answer does not pass the simple
test Av = lambda v for some lambda. So you cannot reproduce incorrect
answers in R!
Your example is unusual: A is of rank 1.
On 9 Jun 2003, Christoph Lehmann wrote:
Hi, dear R pros
I try to understand eigen(). I have seen, that eigen() gives the
eigenvectors normalized to unit length.
What shall I do to get the eigenvectors not normalized to unit length?
Multiply them by any randomly chosen non-zero scalar!
E.g. take the example:
A
[,1] [,2]
V1 0.7714286 -0.2571429
V2 -0.4224490 0.1408163
Calculating eigen(A) by hand gives the eigenvectors (example from
Backhaus, multivariate analysis):
0.77143 and 0.25714
-0.42245 0.14082
The second is not an eigenvector of A: try it! They look like rounded
versions of A with a sign error.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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