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  "body": "\n## Problem\n\nLet $H$ be a complex, separable, infinite-dimensional [[Hilbert space|HM2HHE3J51]] and let $T \\colon H \\to H$ be a bounded linear operator.\nDoes $T$ always have a nontrivial closed [[invariant subspace|0ACX5QT8TE]], that is, a closed linear subspace $M$ with $M \\ne \\set{0}$, $M \\ne H$ and $T(M) \\subseteq M$?\n\n## Remarks\n\nThe problem is open.\nThe hypotheses rule out easy cases.\nIn a finite-dimensional complex space of dimension at least $2$, an eigenvector spans an invariant line.\nIn a non-separable space, the closed span of $x, Tx, T^2 x, \\ldots$ for any $x \\ne 0$ is a nontrivial invariant subspace.\n\nPositive answers are known for many classes of operators.\nCompact 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]].\nNormal operators have them by the spectral theorem, and subnormal operators have them by a theorem of Scott Brown.\nBernstein and Robinson settled the case of polynomially compact operators in 1966, a question that Halmos had posed for operators whose square is compact.\n\nFor general Banach spaces the answer is negative.\nPer Enflo constructed a bounded operator on a separable Banach space with no nontrivial closed invariant subspace, and Charles Read gave such an operator on $\\ell^1$.\n\n## Explanations\n\n### L1 human\n\nAn invariant subspace of an operator is a smaller space that the operator never leaves.\nIn finite dimensions over the complex numbers, the line through an eigenvector is one.\nThe problem asks whether every bounded operator on an infinite-dimensional separable Hilbert space has such a proper closed piece.\nDespite many partial results, nobody knows the answer.\n",
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