Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
no one ? ;( $ $ $ $ – pink_pyjamas Jul 29 '12 at 16:59
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 $r$.

share|cite|improve this answer
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 $r+1=n$. – pink_pyjamas 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! – pink_pyjamas Jul 31 '12 at 8:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.