Finding the ranks of generalized eigenvectors given a Jordan normal form

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Let $A$ be a $7\times 7$ matrix over $\mathbb{C}$ with minimal polynomial $m(t) = (t-1)^3(t-2)^2$.

One possible Jordan canonical form of $A$ that I found, up to similarity, is $$\begin{pmatrix} 1 & 1 & 0 & & & & \\ 0 & 1 & 1 & & & & \\ 0 & 0 & 1 & & & & \\ & & & 2 & 1 & \\ & & & 0 & 2 & \\ & & & & & 2 & \\ & & & & & & 2 \end{pmatrix}$$ where the empty spaces are filles with zeros.

Question How would one go about finding the ranks of $(A-I)^k$ and $(A-2I)^k$ for $k \in \mathbb{Z}_+$?

My initial guess is that the rank of $(A-I)^k$ is 3 since that $3\times 3$ block is an invertible matrix, while the rank of $(A-2I)^k$ is $2$ since the $2\times 2$ block is invertible as well.

I know that a generalized eigenvector $v$ has rank $r$ if $(A-\lambda I)^r v = 0$ but $(A - \lambda I)^{r-1} v \neq 0$; but we don't know what $A$ is, so I can't use this here.

However, I did notice that if we let $B = \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}$, then $(B-I)^i \neq \mathbf{0}$ for $i = 1,2$, but $(B-I)^3 = \mathbf{0}$, where $\mathbf{0}$ is the $3\times 3$ zero matrix. But I don't know if this observation has anything to do with this problem or that it's just a coincidence.

Thank you for your help