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Let $X$ be a Banach or Hilbert space and $A$ be a bounded linear operator on $X$, and fix an element $x \in X$. Then I want to know that are there any good ways or theories to deal with the convergence of the following sequence $\{ y_n\}$: \begin{align} y_n = \frac{A^nx}{||A^nx||}. \end{align} I'm sorry for an ambiguous question. I welcome any kind of comments.

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  • $\begingroup$ If your space $X$ is reflexive (satisfied by Hilbert spaces for example), then the unit ball is compact in the weak topology. In this topology you will thus always find possible limits. $\endgroup$ – s.harp Dec 5 '18 at 9:46
  • $\begingroup$ Such a weak limit might be zero, which might come unexpected. $\endgroup$ – daw Dec 5 '18 at 11:07
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It does not always converge. As a counter example suppose a finite dimensional space $X = \mathbb{R}^2$, and \begin{equation} A = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix},\quad x = \begin{bmatrix}1\\1\end{bmatrix} \end{equation} Of course $A$ is bounded. Then

\begin{equation} y_n = \begin{cases} \displaystyle{\frac{1}{\sqrt{5}}}\begin{bmatrix}2\\-1 \end{bmatrix}, & n=2m+1, \\ \displaystyle{\frac{1}{\sqrt{2}}}\begin{bmatrix}1\\1 \end{bmatrix}, & n=2m. \end{cases} \end{equation}

In finite dimensional spaces, this is indeed the power iteration that converges to the eigenvector corresponding to the largest eigenvalue (in magnitude) of matrix $A$, and the convergence rate is determined by the ratio of first two largest eigenvalues in magnitude, $|\lambda_1 / \lambda_2|$.

I assume more than boundedness is needed for convergence of $y_n$. Perhaps if $A$ is compact, you may show similar argument for rate of convergence based on the singular values.

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  • $\begingroup$ You are right @LutzL, I'll edit the answer. $\endgroup$ – Sia Dec 12 '18 at 9:10
  • $\begingroup$ I think it is clear that the sequence need not converge, for example even in one dimension you have the linear map $\Bbb R\to \Bbb R$, $x\mapsto -x$ for which the sequence will always alternate between two limit points. I think the interesting question is: When does it have limit points? In finite dimensions it is clear that the sequence always admits limit points, as the unit sphere is compact. $\endgroup$ – s.harp Dec 12 '18 at 11:42

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