A question on Hilbert spaces concerning kernel and partial isometry Let $\mathcal{H}$ be an infinite-dimensional Hilbert space, and let $T:\mathcal H\to \mathcal H$ be a continuous linear operator with $\operatorname{Ker} T\neq\{0\}$ and closed range $\mathcal{R}(T)$. Suppose that $T$ is a partial isometry, by which we mean that $T$ is an isometry on the orthogonal complement of its kernel, namely on $(\operatorname{Ker} T)^\perp$. 
Further, assume that $\operatorname{Ker}T\subset \mathcal{R}(T)\neq \mathcal{H}$. Let $I$ denote the identity operator on $\mathcal H$. 
Does there exist $x\in(\operatorname{Ker}(T-I))^\perp$ such that $((T-I)y, x)=0$ for all $y\in(\operatorname{Ker}T)^\perp$, with $(\cdot,\cdot)$ denoting the inner product on $\mathcal H\,$?
Any answer or hint will be appreciated.
 A: Let's focus on the required orthogonality condition
$$0\:=\:((T-I)y,x)\:=\:(y,(T^*-I)x)\quad\forall\,y\in(\operatorname{Ker}T)^\perp,$$
it says that $\,(T^*-I)x\in\operatorname{Ker}T,$ hence the elements $x$ we are seeking have to satisfy
$$TT^*x\:=\:Tx\tag{1}\,.$$
Remark:
This equation is equivalent to the symmetric version $\,T^*Tx=T^*x\,$ since partial isometries are also characterised by $\,T^*TT^*=T^*.$
$(1)$ is satisfied by eigenvectors to the eigenvalue $1$, noting that 
$\,\operatorname{Ker}(T-I)=\operatorname{Ker}(T^*-I)\,$ generally holds for partial isometries, but they do not have to exist.
There may exist non-zero solutions within the orthogonal complement $(\operatorname{Ker}(T-I))^\perp\,$: Looking at the assumptions
$\,\{0\}\neq\operatorname{Ker}T\subset\mathcal{R}(T)\neq\mathcal{H}\,$
and using $\,\mathcal R(T)=(\operatorname{Ker}T^*)^\perp$
one has 
$$\,\{0\}\neq\operatorname{Ker}T\ \perp\ \tag{2}
\operatorname{Ker}T^*\neq\{0\}\,.$$
This yields the orthogonal decomposition 
$$\mathcal{H}\:=\:\operatorname{Ker}T\oplus\operatorname{Ker}T^*\oplus\mathcal S\quad\text{with }\ \mathcal S=\mathcal R(T)\cap\mathcal R(T^*)\,,$$
and $\,\mathcal S=\{0\}\,$ may occur (as in the example below).
Insert $\,x=k+k_*+s\,$ as ansatz into $(1)$, with
$\,k\in\operatorname{Ker}T,\ k_*\in\operatorname{Ker}T^*,
\ s\in\mathcal S$, to obtain 
$$k+s = TT^*(k+k_*+s) = T(k+k_*+s) = Tk_*+Ts\tag{3}\,.$$
Assuming $\,Ts=s\,$ (thus leaving the most general setting by now) gives the condition $\,Tk_*=k$, and $x=k+ k_*$ will do the job.
The set of all these $x$ forms a closed subspace of $\operatorname{Ker}T\oplus\operatorname{Ker}T^*$ (which may be trivial ... )
An example of a partial isometry admitting such elements:
Let $e_j$ denote the elements of an ONB in a countably-infinite $\mathcal H$ and define
$$Te_j = \begin{cases}
e_{j+1} & \text{$j\,$ is odd} \\
0 & \text{$j\,$  is even}\end{cases}\quad\implies\;
T^*e_j = \begin{cases}
e_{j-1} & \text{$j\,$ is even} \\
0 & \text{$j\,$  is odd}\end{cases}$$
Then


*

*$\operatorname{Ker}T\:=\:\operatorname{span}\{e_2,e_4,\ldots\}
\:=\:\mathcal{R}(T)$

*$\operatorname{Ker}T^*\,=\:\operatorname{span}\{e_1,e_3, e_5,\ldots\}
\:=\:\mathcal{R}(T^*)$

*$\mathcal S=\{0\}\,$ and $\,\operatorname{Ker}(T-I) = \operatorname{Ker}(T^*-I) = \{0\}$
and all $\,x=(T^*+I)k\,$ do the job with $k$ running through $\operatorname{Ker}T$.
Addendum:
Due to $(2)$ the corresponding orthogonal projectors satisfy
$$(I-TT^*)(I-T^*T) = 0 = (I-T^*T)(I-TT^*)$$
whence $T(T^*)^2T = T^*T^2T^*$ which is the orthogonal projector onto 
$\mathcal S$. This might prove helpful in the search for further solutions of $(3)$ with $\,Ts\neq s$.
