Let $G$ be an algebraic torus ($\cong(\mathbb C^*)^n$) acting linearly on a finite dimensional $\mathbb C$ vector space $V$. For a character $\chi$ of $G$, we define $V_{\chi}=\{v\in V|\forall g\in G, g\cdot v=\chi(g)v\}$.
I want to show that $V=\oplus_{\chi}V_{\chi}$.
I am following the book 'Linear algebraic groups' by Springer Theorem $3.2.3$, page $44$.
Let $X$ denotes the set of all characters of $G$ and $\phi:G\rightarrow GL(V)$ be the rational representation. This can be seen as a map to $\mathbb C^{n^{2}}$, therefore we have $\phi(g)_{i,j}\in\mathbb C[G]$ and we may write $\phi(g)_{i,j}=\sum_{\chi}a(i,j)_{\chi}\chi$, where $a(i,j)_{\chi}$ are constants. Thus we have \begin{align}\phi=\sum_{\chi}\chi A_{\chi}\end{align} for some $A_{\chi}\in GL(V)$(why $A_{\chi}$ is in $GL(V)$?). Note that only finitely many $A_{\chi}$ are non zero. Then by Dedikind's lemma we have $A_{\chi'}A_{\chi''}=\delta_{\chi',\chi''}A_{\chi'}$. We also have $\sum_{\chi}A_{\chi}=Id$. If we set $V_{\chi}=imA_{\chi}$, then $V=\oplus_{\chi}V_{\chi}$.
My question is why this $V_\chi:=imA_{\chi}=\{v\in V|\forall g\in G, g\cdot v=\chi(g)v\}$. In the book it is also given that $g\in G$ acts on $V_{\chi}$ as $\chi(g).Id$. I also don't understand this.
Any help is highly appreciated.
Thank you.