Is there a non-trivial subspace that fixes all compact operators?

Question

Let $B_0(\mathcal{H})$ be the set of all compact operators on the Hilbert space. Does there exist a non-trivial subspace $\mathcal{H'} \leq \mathcal{H}$ such that

$$T\mathcal{H}' \subseteq \mathcal{H'}, \forall T \in B_0(\mathcal{H})$$

I guess not.

Suppose such a subspace $\mathcal{H}'$ exists. Then $T\mathcal{H'}\subseteq \mathcal{H'}$ for all finite-rank operators as well. Choose $\xi \in \mathcal{H'}\setminus \{0\}$. Let $\eta \in \mathcal{H}\setminus \mathcal{H}'$. Then all we must do is construct an operator $T \in B_0(\mathcal{H})$ with $T \xi = \eta$.

I now we can use Hahn-Banach to do this when $T$ must just be bounded. Can we make sure $T$ is also compact? Maybe by constructing $T$ such that it has finite rank?

Answer 1

Fix $\xi_0\in H'$, $\xi_1\in H\setminus H'$ with $\|\xi_0\|=\|\xi_1\|=1$, and define $T:H\to H$ by $$T\eta=\langle \eta,\xi_0\rangle\xi_1.$$ This is finite-rank, hence compact, and $T\xi_0=\xi_1\notin H'$.