Let $H$ denote a Hilbert space and let $\cal A$ be a subalgebra of the algebra ${\cal B}(H)$ of all bounded operators on $H$ such that $\cal A$ consists of compact operators only and such that each vector $v\in H$ lies in the closure of ${\cal A}v$. 


Is it true that there must exist an irreducible subspace for $\cal A$? 



Existence of irreducible subspace
Well, if ${\cal A}$ can be non self-adjoint then I think there are easy counterexamples. Let $(e\_n)$ be the standard basis of $l^2({\bf N})$ and let ${\cal A}$ be the closed span of the operators $e\_n \otimes e\_m$ for $n \leq m$ (where $(e\_n \otimes e\_m) v = \langle v, e\_n\rangle e\_m$). The condition that any $v$ lie in the closure of ${\cal A}v$ holds because $\sum\_1^N e\_i \otimes e\_i \to I$ strongly.


It's easy to see that the closed invariant subspaces for ${\cal A}$ are precisely the subspaces ${\rm span}\{e\_n: n\geq N\}$ for $N \in {\bf N}$, of which there is no nonzero minimum.


