Rate of decline of rank of matrix powers Questions such as this one (What is the limit of the rank of the power of a matrix?) address the limit of the rank of matrix powers, but do not mention at what rate this limit is approached.  Is there a closed form expression for this rate?  If not in general, then for certain classes of matrices (e.g. Toeplitz)?

If the rank decreases at all, it has to start decreasing right away, and the rate can only slow down. 
For any matrix $A$, consider its Jordan form. The blocks for nonzero eigenvalues are invertible, and so their contribution to the rank is constant. The change in rank can only come from the nilpotent part, that is the Jordan blocks corresponding to the zero eigenvalue. These blocks lose rank by $1$ with every increase in power. So the loss of rank will initially be the number of Jordan blocks with eigenvalue zero; this will continue until some block is exhausted (its power reaches zero), where it will stop contributing to the decrease in rank. 
For instance, if the zero part of the Jordan form is 
$$
\begin{bmatrix}0&1\\0&0\end{bmatrix}
\oplus
\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}
\oplus
\begin{bmatrix}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{bmatrix},
$$
then the rank will go as 
$$
\begin{vmatrix}
\hline\text{Power}&|&\text{Rank}\\
\hline 1&|&7\\
\hline 2&|&4\\
\hline 3&|&2\\
\hline 4&|&1\\
\hline
\end{vmatrix}
$$
With the square, the three blocks lose rank; with the cube, the first block is gone, so only the last two contribute; and, with the fourth power, only the last block contributes. 
As in the answer you quote (which I read before noticing it was mine), one can achieve a "floor" for the rank by adding a block with an invertible matrix of the desired size. 
