The set of irreducible operators is dense
Heydar Radjavi, Peter Rosenthal · Proceedings of the American Mathematical Society · 1969
Let 3C be a separable (finite or infinite-dimensional) complex Hubert space.P. R. Halmos [l] has shown that the set of irreducible operators on JC, (i.e., operators with no nontrivial reducing subspaces), is uniformly dense in the space of bounded operators on 3C.In this note we give a very simple proof of Halmos's theorem.Let A be any bounded operator on 3C and let «>0.By the spectral theorem there exists a Hermitian operator D whose matrix is diagonal with respect to an o.n.basis }e"} such that