Let $G$ be a connected reductive group defined over a perfect field $F$.  The *split component* $A$ of $G$ is the unique maximal $F$-split subtorus of the radical of $G$.  For an algebraic group $H$ over $F$, let $X(H)$ denote the abelian group of rational characters $H \rightarrow \mathbf{G}_m$, and let $X(H)_F$ be the subgroup of $X(H)$ of characters which are defined over $F$.  The book I'm reading ("Eisenstein Series and Automorphic L-functions" by Freydoon Shahidi, pg. 9) mentions that 

$$\textrm{Hom}_{\mathbb{Z}}(X(G)_F,\mathbb{R}) \cong \textrm{Hom}_{\mathbb{Z}}(X(A),\mathbb{R})$$

as real vector spaces.  I'm having trouble seeing this.  I have worked a couple of examples to try and see what is going on:

**Example 1: $G = \textbf{GL}_3$**

In this case, $X(G) = X(G)_F \cong \mathbb{Z}$, since the rational characters of $G$ consist of powers of the determinant function, which is defined over $F$.  Here we have $A = Z(G)$, the center of $G$, which is a one dimensional torus.  Hence $X(A) \cong \mathbb{Z}$.  $\blacksquare$

**Example 2: $F = \mathbb{R}$, $G = U(2,1)$**

As an algebraic group over $\mathbb{C}$, $G$ is equal to $\textrm{GL}_3$, but the $\mathbb{R}$-structure on $G$ is different: if $x \in G$, and $\overline{x}$ denotes the entrywise application of complex conjugation to $x$, then 

$$G(F) = \{ x \in G : w ^t \overline{x}^{-1}w = x \}$$

where $^t \overline{x}^{-1}$ denotes the conjugate inverse transpose of $x$, and $w = \begin{pmatrix} & & 1 \\ & -1 & \\ 1 & & \end{pmatrix}$.

Here a maximal $F$-split subtorus of $G$ is $$A_0 = \{ \begin{pmatrix} t & & \\ & 1 & \\ & & t^{-1} \end{pmatrix} : t \in \mathbb{C}^{\ast} \}$$ and so the split component of $G$ is trivial.  Hence $X(A) = 0$.  

On the other hand, the nontrivial element $\sigma$ of the Galois group $\Gamma = \textrm{Gal}(\mathbb{C}/\mathbb{R})$ acts on $G$ as $\sigma(x) = w ^t \overline{x}^{-1} w$, and a morphism of $F$-varieties $f: G \rightarrow Y$ is defined over $F$ if and only if $f$ preserves the action of $\sigma$.  Since $\sigma$ acts on $\textbf{G}_m$ as $t \mapsto \overline{t}$, we see that the determinant function $D$ is not defined over $F$: 

$$D(\sigma(x)) = D(w ^t \overline{x}^{-1}w) = \overline{D(x))}^{-1}$$ $$\sigma(D(x)) = \overline{D(x)}$$
hence also $X(G)_F = 0$.  $\blacksquare$

Any insight as to why we always have $\textrm{Hom}_{\mathbb{Z}}(X(G)_F,\mathbb{R}) \cong \textrm{Hom}_{\mathbb{Z}}(X(A),\mathbb{R})$ would be appreciated.