The R 2.0.0 pat documentation for "eigen" refers to "http://www.netlib.org/lapack/lug/lapack_lug.html", and the description there for eigen analysis of a non-symmetric matrix says, "This problem can be solved via the Schur factorization of A, defined in the real case as
A = ZTZT,
where Z is an orthogonal matrix and T is an upper quasi-triangular matrix with 1-by-1 and 2-by-2 diagonal blocks, the 2-by-2 blocks corresponding to complex conjugate pairs of eigenvalues of A."
Thanks,
Spencer GravesDouglas Bates wrote:
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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