TITLE: Intuition for the invariance lemma
QUESTION [6 upvotes]: The invariance lemma, as stated below, is commonly used to prove Lie’s theorem.

Lemma (Invariance lemma). Let $\mathbb{k}$ be a field of characteristic zero.
Let $\mathfrak{g}$ be a $\mathbb{k}$-Lie algebra, let $M$ be a finite-dimensional representation of $\mathfrak{g}$ and let $I$ be an ideal of $\mathfrak{g}$.
Let $\lambda$ be an element of the dual space $I^*$ and let
$$
  M_\lambda
  :=
  \{
    m \in M
    \mid
    \text{$x m = \lambda(x) m$ for all $x \in I$}
  \}
$$
be the corresponding $I$-weight space of $M$.
This weight space is already a $\mathfrak{g}$-subrepresentation of $M$.

Every time I’m reading through my old notes on representation theory of Lie algebras I’m wondering where this lemma comes from.
I have no intuition for why it should be true, and if I hadn’t seen a proof of it I would probably even suspect it to be false.

What is the intuition behind the invariance lemma?

REPLY [2 votes]: This is a slight extension of the idea that for commuting endomorphisms $S,T$ of a vector space $V$, the $\lambda$-eigenspace $V_\lambda$ of $T$ is stabilized by $S$. Maybe good to recall the one-line proof: for $v\in V_\lambda$,
$$
T(S v) \;=\; (TS)v \;=\; (ST)v \;=\; S(Tv)\;=\; S(\lambda v) \;=\; \lambda Sv
$$
so $Sv$ is in $V_\lambda$.
In your case, $\lambda:I\to k$, and $\mathfrak g$ does not quite commute with $I$, but stabilizes it. But the proof is essentially the same.