Operator theory Invariant subspaces and dilations Problem
Invariant subspace problem
Builds on Hilbert spaceInvariant subspace
Problem
Let be a complex, separable, infinite-dimensional Hilbert space and let be a bounded linear operator. Does always have a nontrivial closed invariant subspace, that is, a closed linear subspace with , and ?
Remarks
The problem is open. The hypotheses rule out easy cases. In a finite-dimensional complex space of dimension at least , an eigenvector spans an invariant line. In a non-separable space, the closed span of for any 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. 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 .
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About this page
| Tag | 5RF055GK4N |
|---|---|
| URI | m2://5RF055GK4N |
| Kind | problem |
| Writing | W0 stub |
| MSC | 47A15 |
| Status | This copy of the site has no build data. Run m2 status 5RF055GK4N. |
| About | Invariant subspace |
| Uses | Hilbert space, Invariant subspace |
| Same as | |
| Changed | 2 November 2026 |
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