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\begin{document}
\pagestyle{plain} \pagenumbering{arabic}

\begin{itemize}
\item  When $\prod$ is irreducible, this implies that $\exists$ an $n>0$ s.t. for each $\prod^n$ has an element $\prod_{i,j}>0$.  For any $i$ and any $j$.  

\item $\prod_{m \times m}$ is idempotent $\Rightarrow$ $\prod\,^2 = \prod$
\begin{eqnarray}
\prod \cdot \prod\,^2 &=& \prod \cdot \prod		\nonumber
\\
\prod\,^3 &=& \prod\,^2 		\nonumber
\\
\vdots \nonumber
\\
\prod\,^n &=& \prod\,^{n-1} 		\nonumber
\\
 &=& \prod 	
\end{eqnarray}

Therefore $\prod_{i,j}>0$ for $n=1$, $\prod$ is \underline{aperiodic}.


\item Since $\prod$ is idempotent, consider the $j^{th}$ column of two rows, $m$ and $n$.

\begin{eqnarray}
\prod_{m,j}		& = & \sum^{N}_{i=1}{\prod_{m,i}\cdot \prod_{i,j}}\nonumber
\\
1 & = & \sum^N_{i=1}\frac{{\prod_{m,i}\cdot \prod_{i,j}}}{\prod_{m,j}} \nonumber
\\
1 & = & \frac{\prod_{m,1} \cdot \prod_{1,j} + \prod_{m,2} \cdot \prod_{2,j}  + \cdots + \prod_{m,j} \cdot \prod_{j,j} + \cdots + \prod_{m,N} \cdot \prod_{N,j}}{\prod_{m,j}} \nonumber
\end{eqnarray}

\begin{eqnarray}
\prod_{n,j}		& = & \sum^{N}_{i=1}{\prod_{n,i}\cdot \prod_{i,j}}\nonumber
\\
1 & = & \sum^N_{i=1}\frac{{\prod_{n,i}\cdot \prod_{i,j}}}{\prod_{n,j}} \nonumber
\\
1 & = & \frac{\prod_{n,1} \cdot \prod_{1,j} + \prod_{n,2} \cdot \prod_{2,j}  + \cdots + \prod_{n,j} \cdot \prod_{j,j} + \cdots + \prod_{n,N} \cdot \prod_{N,j}}{\prod_{n,j}} \nonumber
\end{eqnarray}


From above

\begin{eqnarray}
\sum^N_{i=1}\frac{{\prod_{m,i}\cdot \prod_{i,j}}}{\prod_{m,j}} & = & 1 - \prod_{j,j} \nonumber
\\
\sum^N_{i=1}\frac{{\prod_{n,i}\cdot \prod_{i,j}}}{\prod_{n,j}} & = & 1 - \prod_{j,j}  \nonumber
\end{eqnarray}

Therefore $\prod_{m,j} = \prod_{n,j}$, thus all rows of the $\prod$ are identical
\end{itemize}

\end{document} 