Exercise 5.A.15

Proof. (a) Suppose $\lambda$ is an eigenvalue of $T$ and $v$ a corresponding eigenvector. Then

Hence $\lambda$ is an eigenvalue of $S^{-1}\mathit{TS}$ and $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(S^{-1}v)$ a corresponding eigenvector.

(b) If $v$ is an eigenvector of $T$, then $S^{-1}v$ is an eigenvector of $S^{-1}\mathit{TS}$ and if $v$ is an eigenvector of $S^{-1}\mathit{TS}$, then $\mathit{Sv}$ is an eigenvector of $T$. □