Operators on Hilbert Spaces...
Solution 1:
This is a good example to show why isometry is not unitary, because $T$ is not surjective.
The existence and uniqueness of $T$ s.t. $Te_i=e_{i+1}$ is clear. Now $T(a_1,a_2,a_3,\cdots)=(0,a_1,a_2,\cdots)$, so it is an isometry. From $(Te_i,e_j)=(e_i,T^*e_j)$ we know that $T^*(a_1,a_2,a_3,\cdots)=(a_2,a_3,a_4,\cdots)$, the backward operator. Now it's easy to see $T^*T=1$ while $TT^*\ne 1$.