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I am able to show (by proving 3 separate cases) that $Ax=\lambda x$ for nonzero $x$ and invertible $A$ implies that $A^nx=\lambda ^n x$ for all integers $n$ greater than or equal to $-1$. I was trying to extend this theorem to the rest of the negative integers, but I ran into a hitch because $A$ invertible doesn't imply $A^n$ invertible. So it seems my original proof was as general as it can be. Is my reasoning correct though? Thanks.

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    No problem, I make many mistakes frequently :-).2012-08-04

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If $A$ is invertible and $A x = \lambda x$ for some $x\neq 0$, then $\lambda \neq 0$, and $A^{-1} x = \frac{1}{\lambda} x$. It follows that $A^{-n} x = \lambda^{-n} x$ for all integers $n$.

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$A$ invertible does imply $A^n$ invertible, since $A \text{ invertible }\iff \det(A)\neq 0\iff \det(A)^n\neq 0\iff \det(A^n)\neq 0\iff A^n \text{ invertible }$ and this works over general rings as well as fields when we replace "$\det(A)\neq 0$" with "$\det(A)$ a unit". This is crucial in order to make $A^{-n}$ well-defined. Note that (using what you've already proven) $Ax=\lambda x\implies A^{-1}x=\lambda^{-1}x\implies (A^{-1})^nx=(\lambda^{-1})^nx\implies A^{-n}x=\lambda^{-n}x$ and so we can conclude that $Ax= \lambda x\implies A^nx=\lambda^n x$ for all integers $n$.

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    @AlexBecker Thanks:)2012-08-04