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An interesting fact popped out of a paper I'm writing: if the invariant subspace problem for Hilbert space operators has a positive solution, then every $A \in B(\mathcal{H})$ can be made "upper triangular" in the sense that there is a maximal chain of closed subspaces of $\mathcal{H}$, each of which is invariant for $A$.

The proof is quite easy, almost trivial, yet I had never heard of it before. (It isn't mentioned in the answers to this question, for example.) It was a surprise to notice that the ISP has such a strong consequence for the structure of arbitrary operators because one thinks of is as merely the first, most basic question on that topic.

Surely this is known?

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    $\begingroup$ If $A$ does not have invariant subspace, the maximal chain also exists and contains two subspaces, right? $\endgroup$
    – Fedor Petrov
    Jan 30, 2019 at 20:03
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    $\begingroup$ Not a maximal chain of invariant subspaces: a maximal chain of subspaces, and each one of them is invariant. $\{0\}, \mathcal{H}$ is not a maximal chain because there are lots of intermediate subspaces which could be added. $\endgroup$
    – Nik Weaver
    Jan 30, 2019 at 20:05
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    $\begingroup$ Ah, I see. It could be read both ways and I have of course chosen the wrong one. $\endgroup$
    – Fedor Petrov
    Jan 30, 2019 at 20:17
  • $\begingroup$ Well, I puzzled a little over how to say it clearly and that's the best I could come up with ... $\endgroup$
    – Nik Weaver
    Jan 30, 2019 at 20:17
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    $\begingroup$ Nik, the observation you mention is referred to implicitly in both Don Hadwin's and my answers in the link you give. Ringrose used the observation to great advantage when the operator is compact. The structural theorem he proved for compact operators as a consequence of the invariant subspace theorem is not a conditional result because the invariant subspace theorem is known for compact operators. $\endgroup$
    – Bill Johnson
    Jan 30, 2019 at 23:22

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Under the assumption that the invariant subspace problem has a positive solution, this is an immediate consequence of Lemma 7.1.11 (The Triangularization Lemma) in Radjavi and Rosenthal's book Simultaneous Triangularization:

Say that a property $P$ of families of operators is inherited by quotients if whenever $N \subseteq M$ are invariant subspaces for a family of operators with $P$, then the family of induced operators on the quotient $M/N$ also has $P$.

Then if $P$ is a property of families of operators that is inherited by quotients, and if every family of operators satisfying $P$ (and acting on a space of dimension greater than 1) has a non-trivial closed invariant subspace, then every family satisfying $P$ is triangularizable in the sense that there is a maximal chain of closed subspaces, each of which is invariant for the family.

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  • $\begingroup$ Welcome to the time-sink, I mean hellmouth, I mean, erm, best use of one's time when not doing admin $\endgroup$
    – Yemon Choi
    Feb 8, 2019 at 3:10
  • $\begingroup$ What is meant by a "quotient" in this context? $\endgroup$
    – Nik Weaver
    Feb 8, 2019 at 3:26
  • $\begingroup$ Does it mean "quotient by an invariant subspace"? Then the result seems false. For instance, say P is the property "is isomorphic to $M_1$ or to the subalgebra of $M_3$ consisting of all matrices of the form $\left[\begin{matrix}*&*&*\cr *&*&*\cr 0&0&*\end{matrix}\right]$". The only invariant subspace of the latter algebra is $\mathbb{C}^2 \oplus 0$, the quotient by which leaves $M_1$. So this property is inherited by quotients (in the only sense I can think of), but the algebra isn't triangularizable. Or does "property" mean something different? $\endgroup$
    – Nik Weaver
    Feb 8, 2019 at 3:52
  • $\begingroup$ @NikWeaver I'm far from fluent in the language here, but I think "quotient" refers to a single operator and not to an algebra of operators, and it means something like: if $T$ is your operator and it has invariant subspaces $V\subset W$ then the "quotient" operator is the one induced on $W/V$. (But Matt should correct me if I have misremembered or misunderstood) $\endgroup$
    – Yemon Choi
    Feb 8, 2019 at 4:27
  • $\begingroup$ Thanks Yemon! You are correct, and I have clarified my answer. $\endgroup$
    – Matt Kennedy
    Feb 8, 2019 at 4:43

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