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