eigen(A)
will return a matrix with eigenvectors that are independent of each other (thus forming a base and the matrix being invertible). This seems not to be the case in the following example
A=matrix(c(1,2,0,1),nrow=2,byrow=T)
eigen(A) ->ev
solve(ev$vectors)
note that I try to get the upper triangular form with eigenvalues on the diagonal and (possibly) 1 just atop the eigenvalues to be used to solve a linear differential equation
x' = Ax, x(0)=x0
x(t) = P exp(D t) P^-1 x0
where D is this upper triangular form and P is the "passage matrix" (not sure about the correct english name) given by a base of eigenvectors. So the test would be
solve(ev$vectors) %*% A %*% ev$vectors - D
should be 0
Thanks for any help, Christian.
ps: please copy reply also to my address, my subscription to the R-help list seems to have delays
--
***********************************************************
http://cognition.ups-tlse.fr/vas-y.php?id=chj [EMAIL PROTECTED]
Christian Jost (PhD, MdC)
Centre de Recherches sur la Cognition Animale
Universite Paul Sabatier, Bat IV R3
118 route de Narbonne
31062 Toulouse cedex 4, France
Tel: +33 5 61 55 64 37 Fax: +33 5 61 55 61 54
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