\beta} \omega^2 & {\overline \beta} \omega & 0 & - \alpha & - \alpha
\omega^2 & -\alpha \omega & 0 \\ 0 & 0 & 0 & \sqrt{3} \beta & 0 & 0&
0& \sqrt{3}\alpha \end{array}\right).$$ The matrix $Q_8$ maps the sixth row to the second row leading to the first two rows $$\left( \begin{array}{cccc|cccc} \alpha & \alpha &
\alpha & 0 & \beta & \beta & \beta & 0 \\ \beta & \beta \omega & \beta
\omega^2 & 0 & \alpha & \alpha \omega & \alpha \omega^2 & 0
\end{array} \right)$$ with two zero columns that do not change the POVM due to zero probability. For convenience, we shift the qubits according to the mapping ${|x_1 \ldots x_{l-1} x_l\rangle} \mapsto {|x_l x_1 \ldots x_{l-1}\rangle}$. We denote this permutation by $R$. After this reordering of qubits the matrix $T_r$ takes the simple form $$RT_rR^\dagger = A \otimes I_{r/4} := 
\sqrt{m}\left( \begin{array}{cccc} \alpha & \beta & 0 &
0 \\ {\overline \beta} & -\alpha & 0 & 0 \\ 0 & 0 & 
\alpha & -{\overline \beta} \\ 0 & 0 & \beta & \alpha 
\end{array} \right) \otimes
I_{r/4} .$$ By combining this equation with Equation (\[Gleichung\]) we get the factorization 