The Invariant Subspace Problem and Lomonosov’s Theorem (Part 3 of 3).

Lomonosov’s Theorem is an amazing theorem which established that if an operator T on a Banach space commutes with a nonzero compact operator, then T 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 K 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 T is an operator that is not a multiple of the identity, and K is a nonzero compact operator that commutes with T, then T has a nontrivial hyper invariant subspace.

Suppose T admits no nontrivial hyper-invariant subspace. This is equivalent to saying that the lattice of its commutant, denoted Lat(\{T\}') is equal to \{(0), X\}. Since \{T\}' is an algebra (do your homework from a previous post!), it follows from Lomonosov’s Lemma that there exists some A that commutes with T so that N=\ker(AK-I)\supsetneq (0). Notice that N is an invariant subspace for AK. Since if x\in N then AK(x)=AK(x)-x+x=(AK-I)x+x=0+x=x\in N. So AK(N)\subseteq N. In fact, AK is the identity on N. We showed in the previous proof that AK is a compact operator. It’s left to the reader to show that if the identity is a compact operator on some Banach space X, then X is finite dimensional. Once this exercise is completed, you’ll be convinced that since AK is compact and the identity on N, that N is finite dimensional. Moreover, since both A and K commute with T, it follows that for each x\in N that Tx=T(AK(x))=AK(Tx). So AK(Tx)-Tx=0, i.e. (AK-I)(Tx)=0 and so Tx\in \ker(AK-I). Thus Tx\in N. Huzzah! We see that T(N)\subseteq N and so N is invariant for T. But wait, there’s more! Since N is finite dimensional we can restrict T|_N and extract an eigenvalue \lambda for T\mid_N . Let M=\ker(T-\lambda). Not M is certainly not (0) since $\lambda$ is an eigenvalue. However, if M=X, then T=\lambda I which is a contradiction since T is not a multiple of the identity by assumption. If T'\in \{T\}' commutes with T, then for all x\in M we have that (T-\lambda)(T'x)=T'(T(x))-\lambda T'(x)=T'(T-\lambda(x))=T'(0)=0. Thus T'(x)\in M and so T(M)\subseteq M. So M is hyper-invariant for T.

Corollary (Aronszajn-Smith 1954)

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

Proof: Suppose p(T) is nonzero. Since p(T) is compact, and no linear combination of the identity is compact, it follows that T is not a multiple of the identity. Moreover, Tp(T)=p(T)T and so T commutes with a nonzero compact operator. The result follows from Lomonosov’s Theorem. If p(T) is nonzero, then write p(T)=\sum_{i=1}^{n}a_iT^i. Then T^n=a_n^{-1}(\sum_{i=1}{n-1}a_iT^i). Thus T is invariant on M=span(x,Tx,\dots,T^{n-1}x) for each x\ne 0. Then M is nonzero and not all of X since M 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. T=\lambda I+K for some \lambda\in \mathbb{C} and compact K. But then K must commute with T, 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.

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About Adam Azzam

This Fall I will be a first year mathematics PhD student at UCLA. I enjoy doing analysis - particularly of the functional variety.
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One Response to The Invariant Subspace Problem and Lomonosov’s Theorem (Part 3 of 3).

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