In the book "Invitation to $C^*$-algebras" by Arveson, I read that:
Let $\mathcal{A}$ be a $C^*$-subalgebra of $B_0(\mathcal{H})$ (= compact operators on the Hilbert space $\mathcal{A}$). The book then reads:
By cutting down to an $\mathcal{A}$-invariant subspace, if necessary, we may assume that $\mathcal{A}$ has trivial null-space.
What exactly does this mean and how can we achieve it?
I think it means that we can find a subspace $\mathcal{K} \leq \mathcal{H}$ such that $T\xi \in \mathcal K$ for all $\xi \in \mathcal{K}$ and for all $T \in \mathcal{A}$ and that the restricted operators $T\vert_\mathcal{K}: \mathcal{K} \to \mathcal{K}$ have trivial nullspace, i.e. the common intersection of all their kernels is $0$.
Is this interpretation correct? And how can we prove such subspace exists?
The space is $$\mathcal K=\overline{\{A\xi:\ A\in\mathcal A,\ \xi\in \mathcal H\}}.$$ It is obviously invariant for $\mathcal A$. And, for any $A\in\mathcal A$, if $A\ne0$ there exists $\xi$ such that $A\xi\ne0$. Then $A^*(A\xi)\ne0$ (since $\ker A^*A=\ker A$). So the intersection of the kernels of all elements in $\mathcal A$ is $\{0\}$.