Can someone explain, or even prove the lemma? I've thought about it for about 3 hrs but no idea about the lemma.
$\begin{align*} R_1 &=\left(\begin{array}{rrrr} -1 & 0 & 0 & 0\\ 2/3 & 1 & 0 & 0\\ 2/3 & 0 & 1 & 0\\ 2/3 & 0 & 0 & 1 \end{array}\right) &\qquad R_2&=\left(\begin{array}{rrrr} 1 & 2/3 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 2/3 & 1 & 0\\ 0 & 2/3 & 0 & 1 \end{array}\right)\\ R_3 &= \left(\begin{array}{rrrr} 1 & 0 & 2/3 & 0\\ 0 & 1 & 2/3 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 2/3 & 1 \end{array}\right), & R_4&=\left(\begin{array}{rrrr} 1 & 0 & 0 & 2/3\\ 0 & 1 & 0 &2/3\\ 0 & 0 & 1 & 2/3\\ 0 & 0 & 0 & -1 \end{array}\right). \end{align*}$
Lemma. Let $M_1,M_2,\ldots,M_d$ be a sequence of the matrices $R_i$, where $2/3$ is replaced by $a$. Assume $M_i\neq M_{i+1}$, $i=1,2,\ldots,d-1$. Then each of the sixteen terms in the product matrix $M_1M_2\cdots M_d$ is a polynomial in $a$ with integer coefficients and lead coefficient $1$ or $-1$ (or else the $0$ polynomial). Moreover, if $M_1=R_i$ and $M_d=R_j$, then:
- The entry in the $i$th row and $j$th column of $M_1M_2\cdots M_d$ has degree $d-1$. Other entries in the $i$th row have degree less than $d-1$.
- The other three entries in the $j$th column have degree $d$. The entries not on the $i$th row and not on the $j$th column have degree less than $d$.