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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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3 Answers3

5

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 $m$).

Tobias Kildetoft
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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 $A^2=0$? – Julien Feb 08 '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 08 '13 at 15:48
  • Oh boy, what a trick. I had not seen that it was the same $n$. Sorry. And +1. – Julien Feb 08 '13 at 15:51
5

Coming back to this question after a few years, I've found a simpler proof, using only basic linear algebra knowledge.

First, if $\operatorname{rank}(A)=n$, use the facts:

  • Matrix is full rank iff it is invertible
  • Product of invertible matrices is invertible

so $\operatorname{rank}(A^{k})=n$ for any natural $k$.

Otherwise, use induction to show the following:

if $rank(T^k) = rank(T^{k+1})$ for some positive integer $k$, then $rank(T^k) = rank(T^m)$ for all positive integer $m \geq k$.

Finally, we have to show that if $n \gt \operatorname{rank}(A)$, then $rank(A^k) = rank(A^{k+1})$ for some $k\le n$. $$ rank(A^k) = \dim(\operatorname{im}(A^k)) $$$$ \operatorname{im}(A) \supseteq \operatorname{im}(A^2) \supseteq \operatorname{im}(A^3) \supseteq \dots $$

$$ n \gt \operatorname{rank}(A) \ge \operatorname{rank}(A^2) \ge \operatorname{rank}(A^3) \ge \dots \ge \operatorname{rank}(A^n) \ge \operatorname{rank}(A^{n+1}) \ge 0 $$ There are n possible values ($0,\dots,n-1$) for n+1 ranks, so there are at least two ranks that are equal.

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4

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

Andreas Caranti
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