I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $(E, |\cdot|)$ be a real Banach space. Let $T \in \mathcal L(E)$, i.e., $T:E \to E$ is a bounded linear operator. Let $I:E \to E$ be the identity map. Let $\rho(T)$ be the resolvent set of $T$. Let $\sigma(T)$ be the spectrum of $T$, i.e., $\sigma(T) := \mathbb R \setminus \rho(T)$.
- Assume that $T^2 = I$. Prove that $\sigma(T) \subset \{-1, 1\}$ and determine $(T-\lambda I)^{-1}$ for $\lambda \neq \pm 1$.
- More generally, assume that there is an integer $n\ge 2$ such that $T^n=I$. Prove that $\sigma(T) \subset \{\pm 1\}$ and determine $(T-\lambda I)^{-1}$ for $\lambda \neq \pm 1$.
- Assume that there is an integer $n\ge 2$ such that $T^n=0$. Prove that $\sigma(T) = \{0\}$ and determine $(T-\lambda I)^{-1}$ for $\lambda \neq 0$.
There are possibly subtle mistakes that I could not recognize in below attempt (2). Could you please have a check on it? I'm also happy to see other approaches.
Let $\lambda \neq \pm 1$.
1.
We have $$ (T-\lambda I)(T+\lambda I) = (T+\lambda I)(T-\lambda I) =T^2-\lambda ^2 I = (1-\lambda^2) I. \tag{1} $$
It follows from $(1-\lambda^2) I$ being bijective that both $(T-\lambda I)$ and $(T+\lambda I)$ are bijective. Hence $(T-\lambda I)$ is bijective and thus $\lambda \in \rho(T)$. So $\sigma(T) \subset \{-1, 1\}$.
It follows from (1) that $$ (1-\lambda^2) (T-\lambda I)^{-1} (T+\lambda I)^{-1} = I. $$
On the other hand, $$ (T+\lambda I) (T+\lambda I)^{-1} = I. $$
Hence $$ (1-\lambda^2) (T-\lambda I)^{-1} = T+\lambda I, $$ and thus $$ (T-\lambda I)^{-1} = \frac{T+\lambda I}{1-\lambda^2}. $$
2.
Consider the polynomial $p(t) := t^n-\lambda^n$ for $t \in \mathbb R$. Then $$ p(t)= (t-\lambda) \sum_{k=0}^{n-1} t^k \lambda^{n-k-1}. $$
We have $p(T)$ is bijective because $$ p(T) = T^n-(\lambda I)^n = (1-\lambda^n) I \tag{1}. $$
On the other hand, we have $$ (T-\lambda I) \sum_{k=0}^{n-1} T^k \lambda^{n-k-1} \overset{(2)}{=} p(T) \overset{(3)}{=} \left ( \sum_{k=0}^{n-1} T^k \lambda^{n-k-1} \right ) (T-\lambda I) \tag{4} $$
It follows from (2) that $(T-\lambda I)$ is surjective and $\sum_{k=0}^{n-1} T^k \lambda^{n-k-1}$ is injective. It follows from (3) that $\sum_{k=0}^{n-1} T^k \lambda^{n-k-1}$ is surjective and $(T-\lambda I)$ is injective. Then both $(T-\lambda I)$ and $\sum_{k=0}^{n-1} T^k \lambda^{n-k-1}$ are bijective. In particular, $\lambda \in \rho(T)$. Then $\sigma(T) \subset \{\pm 1\}$.
It follows from (2) that $$ (T-\lambda I) \frac{1}{1-\lambda^n} \sum_{k=0}^{n-1} T^k \lambda^{n-k-1} = I. $$
Then $$ (T-\lambda I)^{-1} = \frac{1}{1-\lambda^n} \sum_{k=0}^{n-1} T^k \lambda^{n-k-1}. $$
3.
This is the same as the previous question with $p(T) = -\lambda^n I$.
