# Reducing subspaces for compact operators

It is well known that any compact operator in $\mathcal{B}(l_2)$ has an invariant subspace. What about reducing subspaces (subspaces that are invariant for both the operator and its adjoint). Does any compact operator have a reducing subspace?

Thank you.

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Could you give a reference for the well known result? – Norbert Oct 28 '12 at 7:03
A much stronger result of Lomonosov: Every operator commuting with a non-zero compact operator admits an invariant subspace. This holds in general Banach spaces. – Theo Oct 28 '12 at 7:35
Let $K\in\mathcal{K}(\mathcal{B}(\ell_2))$, then $K^*\in\mathcal{K}(\mathcal{B}(\ell_2))$. Let $X$ and $Y$ be their invaiant subspaces, then $Z=X\cap Y$ is what we needed. – Norbert Oct 28 '12 at 10:26
@Norbert We have to check that the intersection is not trivial. – Davide Giraudo Oct 28 '12 at 10:26
@DavideGiraudo Yes you are right, but one can try this way – Norbert Oct 28 '12 at 10:28

(I assume that the intention of the question was to ask whether every compact operator has reducing subspaces, as it is trivial to construct some compact operators with reducing subspaces)

There exist compact operators with no nontrivial reducing subspaces.

Let $\{\xi_j\}$ be an orthonormal basis and $\{e_{kj}\}\subset \mathcal B(\ell^2)$ the corresponding matrix units. Define an operator $x$ by $$x=\sum_{k=1}^\infty\frac1k\,e_{k+1,k}$$ ("weighted shift"). Then $x$ is compact. Suppose that $V$ is a non-zero reducing subspace of $x$. Then $V$ is invariant under the selfadjoint and positive operator $x^*x=\sum_k\frac1k\,e_{kk}$, and it will also be invariant under $f(x^*x)$ for any continuous function $f$. In particular, $e_{kk}V\subset V$ for all $k$. As $V$ is nonzero, there exists $k$ such that $e_{kk}V$ is nonzero. Note that $e_{kk}V$ is either $0$ or $\mathbb C\xi_k$. So there exists some $k$ such that $\xi_k\in V$. But, as $V$ is invariant for both $x$ and $x^*$, we get $$\xi_{k+n}=\frac{(k+n-1)!}{k!}x^{n-1}\xi_k\in V;$$ similarly, $$\xi_{k-n}=\frac{(k-1)!}{(k-n-1)!}\,(x^*)^{n}\xi_k\in V.$$ This shows that $V$ contains every element in the basis, so $V=\ell^2$. So, $x$ admits no non-trivial reducing subspace.

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Thank you. I also found a more general example. Donoghue operators are weighted shifts with weights in $l_2$ and monotone decreasing to $0$. It is known that they only have finite dimensional non-trivial invariant subspaces, while its adjoint only have finite codimensional non-trivial invariant subspaces. Thus a Donoghue operator cannot have a non-trivial reducing subspace. – Theo Oct 28 '12 at 23:36