M Theory Lesson 209

Recall that the $6\times 6$ operator $2K$, with basic permutations as eigenvalues, is of the form for circulants $\left(1\right)$ and $\left(2\right)$. What is the eigenvector? Let $K$ act on an object $(X,Y)$. Then one can solve the eigenvalue equation for $\lambda =\left(231\right)$ 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, $\left(312\right)$.