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I am interested in

$$\lim_{k \to \infty} \operatorname{rank} \left( \mathbf{A}^k \right)$$

for a $n \times n$ matrix $\mathbf{A}$. I know that

  • for a nilpotent matrix, $\mathbf{A}^k=0$ when $k$ is big enough, which means $\operatorname{rank} (\mathbf{A}^k) \to 0$ when $k \to \infty$.

  • for a nonsingular matrix, $\mathbf{A}^k$ is also nonsingular and $\operatorname{rank} \left( \mathbf{A}^k \right)$ is the number of the columns/rows.

Is there any general theorem that tells us the result for a general square matrix?

Rodrigo de Azevedo
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xmfbit
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    Appealing to Jordan Normal Form gives that (1) it is simply the number of nonzero eigenvalues of $A$, and (2) the sequence stabilizes after exactly $p$ steps, where $p$ is the size of the largest Jordan block of eigenvalue $0$. – Travis Willse Nov 19 '14 at 15:30
  • Thanks! Inspired by your hint, I found how to calculate the $k$th power of the Jordan Normal Form in another question Why does the n-th power of a Jordan matrix involve the binomial coefficient?.And I think it can solve the problem. – xmfbit Nov 20 '14 at 04:35
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    You're welcome, I'm glad you found it helpful! Note that since Jordan blocks $J$ are upper triangular, so are their powers $J^k$. So, it's enough to know (1) for Jordan blocks $J$ of nonzero eigenvalue $\lambda$ what happens on the diagonal (namely, that $J^k$ has diagonal entries $\lambda^k$), i.e., we don't have to know what happens to the entries above the diagonal, and (2) for Jordan blocks $J$ of eigenvalue $0$, $J^m = 0$ for large enough $m$. – Travis Willse Nov 20 '14 at 07:33

1 Answers1

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Since the image of $A^{k+1}=A^k\,A$ is included in the image of $A^k$, we have that the function $k\longmapsto \text{rank}(A^k)$ is non-increasing. As the only possible values are $\{1,\ldots,n\}$ the numbers $\text{rank}(A^k)$ have to stabilize in a finite number of steps.

If the rank of $A^{k+1}$ is equal to the rank of $A^k$, this forces the image of $A^{k+1}$ to be equal to that of $A^k$, and thus the rank stabilizes.

So, the possible ranks will be a strictly decreasing sequence of positive integers until it stabilizes.

Here is an example that shows that one can start at any rank $h$, and decrease until rank $m$. Let $B$ be nonsingular $m\times m$ and $C$ an $(h-m+1)\times (h-m+1)$ Jordan block with zero diagonal. Consider $$ A=\begin{bmatrix}B&0\\0& C\end{bmatrix}. $$ Then $$\text{rank}(A^k)=\begin{cases}h-k,&\ k\leq \ h-m\\ m,&\ k>r\end{cases}$$

Martin Argerami
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