Re: [R] eigenvalues of a circulant matrix

From: Globe Trotter <itsme_410_at_yahoo.com>
Date: Mon 02 May 2005 - 23:35:55 EST


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?


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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Mon May 02 23:45:35 2005

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