How can I maintain linear independence through a commutator?

Question

Consider a Lie algebra $\mathcal{L}$, a linearly independent generating set $\mathcal{G}$, and an element $X \in \mathcal{L}$.

Edit: Note that $\mathcal{G}$ is not necessarily a basis; the generation is via the Lie bracket.

What are the conditions on $X$ such that $\{[X,g_i]\; \big| \; g_i \in \mathcal{G}\}$ is also linearly independent?

Or is the question too general and the answer must depend on $\mathcal{G}$? In my specific case I am looking at the Lie algebra $\mathfrak{sp}(2n,\mathbb{R})$ if that helps. Also maybe there is a more general answer for any given set of matrices but this is my specific problem.

Answer 1

Since $\mathcal{G}$ is a basis, we can write $X\neq 0$ as a non-trivial linear combination of elements of $\mathcal{G}$. Then $ad(X)(\mathcal{G})=\{[X,g_i] \mid g_i\in \mathcal{G}\}$ represents $0=[X,X]$ also non-trivially. Hence this set cannot be linearly independent. In formulas: if $X=a_1g_1+\cdots +a_ng_n$, where not all $a_i=0$, then \begin{align*} a_1[X,g_1]+\cdots a_n[X,g_n] & = [X,a_1g_1+\cdots +a_ng_n]\\ & = [X,X]\\ & = 0. \end{align*}