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Let $T:E \to E$ be a bounded linear operator, $E$ infinite dimensional Banach space, such that $T^n =I$, for $n\ge2.$ Show that $\sigma(T)\subset\{-1,1\}.$

My idea is show that $\|T\|=1$ initially, so $\sigma(T)\subseteq [-1,1],$ and by contradiction argument show that if $|\lambda|< 1$ then $\lambda \in\rho(T).$

I tried to compute the $\|T\|$ but I found just that $\|T\|^n \ge 1$.

Someone has can help?

Thanks.


This is part of a exercise of Brezis' book: Functional Analysis, Sobolev Spaces and PDE, question $6.16$.

As @SoumyaSinha noted, $E$ must be a infinite dimensional Banach space.

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  • $\begingroup$ Are you sure this is true? Seems like it fails if $E$ is finite dimensional. $\endgroup$ – Hmm. Jun 13 '16 at 19:16
  • $\begingroup$ Yes, I'm sure. This is a part of a Brezis' exercise. $\endgroup$ – Irddo Jun 13 '16 at 19:18
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    $\begingroup$ Consider $E$ to be a f.d.v.s. Then $\sigma(T)$ is nothing but the set of eigenvalues of $T$. However, if $T^n=I$, it is not necessary that only $1,-1$ are eigenvalues. $\endgroup$ – Hmm. Jun 13 '16 at 19:20
  • $\begingroup$ Ok, I will add this hypothesis. Thanks @SoumyaSinha. $\endgroup$ – Irddo Jun 13 '16 at 19:24
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If $\lambda^n \ne 1$, then you can directly verify that $T-\lambda I$ is invertible by showing \begin{align} I&=(T-\lambda I)\left[\frac{1}{1-\lambda^n}(T^{n-1}+\lambda T^{n-2}+\cdots+\lambda^{n-2}T+\lambda^{n-1}I)\right] \\ &=\left[\frac{1}{1-\lambda^n}(T^{n-1}+\lambda T^{n-2}+\cdots+\lambda^{n-2}T+\lambda^{n-1}I)\right](T-\lambda I). \end{align} If $E$ is a complex space, then any root of $p(\lambda)=\lambda^n-1$ may be an eigenvalue, depending on the specific $T$.

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  • $\begingroup$ Thanks @TrialAndError. Your answer prove that $\rho(T)\subset \Bbb{R}-\{-1,1\}$, right? But for $n=2k-1$, $k\ge 1$, why $-1\notin \rho(T)$? $\endgroup$ – Irddo Jun 14 '16 at 0:55
  • $\begingroup$ @Irddo : If you are working with a real space, then it proves that $\sigma(T)\subseteq \{-1,1\}$ or, equivalently, $\mathbb{R}\setminus\{-1,1\}\subseteq\rho(T)$. For a complex space, $\sigma(T)\subseteq\{ e^{2\pi i k/n }\}_{k=0}^{n-1}$ or, equivalently, $\mathbb{R}\setminus\{e^{2\pi i k/n} \}_{k=0}^{n-1}\subseteq\rho(T)$. $\endgroup$ – DisintegratingByParts Jun 14 '16 at 1:30
  • $\begingroup$ @Irddo : Do you see my comments follow? In particular, can you see by direct demonstration that $\lambda^n \ne 1$ implies $\lambda\in\rho(T)$. So $\sigma(T) \subseteq \{ \lambda : \lambda^n = 1 \}$ and the set $\{ \lambda : \lambda^n = 1 \}$ can change depending on whether the underlying field is real or complex. $\endgroup$ – DisintegratingByParts Jun 14 '16 at 1:46
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    $\begingroup$ I get it, perfect. Thanks so much. $\endgroup$ – Irddo Jun 14 '16 at 1:49

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