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$A$ is a square matrix. Prove

  1. there exists $m$ such that $\operatorname{rank}(A^m)=\operatorname{rank}(A^{m+1})$
  2. there exists a square matrix $B$ such that $A^m=A^{m+1}B$

I already knew the first one , since $A^{m+1}=A^m\cdot A$ , then $\operatorname{rank}(A^m)\ge \operatorname{rank}(A^{m+1}) $. If there is no $m$ to satisfy $\operatorname{rank}(A^m)=\operatorname{rank}(A^{m+1})$ , then $\operatorname{rank}(A^m)\gt \operatorname{rank}(A^{m+1}) $. And there will be $\operatorname{rank}(A^k)=0$, it follows $\operatorname{rank}(A^{k+1})=0$. Now I really don't know how to deal with the second one. Is there any connection?

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Hint:

  • Show that $A^m$ and $A^{m+1}$ have the same column space, using one inclusion and equality of dimensions.

  • Use multiplication on the right by elementary matrices (which amounts to invertible column operations).

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