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I'm studying Lie algebras, and I'm struggling to prove the following result:

Let $L$ be a solvable subalgebra of $\mathfrak{gl}(V)$, $dimV = n < \infty$. Then $L$ stabilizes some flags in $V$ (in other words, the matrices of $L$ relative to a suitable basis of $V$ are upper triangular).

The instruction to prove this is to use a preceding theorem, namely:

Let $L$ be a solvable subalgebra of $\mathfrak{gl}(V)$, $V$ finite dimensional. If $V \neq 0$, then $V$ contains a common eigenvector for all the endomorphisms in $L$.

My attempt was:

If $dimV = 1$, then the result is true.

Now suppose the result is proven for $dimV < n$. If $dimV = n$, by the preceding theorem, since $L$ is solvable, there is $v \in V - \{0\}$ such that $v$ is eigenvector for all $x \in L$. If $W = Span(v)$, then $W$ is one dimensional, and $\frac{V}{W}$ has dimension $n-1$. I've done this so I could use the induction hypothesis, but I feel it doesn't work, since I don't know for sure if $L$ is a subalgebra of $\mathfrak{gl}(\frac{V}{W})$.

I do think I am in the right track, because of this question. Any help?

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Let $p:V\rightarrow V/W$ be the quotient map. For every $g\in L$, consider the endomorphism $g_p$ of $V/W$ such that $g_p(p(x))=p(g(x))$.

$g_p$ is well defined, if $p(x)=p(y), x-y\in W$ and $p(g(x-y))=p(g(x))-p(g(y))$. The map $f:L\rightarrow gl(V/W)$ defined by $f(g)=g_p$ is a morphism of Lie agebras and we denote by $L'$ its image. By hypothesis there exists a basis $v_1,...,v_{n-1}$ of $V/W$ such that the elements of $L'$ in $(v_1,...,v_{-1})$ are upper triangular matrices.

Let $u_i$ such that $p(u_i)=v_i$ and $w$ a generator of $W$, show that the matrices of the elements of $L$ in $(w,u_1,...,u_n)$ are upper triangular matrices. To see this write $f(g)(v_i)=a_{ii}v_i+a_{ii-1}v_{i-1}+..+a_{i1}v_1$, $p(g(u_i)-a_{ii}u_i+a_{ii-1}u_{i-1}+..+a_{i1}u_1))=0$. This implies that $g(u_i)=a_{ii}u_i+a_{ii-1}u_{i-1}+..+a_{i1}u_1+b_{i0}w$.

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