Is there an operator whose non-zero commutants are always injective? Let $H$ be an infinite dimensional separable Hilbert space.    

Is there an operator $T \in B(H)$ such that, if
  $TA=AT$ with $0 \ne A \in B(H)$, then $A$ injective ?  

Bonus question : what is the set of all such operators ?

In finite dimension, the invariant subspaces of $T$ are exactly the nullspaces of the operators that commute with $T$. So there is no such example in finite dimension $\geq 2$.
If $T$ is normal, it admits a handful of non-injective commuting projections (=reducing subspaces). 
So this is a question about non-normal operators in infinite dimension. 
As often when looking for a counterexample to a finite dimension property, the answer is the shift. Not the bilateral one, which is unitary whence normal. But the unilateral shift
$$
S:(x_1,x_2,x_3,\ldots)\longmapsto (0,x_1,x_2,x_3,\ldots)\qquad x\in \ell^2.
$$
For every nonzero $A\in B(\ell^2)$ commuting with $S$, $A$ is injective.
Proof: the unilateral shift is conveniently realized as (unitarily equivalent to) the multiplication operator by $z$ on the Hardy space $H^2(\mathbb{D})$. In this context, it is well-known that the commutant of $S$ is exactly the algebra of analytic Toeplitz operators on $H^2$, i.e. multiplication operators by a bounded analytic function $\phi\in H^\infty$. Such nonzero multiplication operators are injective, since a nonzero analytic function on a domain has isolated zeros. QED.
Reference: "The commutant of analytic Toeplitz operators", Trans. AMS 1973, by Deddens and Wong.
