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.