TITLE: Bounding the decrease after applying a contraction operator $n$ vs $n+1$ times
QUESTION [7 upvotes]: 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 $r<d$?

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 $0$.

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.

REPLY [5 votes]: 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.