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Suppose $A$ is an invertible $n \times n$ matrix , $n \in \mathbb{N}$ ,which has $v$ as eigenvector and $\lambda$ as eigenvalue. Prove that the $A^m$ where $m \in \mathbb{Z}$ has the same eigenvector $v$. For positive integer, I prove it like this: $(A^m)v=(AA...A)(v)=(AA...A)(Av)=(AA...A)(\lambda v)=...=\lambda^{m}v$. For negative case. I don know how to prove it. Can anyone guide me ?

Idonknow
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1 Answers1

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You have $Av=\lambda v$. Now multiply both sides by $A^{-1}$ on the left.

Chris Eagle
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  • Can I prove it like this ? For $m<0$, let $D=A^{-1}$. Then $(A^m)(v)=(A^{-1})^{-m}(v)=D^{-m}v$. Then since $-m>0$, use the induction for the positive integer, is it okay? – Idonknow Mar 08 '13 at 16:00
  • Once you've proven it for $A^{-1}$, yes, you can do that. @Idonknow – Thomas Andrews Mar 08 '13 at 16:17