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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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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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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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