复Banach空间中可逆算子谱逆等式σ(A⁻¹)=σ(A)⁻¹的证明咨询

Let's walk through this standard result step by step. First, let's recap key definitions to set the stage:

• For a bounded linear operator $A: X \to X$ on a complex Banach space $X$, the spectrum $\sigma(A)$ is the set of all $\lambda \in \mathbb{C}$ such that $\lambda I - A$ is not invertible in $B(X)$ (the space of bounded linear operators on $X$).
• Since $A$ is invertible, $0 \notin \sigma(A)$ (because $0I - A = -A$ is invertible), so every element of $\sigma(A)$ has a well-defined inverse. We define $\sigma(A)^{-1} = { \lambda^{-1} : \lambda \in \sigma(A) }$.

To prove the equality of sets, we need to show mutual inclusion: $\sigma(A^{-1}) \subseteq \sigma(A)^{-1}$ and $\sigma(A)^{-1} \subseteq \sigma(A^{-1})$.

1. Prove $\sigma(A^{-1}) \subseteq \sigma(A)^{-1}$ #

Take any $\lambda \in \sigma(A^{-1})$. By definition, $\lambda I - A^{-1}$ is not invertible in $B(X)$.

First, note $\lambda \neq 0$: if $\lambda = 0$, then $0I - A^{-1} = -A^{-1}$, which is invertible (since $A$ is invertible). So $\lambda$ must be non-zero.

Suppose for contradiction that $\lambda^{-1} \notin \sigma(A)$. This means $\lambda^{-1}I - A$ is invertible. Let's rewrite $\lambda I - A^{-1}$ in terms of this invertible operator:
$$
\lambda I - A^{-1} = \lambda A^{-1}(A - \lambda^{-1}I) = -\lambda A^{-1} \left( \lambda^{-1}I - A \right)
$$

The right-hand side is a product of three invertible operators:

• $\lambda \neq 0$, so scalar multiplication by $\lambda$ is invertible.
• $A^{-1}$ is invertible (by assumption).
• $\lambda^{-1}I - A$ is invertible (by our contradiction hypothesis).

Products of invertible operators are invertible, which means $\lambda I - A^{-1}$ is invertible—this contradicts our initial assumption that $\lambda \in \sigma(A^{-1})$.

Thus, our contradiction hypothesis is false: $\lambda^{-1} \in \sigma(A)$, so $\lambda = (\lambda^{-1}){-1} \in \sigma(A)^{-1}$. This proves the first inclusion.

2. Prove $\sigma(A)^{-1} \subseteq \sigma(A^{-1})$ #

Take any $\mu \in \sigma(A)^{-1}$. By definition, $\mu = z^{-1}$ for some $z \in \sigma(A)$, so $zI - A$ is not invertible.

Suppose for contradiction that $\mu I - A^{-1}$ is invertible. Let's rewrite $zI - A$ in terms of this invertible operator:
$$
zI - A = zA \left( A^{-1} - z^{-1}I \right) = -zA \left( \mu I - A^{-1} \right)
$$

Again, the right-hand side is a product of invertible operators:

• $z \neq 0$ (since $z \in \sigma(A)$ and $A$ is invertible, so $0 \notin \sigma(A)$).
• $A$ is invertible.
• $\mu I - A^{-1}$ is invertible (by our contradiction hypothesis).

This implies $zI - A$ is invertible, which contradicts $z \in \sigma(A)$.

Thus, our contradiction hypothesis is false: $\mu I - A^{-1}$ is not invertible, so $\mu \in \sigma(A^{-1})$. This proves the second inclusion.

Since both inclusions hold, we conclude that:
$$
\sigma(A^{-1}) = \sigma(A)^{-1}
$$

内容的提问来源于stack exchange，提问作者Negreanu