Exercise 5.A.21

Proof. (a) Suppose $\lambda$ is an eigenvalue of $T$ and $v$ a corresponding eigenvector. Then, applying $T^{-1}$ to both sides of $\mathit{Tv}=\mathit{\lambda v}$, we get $v=\lambda T^{-1}v$. Dividing both sides by $\lambda$ shows that $\frac{1}{\lambda }$ is indeed an eigenvalue of $T^{-1}$. The converse is basically the same.

(b) The previous item also showed that if $v$ is an eigenvalue of $T$ it must be an eigenvalue of $T^{-1}$ too. □