 Solution

For the transpose shift, $(Cu{)}_1=0$ and $(Cu{)}_i=u_{i-1}$ for $i\ge 2$. If $Cu=\lambda u$, the first equation and the subsequent recurrence force every component to vanish, both for $\lambda =0$ and for $\lambda \ne 0$. Thus $C$ has no eigenvalues, illustrating that a bounded infinite-dimensional operator need not have one.