Consider finite-dimensional matrices $A$ and $B$, and some subspace $V$ of the embedding vector space.
We know that $[A,B]=0$ and $AV\subseteq V$ implies $BV\subseteq V$ in at least some notable cases: if $V=\ker(A)$ or $V=\operatorname{im}(A)$ and $[A,B]=0$ then $AV\subseteq V$ and $BV\subseteq V$.
However, in the more general case, from $AV\subseteq V$ I can only prove that $A(BV)\subseteq (BV)$.
As a counterexample, suppose $Ae_1=e_1$, $Ae_2=e_2$, $Be_1=e_2$, $Be_2=e_1$. Then, if $V=\mathbb Fe_1$, we have $AV\subseteq V$, $ABV=A\mathbb Fe_2=\mathbb Fe_2\subseteq BV$, but $BV=\mathbb F e_2\not\subseteq V$.
Are there conditions on the invariant subspace $V$ that ensure that we in particular also have $BV\subseteq V$, like is the case for $\ker(A)$ and $\operatorname{im}(A)$?