In page 13 of Lang's $SL_2$ there is a proof that for a $*$-closed algebra $\mathscr A$ of compact operators on a Hilbert space $H$, $H$ is completely reducible.
The proof follows by taking the orthogonal complement $F$ of a maximal orthogonal family of irreducible subspaces and finding a new irreducible subspace.
To find the new irreducible subspace, we cook up a hermitian operator $A$ from an arbitrary operator in the algebra.
$$A = \frac {\tilde{A}+\tilde{A}^*}{2} + i\frac{\tilde{A} - \tilde{A}^*}{2}$$
Then, as in the spectral theorem, we can find an eigenvector $v$ for $A$ in $F$ that generates an irreducible space for $A$.
My problem is seeing how this irreducible space, which Lang denotes by $\overline{Av}$, is irreducible for the entire algebra $\mathscr A$. By the way, why do we take the closure $\overline{Av}$?