the other hand the action of $T$ on $\mathfrak h_1$ is diagonal $$T \cdot X= X \qquad T \cdot Y = Y \qquad T \cdot Z = 2 Z;$$ and the action of $T$ on $K^3$ is twice the identity: $T \cdot A= 2 A$ for all $A \in K^3$.

The Lie algebra ${\mathfrak n }_{3,2}$
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Let $e_1, e_2, e_3, e_4, e_5, e_6$ denote the basis of ${\mathfrak n }_{3,2}$ given in (\[b32\]) above. A derivation of this Lie algebra has a matrix as follows $$\left( \begin{matrix}
 z_1+z_{2} & z_4 &  z_6&0 & 0 & 0 \\
 z_{5} & z_1+z_{3} & z_{8}&0 & 0 & 0 \\
  z_{7}& z_9 &z_1 -z_{2}-z_3 &0 & 0 & 0 \\
 -z_{12} & -z_{16}+z_{17} & z_{18} & 2z_1+z_{2}+z_3& z_{8}&-z_6\\
 z_{14}+z_{15}& z_{10} & z_{16}+z_{17} & z_9 & 2z_1-z_3&z_4\\
 z_{13} &z_{14}-z_{15}&z_{11}-z_{12}&-z_{7}&z_{5}&2z_1-z_{2}
 \end{matrix}
 \right).$$ Let $\mathfrak G$ and $\mathfrak R$ denote respectively the sets of matrices in $\Der({\mathfrak n }_{3,2})$ $$\mathfrak G=\left\{ \left( \begin{matrix}
 z_1+z_{2} & z_4 &  z_6&0 & 0 & 0 \\
 z_{5} & z_1+z_{3} & z_{8}&0 & 0 & 0 \\
  z_{7}& z_9 &z_1 -z_{2}-z_3 &0 & 0 & 0 \\
    0    &      0 & 0    & 2z_1+z_{2}+z_3& z_{8}&-z_6\\
     0 &  0 &  