Bounding the decrease after applying a contraction operator $n$ vs $n+1$ times

Can we upper bound the convergence rate of$$\max_{\textbf{v}: \left\Vert \textbf{v}\right\Vert_2=1} \left\{ \left\Vert \textbf{T}^n \textbf{v}\right\Vert^2_2 - \left\Vert \textbf{T}^{n+1} \textbf{v}\right\Vert^2_2 \right\}~,$$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.