[R] matrix of eigenvalues

Kjetil Brinchmann Halvorsen kjetil at acelerate.com
Tue Oct 19 14:31:51 CEST 2004

Christian Jost wrote:

> I thought that the function
> 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)
I guess eigen tries to get independent eigenvectors, butr that is not 
always possible, and your matrix is a case of that.
Note that all eigenvectors of A are a multiple of (0,1)^T, so there 
cannot be two independent ones.


> 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

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