where we note that the first component is fully fixed by $F_3$. This neatly illustrates the $(1,2)$ block form of transformed magic mixing matrices, which is basically a combination of the diagonal of $1$-circulant eigenvalues and a $2$-circulant codiagonal. For each $R_2$ factor, there is an $F_3$ transform that fixes it. It follows that for a suitable twisted Fourier transform, the whole triple product will remain fixed.
15 years ago


No comments:
Post a Comment
Note: Only a member of this blog may post a comment.