# RE: [R] eigenvalues of a circulant matrix

From: Globe Trotter <itsme_410_at_yahoo.com>
Date: Tue 03 May 2005 - 06:50:23 EST

The example that I submitted earlier in the day. Would you like me to send again?

Thanks!

• "Huntsinger, Reid" <reid_huntsinger@merck.com> wrote:

> For which X?
>
> Reid Huntsinger
>
> -----Original Message-----
> From: Globe Trotter [mailto:itsme_410@yahoo.com]
> Sent: Monday, May 02, 2005 2:34 PM
> To: Huntsinger, Reid; Rolf Turner
> Cc: r-help@stat.math.ethz.ch
> Subject: RE: [R] eigenvalues of a circulant matrix
>
>
> By the way, I just noticed that eigen(X) returns eigenvectors, at least two
> of
> which are NaN's.
>
> Best wishes!
>
> --- "Huntsinger, Reid" <reid_huntsinger@merck.com> wrote:
>
> > When the matrix is symmetric and omega is not real, omega and its
> conjugate
> > (= inverse) give the same eigenvalue, so you have a 2-dimensional
> > eigenspace. R chooses a real basis of this, which is perfectly fine since
> > it's not looking for circulant structure.
> >
> > For example,
> >
> > > m
> > [,1] [,2] [,3] [,4] [,5]
> > [1,] 1 2 3 3 2
> > [2,] 2 1 2 3 3
> > [3,] 3 2 1 2 3
> > [4,] 3 3 2 1 2
> > [5,] 2 3 3 2 1
> >
> > > eigen(m)
> > \$values
> > [1] 11.000000 -0.381966 -0.381966 -2.618034 -2.618034
> >
> > \$vectors
> > [,1] [,2] [,3] [,4] [,5]
> > [1,] 0.4472136 0.000000 -0.6324555 0.6324555 0.000000
> > [2,] 0.4472136 0.371748 0.5116673 0.1954395 0.601501
> > [3,] 0.4472136 -0.601501 -0.1954395 -0.5116673 0.371748
> > [4,] 0.4472136 0.601501 -0.1954395 -0.5116673 -0.371748
> > [5,] 0.4472136 -0.371748 0.5116673 0.1954395 -0.601501
> >
> > and you can match these columns up with the "canonical" eigenvectors
> > exp(2*pi*1i*(0:4)*j/5) for j = 0,1,2,3,4. E.g.,
> >
> > > Im(exp(2*pi*1i*(0:4)*3/5))
> > [1] 0.0000000 -0.5877853 0.9510565 -0.9510565 0.5877853
> >
> > which can be seen to be a scalar multiple of column 2.
> >
> > Reid Huntsinger
> >
> > Reid Huntsinger
> >
> > -----Original Message-----
> > From: r-help-bounces@stat.math.ethz.ch
> > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Huntsinger, Reid
> > Sent: Monday, May 02, 2005 10:43 AM
> > To: 'Globe Trotter'; Rolf Turner
> > Cc: r-help@stat.math.ethz.ch
> > Subject: RE: [R] eigenvalues of a circulant matrix
> >
> >
> > It's hard to argue against the fact that a real symmetric matrix has real
> > eigenvalues. The eigenvalues of the circulant matrix with first row v are
> > *polynomials* (not the roots of 1 themselves, unless as Rolf suggested you
> > start with a vector with all zeros except one 1) in the roots of 1, with
> > coefficients equal to the entries in v. This is the finite Fourier
> transform
> > of v, by the way, and takes real values when the coefficients are real and
> > symmetric, ie when the matrix is symmetric.
> >
> > Reid Huntsinger
> >
> > -----Original Message-----
> > From: r-help-bounces@stat.math.ethz.ch
> > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Globe Trotter
> > Sent: Monday, May 02, 2005 10:23 AM
> > To: Rolf Turner
> > Cc: r-help@stat.math.ethz.ch
> > Subject: Re: [R] eigenvalues of a circulant matrix
> >
> >
> >
> > --- Rolf Turner <rolf@math.unb.ca> wrote:
> > > I just Googled around a bit and found definitions of Toeplitz and
> > > circulant matrices as follows:
> > >
> > > A Toeplitz matrix is any n x n matrix with values constant along each
> > > (top-left to lower-right) diagonal. matrix has the form
> > >
> > > a_0 a_1 . . . . ... a_{n-1}
> > > a_{-1} a_0 a_1 ... a_{n-2}
> > > a_{-2} a_{-1} a_0 a_1 ... .
> > > . . . . . .
> > > . . . . . .
> > > . . . . . .
> > > a_{-(n-1)} a_{-(n-2)} ... a_1 a_0
> > >
> > > (A Toeplitz matrix ***may*** be symmetric.)
> >
> > Agreed. As may a circulant matrix if a_i = a_{p-i+2}
> >
> > >
> > > A circulant matrix is an n x n matrix whose rows are composed of
> > > cyclically shifted versions of a length-n vector. For example, the
> > > circulant matrix on the vector (1, 2, 3, 4) is
> > >
> > > 4 1 2 3
> > > 3 4 1 2
> > > 2 3 4 1
> > > 1 2 3 4
> > >
> > > So circulant matrices are a special case of Toeplitz matrices.
> > > However a circulant matrix cannot be symmetric.
> > >
> > > The eigenvalues of the forgoing circulant matrix are 10, 2 + 2i,
> > > 2 - 2i, and 2 --- certainly not roots of unity.
> >
> > The eigenvalues are 4+1*omega+2*omega^2+3*omega^3.
> > omega=cos(2*pi*k/4)+isin(2*pi*k/4) as k ranges over 1, 2, 3, 4, so the
> above
> > holds.
> >
> > Bellman may have
> > > been talking about the particular (important) case of a circulant
> > > matrix where the vector from which it is constructed is a canonical
> > > basis vector e_i with a 1 in the i-th slot and zeroes elsewhere.
> >
> > No, that is not true: his result can be verified for any circulant matrix,
> > directly.
> >
> > > Such a matrix is in fact a unitary matrix (operator), whence its
> > > spectrum is contained in the unit circle; its eigenvalues are indeed
> > > n-th roots of unity.
> > >
> > > Such matrices are related to the unilateral shift operator on
> > > Hilbert space (which is the ``primordial'' Toeplitz operator).
> > > It arises as multiplication by z on H^2 --- the ``analytic''
> > > elements of L^2 of the unit circle.
> > >
> > > On (infinite dimensional) Hilbert space the unilateral shift
> > > looks like
> > >
> > > 0 0 0 0 0 ...
> > > 1 0 0 0 0 ...
> > > 0 1 0 0 0 ...
> > > 0 0 1 0 0 ...
> > > . . . . . ...
> > > . . . . . ...
> > >
> > > which maps e_0 to e_1, e_1 to e_2, e_2 to e_3, ... on and on
> > > forever. On (say) 4 dimensional space we can have a unilateral
> > > shift operator/matrix
> > >
> > > 0 0 0 0
> > > 1 0 0 0
> > > 0 1 0 0
> > > 0 0 1 0
> > >
> > > but its range is a 3 dimensional subspace (e_4 gets ``killed'').
> > >
> > > The ``corresponding'' circulant matrix is
> > >
> > > 0 0 0 1
> > > 1 0 0 0
> > > 0 1 0 0
> > > 0 0 1 0
> > >
> > > which is an onto mapping --- e_4 gets sent back to e_1.
> > >
> > > I hope this clears up some of the confusion.
> > >
> > > cheers,
> > >
> > > Rolf Turner
> > > rolf@math.unb.ca
> >
> > Many thanks and best wishes!
> >
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