$C^*$-subalgebra of $B_0(\mathcal{H})$ containing all compact adjoint operators is equal to $B_0(\mathcal{H})$.

Question

Let $A$ be a $C^*$-subalgebra of $B_0(\mathcal{H})$, the compact operators on the Hilbert space $\mathcal{H}$. Suppose $A$ contains all compact self-adjoint operators. Can we conclude that $A= B_0(\mathcal{H})?$

I think this sounds reasonable. Maybe some decomposition theorem can help here?

Answer 1

Of course. Every $T\in B_0(\mathcal H)$ can be written as $T = (T+T^*)/2 + (T - T^*)/2$ where $(T+T^*)/2$ and $i (T - T^*)/2$ are compact and self-adjoint.