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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 :

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We want to split the Lie algebra into $\mathfrak{g}=\mathfrak{g}_1\bigoplus \mathfrak{g}_2$ where $\dim\mathfrak{g}_1 =2$ , $\dim\mathfrak{g}_2=1$ and $\mathfrak{g}_1$ is non abelian. Can you define $\mathfrak{g}_1,\mathfrak{g}_2$ using the basis $\{X_1,X_2,X_3\}$?

With the new basis this can be done in an easy way. Define $\mathfrak{g_1},\mathfrak{g_2}$ as $\mathfrak{g}_1= \operatorname{span}\{e_1,e_2\}$ and $\mathfrak{g}_2=\operatorname{span}\{e_3\}.$ Now prove that $\mathfrak{g}=\mathfrak{g}_1\bigoplus \mathfrak{g}_2$ i.e. $\mathfrak{g}=\mathfrak{g}_1+ \mathfrak{g}_2 , \ \mathfrak{g}_1 \cap \mathfrak{g}_2 = \{0\} \ \text{and} \ [\mathfrak{g_1},\mathfrak{g_2}] = \{0\}$. Finally prove that $\mathfrak{g_1}$ is non abelian (i.e. $[\mathfrak{g_1},\mathfrak{g_1}] \neq \{0\}$).

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