# [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.
>
> 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

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

```