Does this sequence of operators in Hilbert space stop at $\text{rank}T+1$ steps? Let $\left(T_{n}\right)_{n}$ be a sequence of operator in a infinitdimensional
Hilberspace $H$, defined as restrictions of an operator $T:H\rightarrow H$,
on smaller and smaller subsets, by the algorithm in this question (were also additional information about $T$ is provided).
There it was shown, that if this sequence is finite, $T$ must have
finite rank. 
My question is: Is the number $n$, for which the algorithm described
here stops, always $\text{rank}T+1$ ? How can we prove that ?
This guess came from the fact, that if $H$ were finite dimensional,
it is not too hard to show, that this sequence stops after exactly
$\text{rank}T+1$ steps. 

As the answer you link shows, if the algorithm stop after $n$ iterations, then the rank of $T$ is $\leq n-1$. Let $r$ the rank of $T$, and $n$ the number of iterations until the algorithm stops. We have $r\leq n-1$ hence $r+1\leq n$. If we have more than $r+2$ steps, we have extracted $r+1$ orthogonal non-zero vectors, associated with non-zero eigenvalues. This contradicts the fact that the rank of $T$ is $r$.
