Let $H$ be an infinite dimensional Hilbert space and let $(e_i)_{i∈\mathbb{N}}$ be an orthonormal basis for $H$. We have to show that there exists exactly one operator $T \in \mathcal{B}(H)$ such that $$Te_i=e_{i+1}, i \in \mathbb{N}$$ and that $T$ is an isometry, so that we have $||Tx||=||x||, x \in H$. We have to find $T^*$ and show that $T^*T=1$ and $TT^* \neq 1$
I think I have understand some procedures to checking operators but I'm not sure how to find them for orthonormal basis. Is this a T,P,U,V or N operator? I think it cannot be an U(unitary) operator while we have that $TT^* \neq 1$. Can anyone help me to understand this
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 $TT^*\ne 1$.