Lie's theorem, different formulations

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Lie's theorem : Let $\mathfrak{g}$ be a solvable Lie algebra, $V$ a finite dimensional complex vector space, and let $π : g → gl(V )$ be a Lie algebra representation. Then there is a simultanous eigenvector $v \in V$ for all the members of $π(\mathfrak{g})$.

I was wondering why this implies that all transformations in $\pi(\mathfrak{g})$ are simultaneously triangularizable (if that word even exist).

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This is completely parallel to the linear algebra proof that that fact than any linear endomorphism over $\mathbb C$ has an eigenvector implies that any linear map over $\mathbb C$ is triangularizable. You start with a joint eigenvector $v_1$ which spans an invariant subspace $V_1\subset V$. Then you take the quotient representation of $\mathfrak g$ on $V/V_1$. This has a joint eigenvector $\tilde v_2$ and you choose a representative $v_2\in V$. Then by construction, the action of any element of $\mathfrak g$ maps $v_2$ to a linear combination of $v_2$ and $v_1$, so $v_1$ and $v_2$ span an invariant subspace $V_2$.

Continuing inductively, you find a basis $\{v_i\}$ of $V$ such that the action of any element of $\mathfrak g$ maps $v_i$ to a linear combination of the vectors $v_j$ with $j\leq i$. But this exactly means that the matrix of any transformation in $\pi(\mathfrak g)$ with respect to this basis is upper triangular.