Banach spaces, invertible operators

Question

Does somebody know how to prove that for $X$ being a complex Banach space and $A \in B(X)$ being an invertible operator $\sigma(A^{−1})=\sigma(A)^{-1}$? $\sigma(A)^{−1}$ is defined like this: $\sigma(A)^{−1}:= \{z^{−1}: z \in \sigma(A)\}$

Answer 1

If $\lambda\notin\sigma(A)$, then either $\lambda=0$ (in which case $\lambda\in\sigma(A)^{-1}$), or $A^{-1}-\lambda^{-1}=-\lambda^{-1}A^{-1}(A-\lambda)$. Since $\lambda$, $A^{-1}$, and $A-\lambda$ are invertible, we see that $A^{-1}-\lambda^{-1}$ is invertible. Thus $(\sigma(A)^{-1})^c\subset\sigma(A^{-1})^c$.

If now $\lambda\notin\sigma(A^{-1})$ then either $\lambda=0$ (in which case $A-\lambda$ is invertible), or $A-\lambda^{-1}=-\lambda^{-1} A(A^{-1}-\lambda)$ is a product of invertibles, hence is invertible. Thus $\sigma(A^{-1})^c\subset(\sigma(A)^{-1})^c$.

Hence $\sigma(A^{-1})^c=(\sigma(A)^{-1})^c$, and therefore $\sigma(A^{−1})=\sigma(A)^{-1}$.