Trace of a Product of Finitely Many Matrices with Cosine Entry Can someone help me prove the following identity?
$$
\mathop{\mathrm{Tr}}\left(\prod_{j=0}^{n-1}\begin{pmatrix}  
2\cos\frac{2j\pi}{n} & -m \\
1 & 0
\end{pmatrix}\right)=
\begin{cases}
2 & \text{if } n=1\pmod{2}\\
2m^{\frac{n}{2}} & \text{if } n=0\pmod{4}\\
-2m^{\frac{n}{2}}-4 & \text{if } n=2\pmod{4}\\
\end{cases},
$$
For the 1st equality, since the product of $2\cos\frac{2j\pi}{n}$'s is 2 when $n$ is odd, we only need to prove that the trace is a constant polynomial in $m$. However, because of the cosine term, the approach of polynomial analysis given in my previous post does not seem to work here. 
Situations are the same for the 2nd and 3rd equalities.
 A: I believe this can be evaluated by expanding cos as a sum of exponentials.
Let $\zeta=\exp(2i\pi/n)$. Consider the set $X$ of $n$-tuples $x_0,\dots,x_{n-1}\in\{+,-,1,-m\}$
where for each $j$ we require $x_j=1$ if and only if $x_{j-1}=-m$. Subscripts are modulo $n$.
Define $w_j(+)=\zeta^j$ and $w_j(-)=\zeta^{-j}$ and $w_j(1)=1$ and $w_j(-m)=-m$.
Expanding $2\cos\frac{2j\pi}{n}=w_j(+)+w_j(-)$ gives the desired trace as $\sum_{x\in X}w(x)$ where $w(x)=\prod_{j=0}^{n-1}w_j(x_j)$.
Cyclicly permuting by moving the end element to the start has the effect of multiplying $w(x)$ by $\zeta^{N_+(x)-N_-(x)}$ where $N_+$ and $N_-$ are the number of $+$'s and $-$'s.
The total weight from sequences with $N_+(x)-N_-(x)\not\in\{-n,0,n\}$ is therefore zero.
Now we need another group action.
Given $x\in X$ that is not all $+$'s, for each maximal run of $+$'s, move the $-$ or $-m,1$ after the run to the beginning. Each replacement multiplies $w(x)$ by $\zeta^{2k}$ where $k$ is the length of the run, so this procedure gives a sequence $x'$ with $w(x')=w(x)\zeta^{2N_+(x)}$.
This shows that the total contribution from all sequences with $(N_+(x),N_-(x))=(a,b)$ is zero whenever $a\not\in\{0,n/2,n\}$.
A similar argument applies when $b\not\in\{0,n/2,n\}$.
The only non-constant terms not accounted for are those containing no $+$'s or $-$'s, i.e. $1,-m,1,-m,\dots,1,-m$ and $-m,1,-m,\dots,1,-m,1$ for even $n$,
which contribute the $2(-m)^{n/2}$.
