TITLE: Action on the highest weight vector of a representation of a semisimple linear algebraic group
QUESTION [9 upvotes]: Let $G$ be a semisimple linear algebraic group, $V$ a $G$-representation and $v \in V$ a vector of highest weight $\lambda$. Is it true, that for any positive root $\alpha \in R^+$ the one dimensional unipotent subgroup $U_{-\alpha}$ acts trivially on $v$ if and only if $\langle \alpha^{\vee}, \lambda \rangle = 0$?
I'm asking, because I want to apply this to the following situation. For any simple root $\alpha$, I denote the corresponding fundamental weight by $\omega$. Let $V=V(\omega)$ be the simple representation of highest weight $\omega$ with highest weight vector $v$. I'm trying to prove, that the stabilizor of the element $[v] \in \mathbb{P}(V)$ is the maximal parabolic that doesn't contain $U_{-\alpha}$.
The only reference I have for representations of semisimple groups is the small chapter in Humphrey's book Linear algebraic groups. I would also be grateful, if someone could give me a bigger reference on that subject.

REPLY [5 votes]: A couple of added remarks, too long for a comment.   For a more comprehensive reference in the algebraic group setting, there is (uniquely) the second edition of Jantzen's book Representations of Algebraic Groups (AMS, 2003).  This large book of course contains far more than you need here but also allows the field to be of arbitrary characteristic.   While the specific features of the finite dimensional irreducible representations (such as dimensions) usually differ in prime characteristic from the classical case, the "highest weight" technology doesn't really change so much.   
On the other hand, if you stay in characteristic 0 the Lie algebra representations provide most of the information you want and are worked out in an example-oriented geometric spirit by Fulton and Harris.    But the question you ask is really centered on roots and weights in the context of parabolic subgroups, where Jantzen gives the most extensive treatment.    This is dealt with much more concisely in Wilberd's comment.