# Existence of irreducible subspace

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 $\cal A$?

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What do you mean by irreducible subspace? Why can't I just take A to be K(H)? –  Yemon Choi Oct 23 '12 at 13:42
A subspace is irreducible, if it is closed, stable under $A$, and does not contain any closed, $A$-stable subspace other than $0$ or itself. If you take all compact operators, than the entire space is irreducible. –  doug Oct 23 '12 at 13:49
Right, for some reason I thought you wanted proper irreducible subspaces, but that was my mistaken reading. Thanks for clarification –  Yemon Choi Oct 24 '12 at 11:55

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.
No, that's exactly the point. I don't want the algebra $A$ to be self-adjoint. –  doug Oct 23 '12 at 20:15