# Has a *subalgebra of $L(H)$ with $1$ a trivial null space?

Like the title: is it true that a self-adjoint unital subalgebra of $L(H)$ closed in the weak operator topology (a Von Neumann algebra) has a trivial null space? Why? Thank you.

• What would the null space be? Mar 10 '14 at 1:11
• A null space of a set $S$ (in my case the set is the self-adjoint unital subalgebra of $L(H)$) is the set of the elements $\xi \in H$ such that $A\xi=0$ for all $A \in S$. Mar 10 '14 at 10:45

If the unit of the subalgebra is the identity operator (that is, the algebra is non-degenerate), then the answer is yes: if $A\xi=0$ for all $A$, then in particular $\xi=I\,\xi=0$.
If the unit of the subalgebra is a projection $P\ne I$, then the null space is $(I-P)L(H)$.