occasional meanderings in physics' brave new world
- Name: Kea
- Location: New Zealand
Marni D. Sheppeard
Monday, July 21, 2008
Recall that the $6 \times 6$ operator $2K$, with basic permutations as eigenvalues, is of the form for circulants $(1)$ and $(2)$. What is the eigenvector? Let $K$ act on an object $(X,Y)$. Then one can solve the eigenvalue equation for $\lambda = (231)$ to obtain provided we do arithmetic mod 7. Try it yourself. The cyclic nature of the linear equations forces the eigenvector to live in such a ring. Choosing $K$ instead, rather than $2K$, we find that the same vector is an eigenvector for the other 1-circulant, $(312)$.