Let $K$ be a field and let $V$ be the set of sequences $\{v\_1,v\_2,\dots\}$ of elements of $K$. If $A=\{a\_1,a\_2,\dots\}$ is also a sequence of elements of $K$, then it defines an endomorphism of $V$ $$f\_A : \{v\_1,v\_2,\dots\} \mapsto \{a\_1 v\_1, a\_2 v\_2,\dots\}.$$


Let $(W,f)$ be a $K$-vector space with an endomorphism, and suppose that there exists an embedding of $(W,f)$ into $(V,f\_A)$ sending $f$ to $f\_A$ and an embedding of $(W,f)$ into $(V,f\_B)$ sending $f$ to $f\_B$, for some $A$ and $B$.


Question: if $A$ and $B$ are disjoint, then is $W$ necessarily $\{0\}$?


If $V$ was instead the set of sequences that are zero almost everywhere, then the answer would be an easy "yes".



Linear map with two "incompatible" representations
I don't think so ... let $K = \mathbb{R}$ and let $A$ and $B$ be any two disjoint sequences such that each one contains infinitely many distinct elements. Then let $\vec{v} = \{1,1,1, \ldots\} \in V$ and define $W = {\rm span}\{\vec{v}, f\_A\vec{v}, f\_A^2\vec{v}, \ldots\}$ and $f = f\_A$. The identity map is the first embedding and the map that takes $f\_A^n\vec{v}$ to $f\_B^n\vec{v}$ is the second embedding. We just have to check that this map extends linearly to all of $W$ and defines an embedding. Both statements follow from the fact that the sequences $\{\vec{v}, f\_A\vec{v}, f\_A^2\vec{v}, \ldots\}$ and $\{\vec{v}, f\_B\vec{v}, f\_B^2\vec{v}, \ldots\}$ are both linearly independent.


