Product of a Finite Number of Matrices with a Cosine Entry Does any one know how to prove the following identity?
$$
\mathop{\mathrm{Tr}}\left(\prod_{j=0}^{n-1}\begin{pmatrix}  
2\cos\frac{2j\pi}{n} & a \\
b & 0
\end{pmatrix}\right)=2
$$
when $n$ is odd. The product sign means usual matrix multiplication, and $a$ and $b$ are arbitrary real numbers.
Since the product of $2\cos\frac{2j\pi}{n}$ is $2$, so we only need to prove that the trace is a constant polynomial in $a$ and $b$.
Because of the cosine term, the approach of polynomial analysis used in my previous post does not seem to work here.
 A: Write $\xi_j=2\cos(2\pi\frac{j}{n})$ for $j=1,\cdots,n$. (Order will turn out not to matter.) We seek
$${\rm Tr}\left[\prod_{j=1}^n (\xi_j{\rm e}_{11}+a{\rm e}_{12}+b{\rm e}_{21})\right].$$
We're denoting by ${\rm e}_{ij}$ the elementary matrices. They satisfy ${\rm e}_{ij}{\rm e}_{kl}=\delta_{jk}{\rm e}_{il}$. The noncommutative matrix monomials that result from expanding (with $\xi_j^\square a^\square b^\square$s in front) look like
$$\color{Blue}{(\cdots)}{\rm e}_{11}\color{DarkOrange}{(\cdots)}{\rm e}_{11}\color{DarkOrange}{(\cdots)}\cdots\color{DarkOrange}{(\cdots)}{\rm e}_{11}\color{Magenta}{(\cdots)}.$$
In any orange submonomial, the only two factors that appear are ${\rm e}_{12}$ and ${\rm e}_{21}$. It must begin with a ${\rm e}_{12}$ so as not not be annihilated by the preceding ${\rm e}_{11}$, it must end with a ${\rm e}_{21}$ so as to not annihilate the upcoming ${\rm e}_{11}$, and no ${\rm e}_{12}$ or ${\rm e}_{21}$ is adjacent to itself (else that'd just be $0$). Hence the orange monomials like a alternating strings of ${\rm e}_{12}$s and ${\rm e}_{21}$s, and can be factored as powers $\color{DarkOrange}{({\rm e}_{12}{\rm e}_{21})^\square}$.
Similar reasoning shows the blue submonomial factors as either $\color{Blue}{({\rm e}_{12}{\rm e}_{21})^\square}$ or $\color{Blue}{{\rm e}_{21}({\rm e}_{12}{\rm e}_{21})^\square}$ while the magenta submonomial factors as either $\color{Magenta}{({\rm e}_{12}{\rm e}_{21})^\square}$ or $\color{Magenta}{({\rm e}_{12}{\rm e}_{21})^\square {\rm e}_{12}}$. If we pick the first two forms for blue/magenta then the entire monomial will simplify to ${\rm e}_{11}$, if we pick the second two forms for blue/magenta then the entire monomial will simplify to ${\rm e}_{22}$, and if we pick the first form for blue and second form for magenta or conversely then the entire monomial will simplify to ${\rm e}_{12}$ or ${\rm e}_{21}$.
We need them both to be the first form or both the second form to yield a nontraceless elementary matrix. In each case, if there are $j$ instances of ${\rm e}_{11}$ in the original monomial, then there are $j+1$ different nonnegative powers $\square$ that appear in our factorization. If blue/magenta are in the first form, we must count the nonnegative integer solutions to $t_0+\cdots+t_j=(n-j)/2$ and if they are in the second form we must count the solutions to $t_0+\cdots+t_j=(n-j-2)/2$. These counts are achieved with textbook stars and bars combinatorics.
Note that our $j$ must be odd (so $n-j$ is even), and in either form the number of ${\rm e}_{12}$s in the whole monomial equals the number of ${\rm e}_{21}$s. We must also collect together all $\prod_{i\in J}\xi_i$s for all subsets $J$ of the index set $\{1,\cdots,n\}$ with size $|J|=j$, giving elementary symmetric polynomials.
Putting this all together, we have
$$\sum_{j\textrm{ odd}}\left[\left(\!\!\binom{(n-j)/2+1}{j}\!\!\right)+\left(\!\!\binom{(n-j)/2}{j}\!\!\right)\right]e_j(\xi_1,\cdots,\xi_n)(ab)^{(n-j)/2}.$$
Thus we want to show $e_j(\xi_1,\cdots,\xi_n)=0$ for $j<n$ odd. (By Vieta's formulas, this is equivalent to showing that $\displaystyle\prod_{\zeta^n=1}\big(x-(\zeta+\zeta^{-1})\big)+2$ is an odd function when $n$ is odd.)
So far I haven't been able to finish but this seems to be the final stretch.
