Now suppose the derived algebra has dimension 1. Then there exits some non-zero $X_1 \in g$ such that $L' = span\{X_1\}$. Extend this to a basis $\{X_1;X_2;X_3\}$ for g. Then there exist scalars $\alpha, \beta , \gamma \in R$ (not all zero) such that
$[X_1,X_2] = \alpha X_1$
$[X_1,X_3] = \beta X_1$
$[X_2,X_3] = \gamma X_1$
Suppose $\alpha \neq 0$. Then construct a new basis, as follows:
$e_1 = X_1$,
$e_2 = \frac{1}{\alpha} X_2$
$e_3 = \alpha X_3 - \beta X_2 + \gamma X_1$
Since $\alpha \neq 0$, by assumption, this is a basis for the Lie algebra g. Let us calculate the Lie brackets for this basis:
$[e_1,e_2] = e_1$
$[e_1,e_3] = 0$
$[e_2,e_3] = 0$
This Lie algebra is seen to be the direct sum of two Lie algebras, two dimensional non abelian lie algebra $\bigoplus$ 1 dimensional lie algebra
my question : 1. For what reason we construct new basis $e_1,e_2,e_3$ in here
2. With explanation above, how can we get conclusion that Lie algebra like that is a direct sum from two dimensional non abelian lie algebra and 1 dimensional lie algebra
source pdf where i get : http://math.ucsd.edu/~abowers/downloads/survey/3d_Lie_alg_classify.pdf