Spectrum of $T\in \mathcal{L}(E)$, such that $T^n=I$

Question

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.

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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.

Answer 1

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 $T$.