An operator $T$ on a separable Hilbert space $H$ is called *unicellular* if any two closed invariant subspaces $M$ and $N$ are comparable; that is either $M\subseteq N$ or $N\subseteq M$. There are many examples of (compact) unicellular operators, for example, the Volterra operator.


I am looking at a weaker property. Namely, the existence of a closed nontrivial invariant subspace $M$ that is *comparable* to all the other closed, invariant subspaces. Obviously, any closed invariant subspace of a unicellular operator satisfies this property. On the other hand, any diagonal operator will not be unicellular and won't satisfy this weaker property either.


In particular, I am interested in compact operators that do not have eigenvalues. My questions are:


1. Are there examples of compact operators without eigenvalues having this weaker property, but are not unicellular? (I suspect the answer is *Yes*)
2. Does any compact operator without eigenvalues have this weaker property? (I suspect the answer is *No*)


Edit: The question is edited for clarity.



Unicellular compact operators
$V\oplus V\in B(L^2(0,1)\oplus L^2(0,1))$, with $V$ being the Volterra operator, is an operator without eigenvalues that does not have your property. The invariant subspaces are $L^2(0,a)\oplus L^2(0,b)$, and for any such space $M=A\oplus B$, we can find another one $N=C\oplus D$, with $C\subsetneq A$, $D\supsetneq B$, unless $M=0, H$.


