Lomonosov’s Theorem is an amazing theorem which established that if an operator on a Banach space commutes with a nonzero compact operator, then admits a nontrivial hyper-invariant subspace.

As I eluded to in a previous post, Compact operators are as close to finite dimensional operators as we can get. Both precisely (every compact operator is a limit of finite-rank operators) and imprecisely the images of finite-rank operators and compact operators are both topologically small. Thus, if there’s any justice in the world, compact operators should admit a nontrivial invariant subspace. Lomonosov’s Theorem proves this and much more. Since a nonzero operator commutes with itself, this trivially proves that every compact operator admits a (hyper-) invariant subspace (a theorem that attracted the attention and talent of von Neumann).

**Corollary (Lomonosov’s Theorem 1973)**

If is an operator that is not a multiple of the identity, and is a nonzero compact operator that commutes with , then has a nontrivial hyper invariant subspace.

Suppose admits no nontrivial hyper-invariant subspace. This is equivalent to saying that the lattice of its commutant, denoted is equal to . Since is an algebra (do your homework from a previous post!), it follows from Lomonosov’s Lemma that there exists some that commutes with so that . Notice that is an invariant subspace for . Since if then . So . In fact, is the identity on . We showed in the previous proof that is a compact operator. It’s left to the reader to show that if the identity is a compact operator on some Banach space , then is finite dimensional. Once this exercise is completed, you’ll be convinced that since is compact and the identity on , that is finite dimensional. Moreover, since both and commute with , it follows that for each that . So , i.e. and so . Thus . Huzzah! We see that and so is invariant for . But wait, there’s more! Since is finite dimensional we can restrict and extract an eigenvalue for . Let . Not is certainly not since $\lambda$ is an eigenvalue. However, if , then which is a contradiction since is not a multiple of the identity by assumption. If commutes with , then for all we have that . Thus and so . So is hyper-invariant for .

**Corollary (Aronszajn-Smith 1954****)**

If is a compact operator on a Banach space , then admits an invariant subspace.

Proof: See above. It admits a hyper-invariant subspace.

**Corollary (Bernstein and Robinson 1966 [Non-Standard Analysis], Halmos 1966 [Standard Analysis])**

If is a linear operator on an infinite dimensional Banach space and is a single variable polynomial such that is compact, then admits a nontrivial invariant subspace

Proof: Suppose is nonzero. Since is compact, and no linear combination of the identity is compact, it follows that is not a multiple of the identity. Moreover, and so commutes with a nonzero compact operator. The result follows from Lomonosov’s Theorem. If is nonzero, then write . Then . Thus is invariant on for each . Then is nonzero and not all of since is finite dimensional.

Before I conclude, I recall that in my first post that two mathematicians constructed a Banach space such that every operator admits a nontrivial invariant subspace. The construction is much more interesting, actually. Argyros and Haydon (2009) constructed a Banach space for which *every operator* is a multiple of the identity plus a compact operator, i.e. for some and compact . But then must commute with , and so every operator admits a nontrivial hyper-invariant subspace. Cool stuff!

It’s been a pleasure to share a piece of math(s) with you. Now that graduate applications are in the past, I will have more time to write and share.

Cool!