Mfrhtyj

Given $Ain F^{n times n}$ 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?

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 $mtimes m$ 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 $mtimes m$ 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 $m$).

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

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 $kle n$.
$$
rank(A^k) = dim(operatorname{im}(A^k))
$$$$
operatorname{im}(A) supseteq operatorname{im}(A^2) supseteq operatorname{im}(A^3) supseteq dots
$$

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