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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.

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no one ? ;( $ $ $ $ –  user36675 Jul 29 '12 at 16:59
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up vote 2 down vote accepted

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 $r$.

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Excellent. But that only shows that $r+1\leq n$, I think ? I'd still need to somehow show that we also have $r+1\geq n$, to get $r+1=n$. –  user36675 Jul 30 '12 at 13:59
    
I think the argument I gave before the last sentence works. But if you are not convinced, please say what seems wrong. –  Davide Giraudo Jul 30 '12 at 14:49
    
Ah, your right! –  user36675 Jul 31 '12 at 8:23
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