## anonymous 3 years ago What is a cyclic matrix?

1. anonymous

An another matrix named centered matrix is the inverse of cyclic matrix ?

2. anonymous

I assume you are referring to cyclic permutation matrices. These permute the rows or columns of a given matrix ( depending on whether they multiply on the left or on the right ) such that all rows/columns are permuted and you will get back to the original matrix after some number of permutations. Permutations simply means re-arrange objects amongst themselves and that can lead to many patterns of use. Taking matrices of size 2 x 2 then $\left[\begin{matrix}1& 0 \\ 0 & 1\end{matrix}\right]\left[\begin{matrix}1& 2 \\ 3 & 4\end{matrix}\right]= \left[\begin{matrix}1& 2 \\ 3 & 4\end{matrix}\right]=\left[\begin{matrix}1& 2 \\ 3 & 4\end{matrix}\right]\left[\begin{matrix}1& 0 \\ 0 & 1\end{matrix}\right]$has the identity matrix producing a 'null' permutation. While this seems trivial there is an important point here ( see later ). Now $\left[\begin{matrix}0& 1 \\ 1 & 0\end{matrix}\right]\left[\begin{matrix}1& 2 \\ 3 & 4\end{matrix}\right]= \left[\begin{matrix}3& 4 \\ 1 & 2\end{matrix}\right]$and $\left[\begin{matrix}1& 2 \\ 3 & 4\end{matrix}\right]\left[\begin{matrix}0& 1 \\ 1 & 0\end{matrix}\right]= \left[\begin{matrix}2& 1 \\ 4 & 3\end{matrix}\right]$emphasising that what a given matrix does under multiplication depends on ordering ie. matrix multiplication does not generally permute. Here a permutation matrix applied to the left side re-arranges rows, and if applied to the right side it re-arranges columns. So if we apply those permutations again : $\left[\begin{matrix}0& 1 \\ 1 & 0\end{matrix}\right]\left[\begin{matrix}3& 4 \\ 1 & 2\end{matrix}\right]= \left[\begin{matrix}1& 2 \\ 3 & 4\end{matrix}\right]$and $\left[\begin{matrix}2& 1 \\ 4 & 3\end{matrix}\right]\left[\begin{matrix}0& 1 \\ 1 & 0\end{matrix}\right]= \left[\begin{matrix}1& 2 \\ 3 & 4\end{matrix}\right]$we arrive back where we started. In fact $\left[\begin{matrix}0& 1 \\ 1 & 0\end{matrix}\right]\left[\begin{matrix}0& 1 \\ 1 & 0\end{matrix}\right]= \left[\begin{matrix}1& 0 \\ 0 & 1\end{matrix}\right]$or $P^{2}=I$ from which we deduce that $P=P^{-1}$You'll note that I produced P by re-arranging the rows/columns of the identity. We've just exhausted the ways of doing such permutations for 2 x 2, and for 3 x 3 there are more ways. In any case for given dimensions the permutation matrices are what is formally called a 'group'. Any two members of the group when combined from another group member, there is an identity member, and each member has an inverse. My understanding of the term 'centered matrix' doesn't seem to apply here ( please advise ).