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eigenvalues of a circulant matrix

2 messages · Globe Trotter, Janusz Kawczak

#
OK, lets redo this again, and ensure that we start with a row that will indeed
lead to a symmetric matrix for the circulant matrix:

x<-scan("kinv")
y<-x[c(109:1,2:108)]

X=toeplitz(y)
Z=y
for (i in 2:216) Z=rbind(Z,y[c((216-i+2):216,1:(216-i+1))])

range(X-Z)
[1] 0 0

eigen(X) is the same as eigen(Z), but we know that Z is a circulant matrix so
the eigenvectors are complex....

Any thoughts/screams?
--- Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote:
#
Again, what's your "kinv"?
On Mon, 2 May 2005, Globe Trotter wrote: