Does anyone have an idea how to prove the following identity?

$$ \mathop{\mathrm{Tr}}\left(\prod_{j=0}^{n-1}\begin{pmatrix} x^{-2j} & -x^{2j+1} \\ 1 & 0 \end{pmatrix}\right)= \begin{cases} 2 & \text{if } n=0\pmod{6}\\ 1 & \text{if } n=1,5\pmod{6}\\ -1 & \text{if } n=2,4\pmod{6}\\ 4 & \text{if } n=3\pmod{6} \end{cases}, $$ where $x=e^{\frac{\pi i}{n}}$ and the product sign means usual matrix multiplication.

I have tried induction but there are too many terms in all of four entries as $n$ grows. I think maybe using generating functions is the way?


Not a complete answer, but a promising start. Just reformulating the problem using recurrences.

Let $$ A_k = \begin{pmatrix} x^{-2k} & -x^{2k+1}\\1 & 0 \end{pmatrix}\\ B = A_0 A_1 \cdots A_{n-1}. $$ I'm going to find the eigenvalues of $B$, so $\operatorname{tr} B = \lambda_1 + \lambda_2$.

Consider a sequence $u_0, u_1, \dots$ and a related sequence $$ F_k = \begin{pmatrix}u_k \\ u_{k+1}\end{pmatrix}. $$ Note that $$ F_k = A_k F_{k+1} $$ is equivalent to $$ u_k = x^{-2k} u_{k+1} - x^{2k+1} u_{k+2}\\ u_{k+1} = u_{k+1}. $$ If $F_n$ is an eigenvector of $B$ then $$ \lambda F_n = B F_n = A_0 A_1 \cdots A_{n-1} F_n = \\ = A_0 A_1 \cdots A_{n-2} F_{n-1} = \cdots = A_0 F_1 = F_0 $$ or $$ \lambda u_n = u_0\\ \lambda u_{n+1} = u_1. $$ Eliminating $\lambda$ one gets the necessary condition $$ \frac{u_n}{u_{n+1}} = \frac{u_0}{u_1}. $$ Let $\beta_k = \frac{u_k}{u_{k+1}}$. Then the condition becomes $$ \beta_n = \beta_0 $$ if $F_n$ is the eigenvector. We already have the recurrence equation for $u_k$, let's derive one for $\beta_k$ (dividing by $u_{k+1}$): $$ u_k = x^{-2k} u_{k+1} - x^{2k+1} u_{k+2}\\ \beta_k = x^{-2k} - \frac{x^{2k+1}}{\beta_{k+1}}. $$ Introducing new $\gamma_k = (-x)^{-k}\beta_k,\; \omega \equiv -x^{-3} = x^{n-3}$, $$ \gamma_k = \omega^k + \frac{1}{\gamma_{k+1}} \tag1. $$ If $\beta_0 = \beta_n$ then $$ \gamma_0 = \beta_0 = \beta_n = (-1)^n x^n \gamma_n = (-1)^{n+1} \gamma_n\\ \gamma_0 = \gamma_{2n}. $$ The condition $\gamma_0 = \gamma_{2n}$ together with recurrence relation $(1)$ gives a quadratic equation for two possible values $\gamma_0, \gamma_0'$, each corresponding to an eigenvalue $\lambda, \lambda'$ of the matrix $B$. The eigenvalues $\lambda, \lambda'$ are related to $\gamma_k, \gamma_k'$ as following $$ \lambda = \frac{u_0}{u_n} = \prod_{k=0}^{n-1} \frac{u_k}{u_{k+1}} = \prod_{k=0}^{n-1} \beta_k = \prod_{k=0}^{n-1} (-x)^k \gamma_k = (-x)^{n(n-1)/2}\prod_{k=0}^{n-1} \gamma_k $$

  • $\begingroup$ Thank you very much, uranix! But I am not clear how do you get $\gamma_0$=$\gamma_{2n}$, and how to calculate the product of $\gamma_k$'s in the end. It seems that relation (1) results in a continued fraction. $\endgroup$
    – David Sun
    Aug 8 '15 at 6:11
  • $\begingroup$ @DavidSun It's not a continuous fraction, generalized cf at best. Tried that way, but no luck (just getting another interpretation of the same matrix product). $\beta_k$ is $n$-periodic sequence and $\gamma_k = (-1)^{n+1} \gamma_{n + k}$, thus $\gamma_k$ is $2n$ periodic at worst. I see no way to evaluate the product yet $\endgroup$
    – uranix
    Aug 8 '15 at 7:40
  • 1
    $\begingroup$ This problem has been completely solved by using polynomial analysis. The solution is provided under my cross-post Product of a Finite Number of Matrices Related to Roots of Unity. Thanks for your effort all the same, uranix! $\endgroup$
    – David Sun
    Aug 9 '15 at 17:16

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