Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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.


I think the last question is merely about linear algebra.

Thanks very much.

share|cite|improve this question
The statement in the second (thus third) grayed box is false. Let $G$ be the subgroup of $GL(2, \mathbb{C})$ consisting of all the $2 \times 2$ matrix of the form $\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ or $\begin{pmatrix} 0 & c \\ d & 0 \end{pmatrix}$ with $a,b,c,d \in \mathbb{C}^*$ does not have the form $\begin{pmatrix} * & * \\ 0 & * \end{pmatrix}$, but its derived group $T(2, \mathbb{C})$ has the form $\begin{pmatrix} * & * \\ 0 & * \end{pmatrix}$. – ShinyaSakai Dec 14 '11 at 18:08
Passing from the orginal question to the second grayed box, I miss not only the algebraic group part, but also the fact that no vector in $\langle v_m \cdots v_n \rangle$ is an eigenvector of $G'$, adding which, the counterexample is no longer a counterexample... I should reconsider... – ShinyaSakai Dec 14 '11 at 18:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.