Can we upper bound the convergence rate of
$$\max\_{\textbf{v}: \left\Vert \textbf{v}\right\Vert\_2=1} \left\{ \left\Vert \textbf{T}^n \textbf{v}\right\Vert^2\_2 - \left\Vert \textbf{T}^{n+1} \textbf{v}\right\Vert^2\_2 \right\}~,$$
where $\textbf{T}\in \mathbb{R}^{d \times d}$ is a contraction operator
($\left\Vert\textbf{T}\right\Vert\_2\le1$)
of rank $r<d$?




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For example, $\textbf{T}$ can be *nilpotent* with index $q$ (e.g., $\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]
$) and then the maximal decrease *can* be fixed as long as $n<q$, and afterwards it is $0$.




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I found a Toeplitz operator whose rate is $d/en$:
$$\boldsymbol{T}=\left[\begin{array}{ccccc}
 & 1\\
 & & \ddots\\
 & & & 1\\
1-\epsilon
\end{array}\right]$$ and I have an intuition this should be the upper bound, but I was not able to prove that this is indeed the worst case.



Bounding the decrease after applying a contraction operator $n$ vs $n+1$ times
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 definite. Also $B\_{n+1}=T^\*B\_nT$, so, since $T$ is a contraction, $Tr B\_{n+1}\le Tr B\_n$ (this is obvious if $T$ is diagonal but in general $T=R\_1DR\_2$ where $R\_j$ are orthogonal and $D$ is a diagonal contraction and conjugation by an orthogonal matrix doesn't change either trace, or positive definiteness).


So, $n\, Tr B\_n\le \sum\_{k=1}^n Tr B\_k=Tr A\_1-Tr A\_{n+1}\le Tr A\_1\le r$ and we are done because the trace dominates the norm for positive definite matrices.


