I'm now learning about binary relations. I stumbled upon this question in the book:
Given $A = \{1,3,5,6\}$ and $R$ is a relation over $A$, whose matrix is defined by
$$\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \end{pmatrix}$$
Determine the smallest natural number $n$ such as $M_{R^n} = I_4$ [notice the $n$ next to $R$]
Calculate $M_{R^{2006}}$
First of all, I don't know what does $M_{R^n}$ mean. What is $n$ supposed to represent? I don't know $I_4$ either, although I guess it means something like "index 4", referring to the fourth element in $A$ (6), but I'm just babbling.
Regarding $M_{R^{2006}}$, I guess I can find out by my own once I understand what is $M_{R^n}$ supposed to be.
The book says that the answer for the first question is that $n$ should be $4$. This doesn't quite help me.
$R^n$ here is $R$ "applied" $n$ times, whose relation will then the $n$th power of the given matrix. The notation used is that the given matrix is $M_{R^1}$, the relation applied "only once." $M_R$ is a permutation matrix, and it is not difficult to see that $M_{R^k} = (M_{R^1})^k = I_4$ for some $k$, where $I_4$ is the $4\times 4$ identity matrix. Once you determine that $k$ you can easily calculate $M_{R^m}$ for any $m$.