Let $V$ be a finite dimensional vector space over $\mathbb{Z}\_2$ with a linear map $f\_i : V \to V$ for each $i$ in some finite index set $I$.


Then one can always find some subset $G \subseteq V$ of minimal cardinality such that the set of all elements:


$f\_{i\_1} \circ \dots \circ f\_{i\_n}(g)$ where $g \in G$, $i\_j \in I$ and $n \geq 0$


spans the vector space $V$. Likewise in the dual situation $(V^\*,(f\_i^\*)\_{i \in I})$ there is some respective minimal generating set $G' \subseteq V^\*$.


Here are my questions:


1. How does the cardinality of $G'$ compare to that of $G$?
2. What if each $f\_i$ is assumed to be idempotent?


Any help much appreciated.



Minimal generating sets of monoids acting on finite vector spaces.
Things can differ as much as you like even if all maps are idempotents. Let F be any field. Let M be the monoid of all constant maps on {1,...n} (acting on the left) together with the identity. In otherwords, M has n left zeroes and an identity. Let V be the regular left FM-module. It is a cyclic module generated by the identity of M.


Consider the dual module V\*. It can be identified with the mappings $M\to F$ with right FM-module structure given by (fm)(x)=f(mx). In particular, if m is not the identity map, then fm is a constant map. Thus V\*/constants is an n-dimensional module annihilated by all non-zero elements of M and hence cannot be generated by fewer than n elements (that is, a basis). Thus V\* cannot be generated by fewer than n elements. 


