Does R have a function for the Schur decomposition? The documentation for library(Matrix) describes a function "Schur", but it seems to be missing from the Windows version 0.8-14 (2004-09-14) and 0.8-15 (2004-10-02).

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 Graves

Douglas 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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Spencer Graves, PhD, Senior Development Engineer
O:  (408)938-4420;  mobile:  (408)655-4567

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