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Answer by fedja for Bounding the decrease after applying a contraction...
You can easily get a slightly cruder bound $d/n$ (or, if you want, $r/n$) as follows.Let $A_n=(T^*)^nT^n$ and $B_n=A_n-A_{n+1}$. Then $(B_nv,v)=\|T^nv\|^2-\|T^{n+1}v\|^2\ge 0$, so $B_n$ is positive...
View ArticleBounding the decrease after applying a contraction operator $n$ vs $n+1$ times
Can we upper bound the convergence rate of$$\max_{\textbf{v}: \left\Vert \textbf{v}\right\Vert_2=1} \left\{ \left\Vert \textbf{T}^n \textbf{v}\right\Vert^2_2 - \left\Vert \textbf{T}^{n+1}...
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