# Given a square matrix A of order n, prove $\operatorname{rank}(A^n) = \operatorname{rank}(A^{n+1})$

Given $A\in F^{n \times n}$ prove:

$$\operatorname{rank}(A^n) = \operatorname{rank}(A^{n+1})$$

$\operatorname{rank}(A^{n+1}) \leq \operatorname{rank}(A^n)$ is easy, just from:

How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$?

But how can I prove the other direction? or should I do it otherwise?

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The matrix $A$ must satisfy a polynomial equation of degree $n$(its characteristic polynomial), and hence $A^{n+1}$ is expressible as the sum of some powers of $A$ of lower exponent. Hence the assertion follows. – awllower Feb 8 '13 at 14:06
@awllower But taking the sum of matrices can reduce the rank. – Tobias Kildetoft Feb 8 '13 at 14:08
@awllower This only proves the same direction that was already noted by the OP. – 1015 Feb 8 '13 at 15:14
@julien Thanks for the comments indicating my error. – awllower Feb 9 '13 at 13:45

Note that we can assume the field is algebraically closed, as the rank of the matrix does not change if we look at it as being over a larger field.

Now the matrix is similar to an upper triangular matrix. We can assume that it has a block form consisting of an upper triangular $m\times m$ matrix with only non-zero elements on the diagonal, and a block consisting of a strictly upper triangular $(n-m)\times (n-m)$ matrix. Now both the $n$'th and the $n+1$'st power of such a matrix will simply consist of some $m\times m$ upper triangular block with only non-zero elements on the diagonal (as we kill off the strictly upper triangular block when the power is at least $n-m$). This shows that these two powers have the same rank (namely $m$).

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So you proved in particular that the rank of $A^2$ is always equal to the rank of $A$? What do you do with nonzero matrices $A$ such that $A^2=0$? – 1015 Feb 8 '13 at 15:13
@julien no, this does not show that the rank of $A$ is the same as the rank of $A^2$ unless $n = 1$ in which case it is trivial. – Tobias Kildetoft Feb 8 '13 at 15:48
Oh boy, what a trick. I had not seen that it was the same $n$. Sorry. And +1. – 1015 Feb 8 '13 at 15:51
+1, well done indeed. – Andreas Caranti Feb 8 '13 at 16:04

Using Fitting's Lemma, one can give another version of the fine argument of @Tobias.

The sequence $$\ker(A) \subseteq \ker(A^2) \subseteq \ker(A^3) \subseteq \dots$$ is ascending, and the sequence $$\operatorname{im}(A) \supseteq \operatorname{im}(A^2) \supseteq \operatorname{im}(A^3) \supseteq \dots$$ is descending. Choose the smallest $m$ such that $$\ker(A^m) = \ker(A^{m+i}), \qquad \operatorname{im}(A^m) = \operatorname{im}(A^{m+i})$$ for all $i \ge 0$. Note that if $\ker(A^m) = \ker(A^{m+1})$, then $\ker(A^m) = \ker(A^{m+i})$ for all $i \ge 0$. In particular $m \le n$.

Now Fitting's Lemma states that $$F^n = \ker(A^m) \oplus \operatorname{im}(A^m),$$ and $A$ is nilpotent on the first summand, and invertible on the second one.

Then for any $k \ge m$ (actually, I believe, exactly for these values of $k$) we will have $$\operatorname{rank}(A^k) = \operatorname{rank}(A^{k+1}).$$

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