Application to the case N=2
The matrix
L_{ik} defined in Eq.
(16) is then
\begin{equation*}
\hat{L}=\left(\begin{array} {cc} 1 & -\frac{1}{\sqrt{3}}\\
1 & \frac{1}{\sqrt{3}}\end{array}\right),
\end{equation*}
with an inverse given by
\begin{equation*}
\hat{L}^{-1}=\frac{\sqrt{3}}{2}\left(\begin{array} {cc} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\
-1 & 1\end{array}\right).
\end{equation*}
The weights are given by the matrix elements
2(L_{0k})^{-1} . We have thence
\omega_0=1 and
\omega_1=1 .