## 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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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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### 1 Response to The Invariant Subspace Problem and Lomonosov’s Theorem (Part 3 of 3).

1. Z Norwood says:

Cool!