# Help proving Lie's Theorem

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?

## 1 Answer

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 $$g(u_i)=a_{ii}u_i+a_{ii-1}u_{i-1}+..+a_{i1}u_1+b_{i0}w$$.