Question

Let TεB(X,X) be invertible in B(X , X) , X is a complex Banach space.Show that σ(Inverse of T)={1/λ : λЄσ(T)}

Answer 1

Question 1. Let $T \in B(X, X)$ be invertible in $B(X, X)$, where $X$ is a complex Banach space. Show that $\sigma(T^{-1}) = \{1 / \lambda : \lambda \in \sigma(T)\}$.

Solution. By the definition of the spectrum, we need to prove that $T - \lambda I$ is not invertible if and only if $T^{-1} - \frac{1}{\lambda} I$ is not invertible. Note that $\lambda = 0$ does not belong to $\sigma(T)$, because $T$ is invertible. Thus, our task is to prove that $T - \lambda I$ is invertible iff $T^{-1} - \frac{1}{\lambda} I$ is invertible for all $\lambda \neq 0$.

Let $T$ and $T - \lambda I$ be invertible, $\lambda \neq 0$. Prove that $T^{-1} - \frac{1}{\lambda} I$ is invertible. It is sufficient to show that $T^{-1} - \frac{1}{\lambda} I$ is invertible on the left and on the right. Indeed, note that

$$- \lambda T \left(T ^ {- 1} - \frac {1}{\lambda} I\right) = T - \lambda I.$$

Since $T - \lambda I$ is invertible, this implies

$$- \lambda (T - \lambda I) ^ {- 1} T \left(T ^ {- 1} - \frac {1}{\lambda} I\right) = I.$$

Furthermore

$$\left(T ^ {- 1} - \frac {1}{\lambda} I\right) (- \lambda T) (T - \lambda I) ^ {- 1} = (T - \lambda I) (T - \lambda I) ^ {- 1} = I.$$

Thus, $-\lambda (T - \lambda I)^{-1}T = -\lambda T(T - \lambda I)^{-1} = (T^{-1} - \frac{1}{\lambda} I)^{-1}$.

The converse implication: "if $T^{-1} - \frac{1}{\lambda} I$ is invertible, then $T - \lambda I$ is invertible" is proved similarly.