Definition
Let $T$ be a torus and $R: G \to GL(V)$ a representation. $R(T)$ is a collection of commuting matrices and therefore can be simultaniously diagonalized.
For a character $\lambda \in \mathcal{X}(T)$ define the weight space $$ V_\lambda = \left\{ v \in V \mid \forall \tau \in T : R(\tau)v = \lambda(\tau)v \right\} \ .$$ A weight of a representation $R$ with respect to $T$ is a differential $d\lambda : \mathfrak{t} \to \mathbb{R}$ of a character $\lambda \in \mathcal{X}(T)$ for which $V_\lambda \ne \{0\}$.
The set of weights is denoted by $\Phi_R$.
We have a decomposition $$ V = \bigoplus_{d\lambda \in \Phi_R} V_\lambda \ .$$
Question
I want to show that over $\mathbb{R}$ the set $\Phi_R$ spans the subspace $$ W = \left\{ f \in \mathfrak{t}^* \mid \forall x \in \mathfrak{t} \cap \ker(dR) : f(x) = 0 \right\} \ . $$ If $dR$ is faithful than $\Phi_R$ spans $\mathfrak{t}^*$.
The claim "If $dR$ is faithful than $\Phi_R$ spans $\mathfrak{t}^*$" is a corollary of the general thing I want to prove taking $\ker dR = \{ 0 \}$, and then $f(0)=0$ for all $f \in \mathfrak{t}^*$.
The Example I have in mind is taking $$ (\mathbb{R}^\times)^n = T \ni \tau = \operatorname{diag}(t_1 , ... , t_n) = \operatorname{diag}(e^{x_1}, ... , e^{x_n}) $$ and then $$ \mathfrak{t} = \operatorname{Lie}(T) \ni x = \operatorname{diag}(x_1 , ... , x_n)$$ This is faithful representation $R: T \to GL_n(\mathbb{R})$.
The set of weights: the characters are $\epsilon_1,...,\epsilon_n$ where $\epsilon_i(\tau) = t_i$. Then we take $d \epsilon_i(x) = x_i$.
I do have that if $\epsilon_1,...,\epsilon_n$ are linearly independent characters then $d\epsilon_1,...,d\epsilon_n$ is a basis for $\mathfrak{t}^*$.
I can constuct a non-faithful representation: $$ (\mathbb{R}^\times)^{n+1} \ni (t_1,...,t_n,t_{n+1}) \longmapsto \operatorname{diag}(t_1,...,t_n,1) \in GL_{n+1}(\mathbb{R}) $$ where $t_{n+1} \mapsto 1$.