Hi! This is my first question here, so please excuse me if it is too elementary.

I was wondering if the notion of a simultaneous decomposition into eigenspaces could be generalized in a special way I describe below. Let $V$ be a vector space over an algebraically closed field $k$ and let $T \subset \mbox{End}(V)$ be a finite dimensional subspace consisting of pairwise commuting and diagonizable endomorphisms. Than we have a decomposition

$\begin{align*} V = \bigoplus_{\lambda \in T^{\*}} V_{\lambda}, \end{align*}$

where $V_{\lambda} = \lbrace v\in V \hspace{0.3em}\lvert \hspace{0.3em} xv = \lambda v \mbox{ for all } x\in T \rbrace$ and $T^{\*}$ is the dual space of $T$.

I was wondering now if a very similar thing in another context might be possible as well. Some notations first. Let $V$ be as above and let $f \in \mbox{End}(V)$. Then set $\mbox{Hau}(f,\lambda) = \bigcup_{n\ge 0} \mbox{ker}(f-\lambda\cdot\mbox{id})^n$. It is known that $V = \bigoplus_{\lambda\in k} \mbox{Hau}(f,\lambda)$ if and only if $f$ is locally finite.

Now let $S\subset \mbox{End}(V)$ be an abelian, finitely generated subalgebra such that each $x\in S$ is locally finite. By $S^{\times}$ I denote the set of algebra homomorphisms $S\to k$ (that map $1$ to $1$) and for $\chi\in S^{\times}$ I denote

$\begin{align*} \mbox{Hau}_s(S,\chi) = \bigcap_s\mbox{Hau}(s,\chi(s)), \end{align*}$

where $s$ runs over all $s\in S$. My question now is the following: is it true that

$\begin{align*} V = \bigoplus_{\chi\in S^{\times}}\mbox{Hau}_s(S,\chi) ? \end{align*}$

I have serious difficulties proving it. My attempts so far have been that $S$ must be isomorphic to $k[x_1, \dots, x_l]/I$ for some $l$, and I tied induction over $l$. The above equality seemed basic enough for me to be found in any text on linear algebra - I thought. But I did not find it. I would be very very glad for any pointers to literature or anything else. Or is the statement false in this way?

Thank you very much.

EDIT:

To give you my motivation for such a question: In the representation theory in the context of category $\mathcal{O}$, $\mathcal{O}$ can be decomposed into blocks, parameterised by algebra homomorphisms $\chi: \mbox{Z}(\mathfrak{g})\to k$, where each $M\in \mathcal{O}_{\chi}$ satisfies

$\begin{align*} \forall z\in \mbox{Z}(\mathfrak{g}) \forall v\in M: (z-\chi(z))^n v = 0 \mbox{ for some } n>0 \mbox{ depending on } z. \end{align*}$

Since $M$ is an $\mbox{U}(\mathfrak{g})$-module, we get an algebra homomorphism $\mbox{Z}(\mathfrak{g})\to \mbox{End}(M)$. $\mbox{Z}(\mathfrak{g})$ is known to be isomorphic to a polynomial algebra in finitely many variables. The image of $\mbox{Z}(\mathfrak{g})$ under this morphism would play the role of $S$ in the above paragraph, and if I had the statement I want to prove, it would explain why each $\mbox{U}(\mathfrak{g})$-module decomposes into a direct sum, where each summand belongs to a block...