Math 2.0

Operator theory Invariant subspaces and dilations Problem

Invariant subspace problem

Builds on Hilbert spaceInvariant subspace

Problem

Let HH be a complex, separable, infinite-dimensional Hilbert space and let T ⁣:H→HT \colon H \to H be a bounded linear operator. Does TT always have a nontrivial closed invariant subspace, that is, a closed linear subspace MM with M≠{0}M \ne \set{0}, M≠HM \ne H and T(M)⊆MT(M) \subseteq M?

Remarks

The problem is open. The hypotheses rule out easy cases. In a finite-dimensional complex space of dimension at least 22, an eigenvector spans an invariant line. In a non-separable space, the closed span of x,Tx,T2x,…x, Tx, T^2 x, \ldots for any x≠0x \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. 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 ℓ1\ell^1.

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  1. Hilbert space concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Invariant_subspace_problem
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AboutInvariant subspace
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Changed2 November 2026

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