[R] matrix of eigenvalues

[Douglas Bates]

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$vectors)
> 
> note that I try to get the upper triangular form with eigenvalues on the 
> diagonal and (possibly) 1 just atop the eigenvalues to be used to solve 
> a linear differential equation
> x' = Ax, x(0)=x0
> x(t) = P exp(D t) P^-1 x0
> where D is this upper triangular form and P is the "passage matrix" (not 
> sure about the correct english name) given by a base of eigenvectors. So 
> the test would be
> solve(ev$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

That particular matrix has repeated eigenvalues and a degenerate eigenspace.

 > A <- matrix(c(1,0,2,1),nc=2)
 > A
      [,1] [,2]
[1,]    1    2
[2,]    0    1
 > eigen(A)
$values
[1] 1 1

$vectors
      [,1]          [,2]
[1,]    1 -1.000000e+00
[2,]    0  1.110223e-16