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Verifying that (p=2) satisfies

$$\forall n\in\mathbb{Z}^+.\exists A\in(\mathbb{R}^{n\times n}\setminus\{I_n\}).\forall k\in\mathbb{R}.\forall v\in\mathbb{R}^{n}.\left\|A^kv\right\|_p\!\!=\left\|v\right\|_p$$

is trivial (let each $A$ be a rotation matrix), but is there a clean way to obtain the value two from this condition?

(Note that $A^k$ is ambiguous for some choices of $A$ and $k$. I believe that the solution set is the same regardless of how this ambiguity is resolved, even if such an $A^k$ is taken to be undefined.)

Related, but not (I believe) duplicates:

Probably related:

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Edward
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  • I was assuming $k$ was an integer in my now-deleted answer. Yes, I see you specified $k\in\Bbb R$, sorry. That makes it a totally different question. Of course the definition of $A^k$ for non-integer $k$ is problematic. But one could rephrase the question in terms of the existence of a "one-parameter group" ${A_k:k\in\Bbb R}$ of isometries. What the answer is then I'm not so sure, but it seems probably yes - surely the isometry group is discrete for $p\ne 2$. Probably someone else can give a smoother proof than what I'd concoct... – David C. Ullrich May 02 '16 at 01:43
  • @DavidC.Ullrich Okay. Thank you for the time you took on that first answer. I admit that I'm playing fast and loose with $A^k$ in the write-up; it just seems unlikely to me that the choice in how to resolve the ambiguity could affect the set of solutions. – Edward May 02 '16 at 02:02

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