Skip to content

eigenvalues of a circulant matrix

1 message · Globe Trotter

#
Dear Professor Ripley:

Lets do this professionally, shall we?
--- Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote:
Bellman is an excellent book on the topic and that is what I was alluding to,
that you can calculate the eigendecomposition by hand without any costly
computations, actually.
Sorry about kinv.dat -- in the e-mail that came back to me, it read "kinv",
btw.

I am unclear why you say that "X is not circulant as usually defined" -- do you
think you could clarify? It is true I use a Toeplitz matrix to set this up, but
how does that matter? The end result in this case is still a circulant matrix
that is symmetric, is it not? I would like to know why my result is not
circulant here.
Yes, I know that R calls LAPACK (which now contains EISPACK, btw). But I also
know that LAPACK contains complex eigendecomposition routines in addition to
double precision ones and it would need to be used if there is reason to
believe that the result is complex valued. (In particular ZGESDD would do it.)

The eigendecomposition of a matrix is unique. Whatever you think of Bellman,
the book does show how the eigenvectors of a circulant matrix are given by the
complex roots of unity as given above. We have therefore exhibited an
eigendecomposition without actually going through major computations (which is
good, because statistical computing is best when you use it sparingly). 
Why then does the result differ from that in R, and why by so much? (After all,
the eigendecomposition is unique, or is that only fpr real matrices?)
Unclear, but would like to hear about your views on the actual differences in
this specific example.
In my case, my matrix is symmetric and the result is a circulant matrix.
True, but bugs in software are not exactly rare. Though R does have very few
bugs and which is why I recommend the software to every Tom.

Besides I asked a question here because I was confused....

Many thanks and best wishes!