The reference is Linear Algebraic Groups by A. Borel. Fix a Borel subgroup $B_0$ of a connected linear algebraic group $G$. Let $T$ be a maximal torus of $G$. Then there exists an embedding $G \rightarrow \textrm{GL}(V)$ and a $0 \neq w \in V$ such that $B_0$ consists of the set of $x \in G$ for which $xw$ is a scalar multiple of $w$. Then $G/B_0$ is isomorphic to the orbit $X$ of $[w]$ in $\mathbb{P}(V)$.
We can choose $V$ so that $X$ is not contained in any hyperplane in $\mathbb{P}(V)$. There exits a basis $v_1, ... , v_n$ of $V$ together with characters $\alpha_1, .. , \alpha_n$ of $T$ such that $\alpha_i(t)v_i = tv_i$ for all $i$ and all $t \in T$.
I'm confused on a point in part (3) of the proposition. One can show that there is a unique index $i_{\lambda}$ such that $\langle \alpha_{i_{\lambda}}, \lambda \rangle > \langle \alpha_i, \lambda \rangle$ for all $i \neq i_{\lambda}$. One then shows that $[v_{i_{\lambda}}]$ is an element of $X$ which is fixed by $T$, and corresponds to the desired Borel subgroup $B(\lambda)$.
Now, Borel takes the characters $\beta_i$ in (3) to be the differences $\alpha_{\lambda_i} - \alpha_i$ for $i \neq i_{\lambda}$. I'm confused on the argument for why $\beta_i$ is trivial on $T \cap R(G)$. Borel says the following: "Since $\alpha_i$ are characters of $T$, trivial on $T \cap R(G)$, this proves (3)."
What I don't understand is why the $\alpha_i$ are trivial on $T \cap R(G)$.
I mean, I think I can argue that the $\beta_i$ are trivial on $T \cap R(G)$. It's just that Borel seems to be claiming something stronger.
My proof that $\beta_i$ is trivial on $T \cap R(G)$:
More generally, I can prove that the differences $\alpha_i - \alpha_j$ for all $i, j$ are trivial on $T \cap R(G)$. If $i = j$, there's nothing to prove, so suppose $i \neq j$. Then there exists a point $[v] = [c_1v_1 + \cdots + c_nv_n]$ in $X$ for which both $c_i$ and $c_j$ are nonzero: otherwise, $X$ is contained in the union of two hyperplanes $H_i = \{ [a_1v_1 + \cdots + a_nv_n] : a_i = 0 \}$ and $H_j$, which we assumed not to be the case.
Now, let $t \in T \cap R(G)$. Now $t$ fixes a point of $X$ if and only if $t$ lies in the corresponding Borel subgroup of that point. So $R(G)$ is the set of $g \in G$ which fix every point of $X$. In other words, $tv = \lambda v$ for some $\lambda \neq 0$. Then $$\lambda c_1v_1 + \cdots + \lambda c_nv_n = \alpha_1(t)c_1v_1 + \cdots + \alpha_n(t) c_n v_n $$ so $\lambda c_iv_i = \alpha_i(t)c_iv_i$, and $\lambda c_jv_j = \alpha_j(t) c_jv_j$. This implies $\alpha_i(t) = \lambda = \alpha_j(t)$.
This implies $(\alpha_i - \alpha_j)(t) = 1$, for any $t \in T \cap R(G)$. $\blacksquare$
Is Borel correct that $\alpha_i$ is trivial on $T \cap R(G)$?
I may be missing something, but as far as I understand, Borel constructs the representation of $G$ in $GL(V)$ in such a way that it factors over $G/R(G)$. Thus elements of $R(G)$ act trivially on $V$, so the weights $\alpha_i$ must map $R(G)$ to $1$.