Exercise 5.A.20

Proof. Suppose $\lambda$ is an eigenvalue of $T$. Then

which means that $z_k=\frac{1}{\lambda }{z}_{k+1}$. But $z_{k+1}=\frac{1}{\lambda }{z}_{k+2}$. It easy to see (and easily proven by induction) that $z_k={(\frac{1}{\lambda })}^{n-k}{z}_n$ or $z_n={\lambda }^{n-k}{z}_k$. Choose $k=1$ and we have a formula for any geometric sequence. Hence all real numbers are eigenvalues of $T$ and any of their respective geometric sequences are the corresponding eigenvectors. □