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Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity in a suitable extension of ${\mathbb F}_p$.

The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$ (which turns out to be a discrete Fourier analysis). Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s.

The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the exceptional case that this lcm is odd (unlikely) and the order of the solutions of $t^2-6t+1=0$ divide $\ell$.

In other words, defining the vectors $v_\omega=(1,\omega,\ldots,\omega^{\ell-1})$, we see that the subspaces defined by $y,z\in{\rm Span}(v_\omega,v_{\omega^{-1}})$ are stable under both $A$ and $B$, and we are able to compute the spectrum of $AB$ by studying its restriction to these $4$-dimensional (exceptionally $2$-dim) spaces.

Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity in a suitable extension of ${\mathbb F}_p$.

The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$ (which turns out to be a discrete Fourier analysis). Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s.

The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the exceptional case that this lcm is odd (unlikely) and the order of the solutions of $t^2-6t+1=0$ divide $\ell$.

In other words, defining the vectors $v_\omega=(1,\omega,\ldots,\omega^{\ell-1})$, we see that the subspaces defined by $y,z\in{\rm Span}(v_\omega)$ are stable under both $A$ and $B$, and we are able to compute the spectrum of $AB$ by studying its restriction to these $2$-dimensional spaces. Notice that the action on the spaces associated with $v_{\omega}$ and $v_{\omega^{-1}}$ are conjugate to each other.

Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity. The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$. Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s.

The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the exceptional case that this lcm is odd (unlikely) and the order of the solutions of $t^2-6t+1=0$ divide $\ell$.

Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity in a suitable extension of ${\mathbb F}_p$. 

The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$ (which turns out to be a discrete Fourier analysis). Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s.

The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the exceptional case that this lcm is odd (unlikely) and the order of the solutions of $t^2-6t+1=0$ divide $\ell$.

In other words, defining the vectors $v_\omega=(1,\omega,\ldots,\omega^{\ell-1})$, we see that the subspaces defined by $y,z\in{\rm Span}(v_\omega,v_{\omega^{-1}})$ are stable under both $A$ and $B$, and we are able to compute the spectrum of $AB$ by studying its restriction to these $4$-dimensional (exceptionally $2$-dim) spaces.

Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity. The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$. Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s.

The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies either $\omega=-1$ (then $\ell$ itself is even) or $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the case this lcm is odd (unlikely).

Denote $L=2\ell$, and let me assume that $p$ is an odd prime, not dividing $\ell$. Then there are $\ell$ distinct $\ell$-th roots of unity. The Question can be solved by Fourier analysis over ${\mathbb F}_p^\ell$. Let me split an $L$-vector $x$ into its odd and even parts: $$y_i=x_{2i-1},\qquad z_i=x_{2i}.$$ Then $x':=Ax$ and $x'':=Bx$ are given by $$y'=y+z,\quad z'=y-z,\qquad y_i''=-y_i+z_{i-1},\quad z_i''=z_i+y_{i+1}.$$ Let me define the Fourier transform of an $\ell$-vector $v$ by $$\hat v(\omega)=\sum_iv_i\omega^{-i},$$ where the argument $\omega$ takes values in the group of $\ell$-th roots of unity. The formula above yield $$\hat y'=\hat y+\hat z,\quad\hat z'=\hat y-\hat z,\qquad\hat y''(\omega)=-\hat y(\omega)+\omega^{-1}\hat z(\omega),\quad\hat z''(\omega)=\hat z(\omega)+\omega\hat y(\omega).$$ We infer that $X:=Mx$ is given by $$X(\omega)=\hat M(\omega)x(\omega),\qquad\hat M(\omega):=\begin{pmatrix} \omega-1 & 1+\omega^{-1} \\ -1-\omega & \omega^{-1}-1 \end{pmatrix}.$$ The characteristic polynomial of $\hat M(\omega)$ is $$P_\omega(\lambda)=\lambda^2-(\omega+\omega^{-1}-2)\lambda+4.$$ The spectrum of $M$ is thus the union of the pairs $(\lambda_\omega,\mu_\omega)$ of roots of the $P_\omega$'s.

The blocks $M(\omega)$ are diagonalisable, unless an $\ell$-root of unity satisfies $\omega^2-6\omega+1=0$. The order of $M$ in ${\bf M}_L({\mathbb F}_p)$ is the lcm of the orders of the eigenvalues $\lambda_\omega,\mu_\omega$ in the algebraic closure (actually some finite extension), mulitiplied by $2$ in the exceptional case that this lcm is odd (unlikely) and the order of the solutions of $t^2-6t+1=0$ divide $\ell$.