Linear map with two "incompatible" representations

Question

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".

Answer 1

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.