Lomonosov's invariant subspace theorem

Remarks

Proof idea: let $\mathcal{A}$ be the algebra of operators that commute with $K$, and suppose that it has no common nontrivial closed invariant subspace. Then the orbit $\set{Ay : A \in \mathcal{A}}$ of every $y \ne 0$ is dense. A compactness argument with finitely many $A_i \in \mathcal{A}$ and a partition of unity builds a continuous map of a closed ball into itself. The Schauder fixed-point theorem gives a fixed point. The fixed point is an eigenvector, with eigenvalue $1$, of a compact operator $L$ that commutes with $K$. So $K$ has an eigenvalue $\lambda$ on the finite-dimensional space $\ker(L - I)$. Then $\ker(K - \lambda I)$ is a common invariant subspace, a contradiction.

The infinite dimension is needed in part 1. On $\C^n$ with $n \ge 2$, the identity $K = I$ is compact and commutes with every operator, and the operators on $\C^n$ have no common nontrivial invariant subspace.

Example: the Volterra operator $(Vf)(x) = \int_0^x f(t)\,dt$ on $L^2[0,1]$ is compact, so every operator that commutes with $V$ has a nontrivial closed invariant subspace.

Going deeper

For a student

Before this theorem, the invariant subspace problem was known for compact operators and for polynomially compact operators, with long proofs. Lomonosov's short fixed-point argument covers all of these at once, and in arbitrary complex Banach spaces. It does not settle the general problem: there are operators on Hilbert space that commute with no nonzero compact operator.