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Let $H$ be an infinitedimensional Hilbert space and $T$ a compact selfadjoint operator in it. Consider the following


Let $$ H_{1}=H,\ T_{1}=T $$ and let $\lambda_{1}$ be that eigenvalue of $T_{1}$ whose absolute value equals $\left\Vert T_{1}\right\Vert $ (there is a theorem that tells me that compact selfadjoint operators $T$ always possess an eigenvalue $\lambda$, such that $\left|\lambda\right|=\left\Vert T\right\Vert $) and $f_{1}$ the associated normed eigenvector.

Now let $$ H_{2}=\left\{ f_{1}\right\} ^{\perp},\ T_{2}=T\Bigr|_{H_{2}} $$ (one can check that setting $T_{2}=T\Bigr|_{H_{2}}$ is welldefined) and $\lambda_{2}$ be again the eigenvalue of $T_{2}$ such that $\left\Vert T_{2}\right\Vert $ and $f_{2}$ be again its corresponding normed eigenvector.

Continuing let $$ H_{3}=\left\{ f_{1},f_{2}\right\} ^{\perp},\ T_{3}=T\Bigr|_{H_{3}} $$ and so on...

This algorithm shall terminate of $T_{n}$ is the zero operator for some $n\in\mathbb{N}$.

Now my question is: If $T$ isn't a finite rank operator, is it possible that this algorithm stops after a finite number of steps? If yes, can one please provide me with detailed example of such an operator $T$ (or otherwise a proof that this algorithm never terminates )?

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Right; I corrected my typo now. – pink_pyjamas Jul 30 '12 at 14:12
up vote 4 down vote accepted

If the algorithm stops after $n$ iterations, then you have $T_n$ is the zero operator on $H_n$.

Since $H = \mathbb{sp}\{f_k\}_{k=1}^{n-1} \bigoplus H_n$, if $x \in H$, then $x = \sum_{k=1}^{n-1} \alpha_k f_k + y$, where $y \in H_n$. Then you have $$Tx = \sum_{k=1}^{n-1} \alpha_k T f_k +T y = \sum_{k=1}^{n-1} \alpha_k T f_k +T_n y = \sum_{k=1}^{n-1} \alpha_k T f_k \in \mathbb{sp}\{f_k\}_{k=1}^{n-1}.$$ Hence $T$ is a finite rank operator.

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