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.
 A: (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. 
