Decomposition of a vector space $V$ into a direct sum of eigenspaces of semisimple subgroup of $GL(V)$ Let $G$ be a subgroup of $GL(V)$, where $V$ is a finite dimensional vector space over $k$ (algebraically closed). Suppose $G$ is commutative and $G_s$ is the subgroup of semisimple elements of $G$. I want to prove that 
we can decompose $V$ as 
$$
V = \oplus_{\lambda: G_s \to k^{\times}} V_{\lambda}, 
$$
where
$$
V_{\lambda} = \{v \in V: \forall g \in G_s, g v = \lambda(g) v   \}.  
$$
Any explanation would be appreciated. Thank you very much. 
 A: We're being a little bit imprecise here, but I assume that $G$ means closed subgroup of $\mathrm{GL}(V)$ (or just algebraic, since the closedness assumption is implied by the closed orbit lemma). Let me prove something simple that maybe will clarify the situation for you.
Since $G$ is abelian it is, in particular, solvable and thus we know that $G^\circ$ (the connected component of $G$) is isomorphic to $U\rtimes T$ where $U$ is a unipotent group and $T$ is a torus. Then, $(G^\circ)_s=T$. We claim that $(G^\circ)_s=(G_s)^\circ$. Note that since $G_s\subseteq G$ that $(G_s)^\circ\subseteq G^\circ$ and so evidently $(G_s)^\circ\subseteq (G^\circ)_s$. But we have just seen that $(G^\circ)_s=T$ and thus $(G^\circ)_s$ is connected so that $(G^\circ)_s\subseteq (G_s)^\circ$ as desired.
Thus, we see that $(G_s)^\circ=T$ a torus. Note then that we have the usual short exact sequence
$$1\to (G_s)^\circ\to G_s\to \pi_0(G_s)\to 1$$
where the group $\pi_0(G_s)$ is some finite abelian group. It's easy to see that this sequence splits so that $G_s\cong (G_s)^\circ \times \pi_0(G_s)$. Thus, $G_s$ is the product of a torus and a finite abelian group. The fact that any representation of $G_s$ decomposes as you've described is classical.
