[R] matrix of eigenvalues
Douglas Bates
bates at stat.wisc.edu
Tue Oct 19 14:27:39 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$vectors)
>
> note that I try to get the upper triangular form with eigenvalues on the
> diagonal and (possibly) 1 just atop the eigenvalues to be used to solve
> a linear differential equation
> x' = Ax, x(0)=x0
> x(t) = P exp(D t) P^-1 x0
> where D is this upper triangular form and P is the "passage matrix" (not
> sure about the correct english name) given by a base of eigenvectors. So
> the test would be
> solve(ev$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
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