I have to do this exercise:
Given an abelian Lie algebra $\mathfrak{h}$ on an algebraically closed field $\mathbb{F}$. Let $ \varphi : \mathfrak{h} \to \text{End}_{\mathbb{F}}(V) $ be a representation over the finite dimensional vector space $V$ over $\mathbb{F}$. For every $\lambda \in \mathfrak{h}^*= \text{Hom}_{\mathbb{F}}(\mathfrak{h}, \mathbb{F})$ define \begin{gather} V_\lambda =\{ v \in V | \varphi(h)v=\lambda(h)v, \forall h\in \mathfrak{h} \} \end{gather} Prove that $V= \bigoplus_{\lambda \in \mathfrak{h}^*}V_\lambda$.
I think that this statement isn't true, indeed pick \begin{gather} A=\begin{pmatrix} 1&1\\0&1 \end{pmatrix} \end{gather} and let $\mathfrak{h}=\langle A \rangle$ be the set of all $cA$ with $c \in \mathbb{F}$, then $\mathfrak{h}$ acts naturally on $\mathbb{F}$ and the existence of such a decomposition would imply that $A$ is diagonalizable, which isn't. Am I right?
Such decomposition of $V$ into weight spaces $V_\lambda$ is a possible under some additional assumption. For example, it holds if $\mathfrak{h}$ is a Cartan subalgebra of a semisimple Lie algebra $\mathfrak{g}$, but in this case, there is no such semisimple $\mathfrak{g}$. Indeed, the canonic representation $F^2$ (if this is what you mean) of your $\mathfrak{h}$ is not completely reducible.