Let $e_k$ be an orthonormal basis for a separable hilbert space $H$. Suppose $T$ is such that $Te_j = e_{j-1}$, $Te_1 = 0$.
How do I show that if $B$ is bounded with $\|B\| < 1$, then $$T+B$$ has a one dimensional null space? I'm assuming this follows from some argument about the Neumann Series, since you can show that $$Id - B$$ is invertible given the same condition, and $T$ is essentially the identity with an extra 1d null space.
Consider the other shift operator $S:=e_j\mapsto e_{j+1}$.
Its image is of codimension $1$, its norm is $1$, so $\Vert BS\Vert<1$, and $TS=id$.
Then $(T+B)S=id+BS$ is invertible, so $T+B$ restricted to the image of $S$ is bijective, so it will have a $1$ dimensional kernel when seen on the whole space.
Specifically, let $v:=(T+B)e_1$ and take the unique $u\in \langle e_2,e_3,\dots\rangle=\mathrm{im}(S)$ with $(T+B)u=v$, then the kernel will be spanned by $u-e_1$.