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I look at this question:Spectrum of isometry, but I try to consider the similar question:

Let $X$ be a Banach space. Let $T\in \mathbb{B}(X)$. If $T$ is an isometry and invertible, prove that $\sigma(T) \subset \mathbb{S}:=\{\lambda: |\lambda|=1\}$.

To prove that we need to check that when $\lambda\notin \mathbb{S}$, $T_\lambda:=\lambda I-T$ is invertible. I know that there is a well-known result in $C^*$ algebra. But how to use definition to prove that?

It seems that we can prove that eigenvalues of $T$ is in the unit disk: for non-zero vector $x$ $$ (\lambda I-T)x=0 \mbox{ means } Tx=\lambda x $$ Then $\|Tx\|=|\lambda|\|x\|$. So $|\lambda|=1$. Thus, $\sigma_p(T)\subset \mathbb{S}$

Hermi
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2 Answers2

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If $|\lambda|>1$ the operator $\lambda I-T$ is invertible, because $\|T\|\le 1.$ For $0<|\lambda|<1$ we have $$\lambda I-T = -\lambda T(\lambda^{-1}I-T^{-1})$$ Hence the operator $\lambda I- T$ is invertible, as $\|T^{-1}\|\le 1$ and $|\lambda|^{-1}>1.$

Ryszard Szwarc
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Since $T$ is isometry and invertible, $T^{-1}$ is also an isometry. This implies that norm of $T$ and $T^{-1}$ are both $1$. For any $\lambda \in \sigma(T)$, we have that $|\lambda| \le \|T\| =1$. Since $T$ is invertible, we also have $\lambda^{-1} \in \sigma(T^{-1})$, thus $|\lambda^{-1}| \le \|T^{-1}\|=1$. Combining these two inequalities yields $|\lambda|=1$, which gives the desired result.

Tri
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