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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