A premise: this comes from a physics book (Quantum Field Theory and Critical Phenomena by J. Zinn-Justin, page 323, 4th edition), so the math may need some correcting...
Let $G$ be a Lie group acting on $\mathbf{R}^n$, $v\in\mathbf{R}^n$, and $\{t_a\}$ a basis of the Lie algebra $g$ of $G$ such that the first $p$ elements generate $\mathrm{Stab}_v(G)$, the stabilizer subgroup of $v$. Let $g$ have a representation acting on $\mathbf{R}^n$, too, such that it makes sense to write, if $h\in G$ is $\exp(\epsilon^at_a)$, \begin{equation} \exp(\epsilon^at_a)v\approx v+\epsilon^at_av \tag{1} \end{equation} for small $\epsilon_a$. The author claims that if \begin{equation} \sum_{a=p+1}^{n}\lambda^at_av=0 \tag{2} \label{2} \end{equation} then $\lambda_a=0$ for all $a$, i.e. the $t_av$ are linearly independent (for $a>p$).
I'm trying to prove this claim. I see that the first equation implies that $t_av=0$ for all $a=1,\dotsc,p$, since the result must be equal to $v$. I tried the following: take a linear combination $\sum_{a=1}^{n}\lambda_at^a$ such that \eqref{2} holds, then \begin{equation} \exp\biggl(\sum_{a=1}^{n}\lambda_at^a\biggr)v\approx v+\exp\sum_{a=1}^{n}\lambda_at^av= v+\exp\sum_{a=1}^{p}\lambda_at^av= v \end{equation} which means that $\exp(\sum_{a=1}^{n}\lambda_at^a)\in\mathrm{Stab}_v(G)$ thus $\lambda_a=0$ for all $a\geq p+1$. I think there's something wrong in this proof, but I can't quite put my finger on it...
First of all, you should not use the same notation $n$ for the dimension of $G$ as for the dimension of the vector space ${\mathbb R}^n$ it acts on; hence, I will use $N$ for the dimension of $G$.
Then the claim is true and can be verified as follows. Let $G_v<G$ denote the stabilizer of $v$ in $G$; let $g_v$ denote the Lie algebra of $G_v$. Then $$ g_v=\{t\in g: tv=0\}. $$ We have the linear map $$ A: g_v\to {\mathbb R}^n, t\mapsto tv. $$ Then $g_v$ equals the kernel of this map. The basis vectors $t_a, a=p+1,...,N$ form a basis for a subspace $S$ in $g$ which is complementary to $g_v$, i.e. such that $g=g_v\oplus S$. Now, consider the restriction of the linear map $A$ to $S$: This restriction has zero kernel (since $S\cap g_v=\{0\}$). Hence, $A|_{S}$ is a linear isomorphism to its image. In other words, the vectors $$ A(t_{p+1}),..., A(t_{N}) \in {\mathbb R}^n $$ are linearly independent, as required. qed