Let $G$ be a linear algebraic group, $T$ a subtorus of $G$ of dimension $\geq 1$. Let $\mathfrak g$ be the Lie algebra of $G$. Then the Ad operator $$\textrm{Ad } : G \rightarrow \textrm{GL}(\mathfrak g)$$ is a rational representation of $G$, and the inclusion $T \subseteq G$ induces a rational representation of $T$. Since $T$ is a torus, we know there exist characters $\chi_1, ... , \chi_t$ of $T$, and subspaces $V_i$ of $\mathfrak g$, such that $\mathfrak g$ is the direct sum of the $V_i$, and $\textrm{Ad}$ has the effect $$\textrm{Ad }x(v) = \chi_i(x)v$$ for all $x \in T, v \in V_i$. The characters $\chi_i$ are called the weights of $T$ in $\mathfrak g$. Now just as $T$ acts on $G$ by conjugation, $T$ also acts on $\mathfrak g$ by the Ad operator. And one can show (5.4.7 in Springer), that fixed points correspond to fixed points via the Lie algebra. In other words, $$\mathscr L(Z_G(T)) = \{ X \in \mathfrak g : \textrm{Ad }x(X) = X, \textrm{ for all } x \in T\}$$ Since $T \subseteq Z_G(T)$, $Z_G(T)$ has dimension $\geq 1$, hence $\mathscr L(Z_G(T))$ is a nonzero subspace. It appears that $\mathscr L(Z_G(T))$, given the above description and the fact that it is nonzero, coincides with the weight space corresponding to the trivial character. This seems to show that the trivial character must show up as one of the weights $\chi_1, ... , \chi_t$.
Is this correct? If so, is there a more obvious or intuitive reason why this is true?