---
tag: 5RF055GK4N
kind: problem
name: Invariant subspace problem
msc: [47A15]
uses: [HM2HHE3J51, 0ACX5QT8TE]
about: [0ACX5QT8TE]
sources:
  - url: https://en.wikipedia.org/wiki/Invariant_subspace_problem
same_as:
  wikidata: Q6059532
  wikipedia: Invariant subspace problem
history:
  posed_by: "unknown; it grew out of work of Arne Beurling and John von Neumann"
  first_stated: "mid-20th century"
---

## Problem

Let $H$ be a complex, separable, infinite-dimensional [[Hilbert space|HM2HHE3J51]] and let $T \colon H \to H$ be a bounded linear operator.
Does $T$ always have a nontrivial closed [[invariant subspace|0ACX5QT8TE]], that is, a closed linear subspace $M$ with $M \ne \set{0}$, $M \ne H$ and $T(M) \subseteq M$?

## Remarks

The problem is open.
The hypotheses rule out easy cases.
In a finite-dimensional complex space of dimension at least $2$, an eigenvector spans an invariant line.
In a non-separable space, the closed span of $x, Tx, T^2 x, \ldots$ for any $x \ne 0$ is a nontrivial invariant subspace.

Positive answers are known for many classes of operators.
Compact operators have nontrivial invariant subspaces (Aronszajn and Smith, 1954), and so do operators that commute with a nonzero compact operator, by [[Lomonosov's theorem|JEBP2FYZFM]].
Normal operators have them by the spectral theorem, and subnormal operators have them by a theorem of Scott Brown.
Bernstein and Robinson settled the case of polynomially compact operators in 1966, a question that Halmos had posed for operators whose square is compact.

For general Banach spaces the answer is negative.
Per Enflo constructed a bounded operator on a separable Banach space with no nontrivial closed invariant subspace, and Charles Read gave such an operator on $\ell^1$.

## Explanations

### L1 human

An invariant subspace of an operator is a smaller space that the operator never leaves.
In finite dimensions over the complex numbers, the line through an eigenvector is one.
The problem asks whether every bounded operator on an infinite-dimensional separable Hilbert space has such a proper closed piece.
Despite many partial results, nobody knows the answer.