Does $\operatorname{rank}\left(A^n\right)=\operatorname{rank}\left(B^n\right)$ imply $\operatorname{rank}(A)=\operatorname{rank}(B)$

Question

Let $A,B$ be two $d\times d$ matrices, and $n\in\Bbb N$ s.t. $A^n,B^n\neq 0$. If $\operatorname{rank}\left(A^n\right)=\operatorname{rank}\left(B^n\right)$, then when can we conclude that $\operatorname{rank}(A)=\operatorname{rank}(B)$?

Thanks.

Answer 1

Let $A = \begin{bmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$. Then $\text{rank}(A^3) = \text{rank}(B^3)=1$ but $\text{rank}(A) \ne \text{rank}(B)$.

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Edit after user26857's comment:

By the rank-nullity theorem, we can think about the nullities of these matrices instead of their ranks.

It suffices to think the about the Jordan decomposition of a matrix $A$. Suppose the Jordan blocks corresponding to eigenvalue $0$ have sizes $m_1, \ldots, m_K$. Then $K$ is the nullity of the matrix. More generally, can check that the nullity of $A^n$ is $$\text{nullity}(A) = \min\{n, m_1\} + \min\{n, m_2\} + \cdots + \min\{n, m_K\}.$$

If we let $m'_1, \ldots, m'_{K'}$ denote the sizes of the Jordan blocks of $B$ corresponding to eigenvalue $0$, then we have the following.

$$\text{nullity}(A) = \text{nullity}(B) \iff K = K' \text{ (same number of Jordan blocks for eigenvalue $0$)}$$ $$\text{nullity}(A^n) = \text{nullity}(B^n) \iff \sum_{k=1}^K \min\{n, m_k\} = \sum_{k=1}^{K'} \min\{n, m'_k\}.$$

Not sure if this is helpful to the OP or not...