The famous Lie-Kolchin theorem in the theory of algebraic groups states:
Let $G$ be a connected solvable subgroup of $GL(V)$, $0 \neq V$ finite dimensional. Then $G$ has a common eigenvector in $V$.
In the proof of this theorem, the author uses induction on the dimension $n$ of $V$ and the derived length $d$ of $G$, the case of $n=1$ or $d=1$ being clear. When $V$ is not irreducible, the induction hypothesis can be applied. So what need to be proved is when $G$ acts irreducibly on $V$, $n$ must be $1$.
Let G' be the derived group of $G$. G' is connected solvable of derived length $d-1$. By induction, G' has a common eighenvector in $V$. Let $W$ be the span of all such vectors in $V$. Then G' acts diagonally on $W$. So, $W$ is $G$-stable.
My question is: Why is $W$ $G$-stable?
More generally, is this statement
Let $G$ be a closed subgroup of $GL(V)$. If $V$ has a basis $\{v_1, \cdots, v_m, v_{m+1}, \cdots, v_n \}$, and $W$ is the subspace spanned by $\{v_1, \cdots, v_m \}$ on which G' acts diagonally, then $W$ is $G$-stable.
true?
Or, even more generally, consider $GL(V)$ as an abstract group, is
Let $G$ be a subgroup (as an abstract group) of $GL(V)$. If every element of G' is of the form $\left( \begin{array}{c:c} D & * \\ \hdashline 0 & * \end{array} \right)$ where $D$ is a diagonal $m \times m$ block ($m$ is less than or equal to the dimension of $V$), and the blocks denoted by a star can consist of any number, then every element of $G$ is of the from $\left( \begin{array}{c:c} * & * \\ \hdashline 0 & * \end{array} \right)$, where the blocks have the same size with the representation of G' elements.
true?
I think the last question is merely about linear algebra.
Thanks very much.