Christian Jost wrote:
I thought that the functionI guess eigen tries to get independent eigenvectors, butr that is not always possible, and your matrix is a case of that.
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$vectors)
Note that all eigenvectors of A are a multiple of (0,1)^T, so there cannot be two independent ones.
Kjetil
note that I try to get the upper triangular form with eigenvalues on the diagonal and (possibly) 1 just atop the eigenvalues to be used to solve a linear differential equation
x' = Ax, x(0)=x0
x(t) = P exp(D t) P^-1 x0
where D is this upper triangular form and P is the "passage matrix" (not sure about the correct english name) given by a base of eigenvectors. So the test would be
solve(ev$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
--
Kjetil Halvorsen.
Peace is the most effective weapon of mass construction.
-- Mahdi Elmandjra______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
