\cr
0 & 0 & B^2
}\right)+
M^{-1}_{\nu_2}
\left(\matrix{
{E^6 \over A^2 B^2} &
{C^2 E^3 \over A^2 B} &
{F^2 E^3 \over A B^2} \cr
{C^2 E^3 \over A^2 B} &
{C^4 \over A^2} &
{C^2 F^2 \over AB} \cr
{F^2 E^3 \over A B^2} &
{C^2 F^2 \over AB} &
{F^4 \over B^2}
}\right)
\label{SRNDmLLpp}\end{aligned}$$ where $$\begin{aligned}
C^2 &=& \lambda_{22}\lambda_{33}-\lambda_{32}\lambda_{23} \\
D^2 &=& \lambda_{33}\lambda_{32}+\lambda_{22}\lambda_{23} \\
E^3 &=& \lambda_{12} (\lambda_{33}^2+\lambda_{23}^2) - 
        \lambda_{13} (\lambda_{33}\lambda_{32}+\lambda_{22}\lambda_{23}) \\
F^2 &=& \lambda_{33}\lambda_{32}+\lambda_{22}\lambda_{23}+\lambda_{12}\lambda_{13}\end{aligned}$$ The diagonalization of the 12 block of $m^{\prime\prime}_{LL}$ in Eq.  is achieved by a $R_{12}$ matrix parameterised by the following $\theta_{12}$ rotation angle : $$s_{12} ={E^3 \over \sqrt{E^6+B^2 C^4}} \qquad
c_{12} ={B C^2 \over \sqrt{E^6+B^2 C^4}}$$ Thus we find : $$m^{\prime\prime\prime}_{LL} \sim
M^{-1}_{\nu_3}
\left(\matrix{
0 & 0 & 0 \cr
0 & 0 & 0 \cr
0 & 0 & B^2
}\right)+
M^{-1}_{\nu_2}
\left(\matrix{
0 & 0 & 0 \cr
0 & {E^6+B^2 C^4 \over A^2 B^2} & {F^2 \sqrt{E^6+B^2 C^4} \over A B^2} \cr
0 &  {F^2 \sqrt{E^6+B^2 C^4} \over A B^2} & {F^4 \over B^2}
\label{SRNDmLLppp}
}\right)$$ It is interesting to note that the $R_{12}$ rotation has not only diagonalized the 12 block of $m^{\prime\prime}_{LL}$ but also put zeros in 