Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Take a periodic one-dimensional lattice of size $N$ with $2k$ nearest neighborers. That is, vertex $i$ is connected to $i+1,i+2,...,i+k$ and $i-1,i-2,...i-k$ (with the understanding that the indices are modulo $N$). Call the associated adjacency matrix $A_k$, and consider the limit of large $N$.

If I'm not mistaken, the eigenvalue spectrum for $A_1$ can be well approximated by a cosine wave with amplitude $2$ and period $2N$.

Main Question

Why does the spectra for $A_2$ have a 'kink'?

Side Questions

Why does the kink in $A_2$ appear at a magnitude of zero?

Why do the kinks in $A_3$ appear touch exactly when the curve hits the spectra for $A_2$ (it looks that after a certain point, one graph is supported by the other).

In general, the spectra for $A_k$ seem to have $k-1$ kinks, can this be shown?

I've included an example of the spectra for $N=2^{11}$

share|improve this question
    
I count $2k$ nearest neighbors. –  joriki Apr 21 '11 at 22:54
    
@joriki - you're correct, there are $2k$ nearest neighbors (mistyped) –  Hooked Apr 21 '11 at 23:10

1 Answer 1

up vote 2 down vote accepted

The eigenvectors of such a circulant matrix are $(v_j)_m=\exp(2\pi \mathrm i jm/N)$. The corresponding eigenvalue spectra are the linear combinations of exponentials corresponding to a row of the matrix. For your $A_k$, these combinations are

$$\lambda_{kl}=\sum_{j=1}^k \cos(2\pi\mathrm i jl)\;,$$

Here are plots of the spectra for $A_2$ and for $A_3$. If you reorder them in descending order, as it seems that you have, then once your sort hits a maximum, you suddenly have three branches instead of just one, and also the eigenvalues are more dense around the maxima; that accounts for the kinks. The second, "inverse" kink for $A_3$ is caused by the minimum in the spectrum.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.