$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\Sub}{Sub} \newcommand{\I}{\mathcal{I}}$ Fix $t\colon V\to V$ over a $k$-vector space, and consider the corresponding module $$ k[x]/\mu_t \stackrel{\mathrm{ev}_t}{\to} \End_{\underline{\mathrm{AbGrp}}} (V) =: M, $$ where $\mu_t$ denotes the minimal polynomial associated to $t$. We divide that one out so we get a faithful module, i.e. we don't have ring elements that act trivially on $V$.
Def Let $M = (R\to \End_V)$ be a module, $r\in R$. Denote by $$ \I_r:=\{ W\leq V \mid W \ \mathrm{is}\ r\mathrm{-Invariant} \}. $$ the subset of $r$-invaraint subspaces. Let further $S\leq R$ and define $$ \I_S := \bigcap_{s\in S} \I_s. $$
LEM $\I_r$ and $\I_S$ is a sublattice of $\Sub(V)$
LEM Let $S\leq S^\prime \leq R$. Then $\I_S\geq \I_{S^\prime}$.
Proof. Let $W\leq V$ invariant under all $s^\prime\in S^\prime$. Then it is most certainly invariant under all $s\in S$.
Q: Fix an operator $t$. What other elements leave the spaces in $\I_t$ invariant? Well, we might want to pick a basis that plays nice with our invariant spaces: Perhaps it should lie in some “designated set” of minimal invariant spaces – where minimality implies that they have pairwise trivial intersection. However, if you look at e.g. the nilpotent operator $e_1 \mapsto e_2 \mapsto 0\in \End(\mathbb R^2)$, you will see that the lattice of invariant spaces is of the form $0 \leq \langle e_1 \rangle \leq \langle e_1, e_2 \rangle = \mathbb R^2 $: Our minimal invariant space does not have a complement, so just choosing elements from minimal invariant spaces need not be enough to form a basis.
Observation Take the finite lattice consitsting of $0\leq 1$ and inbetween them three incomparable elements $x,y,z$. Each of those three elements complement each other, so the notion of a “complement” to an element is not unique. Also note that this may arise as a sublattice of the lattice of e.g. the identity.
NOTE If we have an operator $t$, we know at least that $0, V, \operatorname{Ker} t, \operatorname{Im} t$ are invariant spaces.
Q: When can a lattice be represented as a lattice of invariant subspaces of an operator? For instance, is there a lattice-theoretic notion of “linear independence” of subspace?
NOTE In the finite case, we have the decomposition into generalized eigenspaces, which are pairwise complements.
Q: What properties does an operator have in which no invariant space has a (nontrivial) complement? Does it have to be one-generated in the finite dimensional case?
Q: Can we always make an invariant subspace one dimension larger? How do one-generated invariant spaces look like?
Idea: We don't only have a lattice structure, we have a lattice structure equipped with an endomorphism: $t$ maps invariant spaces to invariant spaces. Is it a lattice homomorphism?
Note A module $R\to \End(V)$ may map non-units to units: Consider $$ \mathbb R [x] \stackrel{\mathrm{ev}_(\rho)}{\to} \mathbb R^{\mathbb Z} $$ where $\rho$ is the right-shift: It surely has the left shift as an inverse, but this is not expressible as a polynomial in the right shift.