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.
Does $A^nx=\lambda ^n x$ apply for $n$ smaller than $-1$ (assuming $A$ is invertible)?
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1"$A$ invertible doesn't imply $A^n$ invertible" If $A'$ is the inverse of $A$, why wouldn't $A'^n$ be the inverse of $A^n$? – 2012-08-04
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0If $A$ is invertible, $A^n$ must be, and the inverse will be given by $(A^{-1})^n$. $A^n x = y$ gives $A^{n-1} x = A^{-1} y$, etc., so $x = (A^{-1})^n y$, hence the result. – 2012-08-04
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0@CliveNewstead: I think he was referring to $n$. I don't know why he has this restriction either. – 2012-08-04
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1If $A$ is invertible, then $\lambda \neq 0$. If $Ax = \lambda x$, then clearly $A^{-1} x = \frac{1}{\lambda} x$, hence $A^{-n} x = \lambda^{-n} x$. – 2012-08-04
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0@copper.hat The n=0 case is trivial, the positive n case was proved by induction, and the -1 case was proved by algebraic manipulation. Oh thanks so much for pointing out what should've been obvious. I feel silly. So then the answer to my question is Yes it applies for all integers n. Cheers – 2012-08-04
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0No problem, I make many mistakes frequently :-). – 2012-08-04
2 Answers
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 $n$.
$A$ 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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0A unit or just an invertiable element in the ring ? – 2012-08-04
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1@Belgi I've always used "unit" to mean "an invertible element in the ring", e.g. "the group of units of a ring". – 2012-08-04
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0@AlexBecker Thanks:) – 2012-08-04