# 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

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.
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