Exercise 5.A.18

Proof. Suppose $(z_1,z_2,\dots )\in C^{\infty }$ and $\lambda \in 𝔽$ are such that

Therefore $\lambda z_{k+1}=z_k$ for each positive $k$. Because $\lambda z_1=0$, if $\lambda =0$, by the previous equation it follows that each $z_k=0$ and, if $\lambda \ne 0$, then $z_1=0$ and by the previous equation $z_{k+1}=0$. Hence all $z$’s are $0$ and $T$ has no eigenvalues. □