This is a part of Wold's Decomposition Theorem as in Theorem 1.1 of https://arxiv.org/pdf/1704.04200.pdf
Let $H$ be an infinite dimensional, complex Hilbert space. Suppose $T: H \rightarrow H$ is an isometry, i.e. $\| Tx\| = \| x\|$ for all $x\in H$.
Show that there is a unitary operator $W$ and a cardinal $\alpha$ such that $U \bigoplus V^{(\alpha)} =W T W^*$, where $V$ is the unilateral forward shift $Ve_n=e_{n+1}$.
I'm looking for the proof and\or constructing a proof by myself.
My biggest struggles are to construct such unitary operator $U$ and find the cardinal $\alpha$.
Any help will be greatly appreciated!
Suppose $TH \ne H$. (Otherwise $T$ is unitary.) Then there exists a unit vector $x\in H$ such that $x\perp TH$. It follows that $\{ x, Tx, T^2x, \cdots \}$ is an orthonormal sequence because $$ \langle T^{n+k}x,T^{n}x\rangle = \langle T^n(T^k x),T^n x\rangle = \langle T^kx,x\rangle=0,\;\; n=0,1,2,3,\cdots,\;\; k=1,2,3,\cdots. $$ The restriction of $T$ to the closed linear span $H_x$ of $\{ x,Tx,T^2x,\cdots \}$ is the unilateral shift. There is one copy of the unilateral shift for every vector $e_{\alpha}$ in an orthonormal basis $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ of $(TH)^{\perp}$, and the corresponding spaces $H_{e_{\alpha}}$ are mutually orthogonal. The number of such mutually orthogonal representations is the same as the cardinality $|\Lambda|$ of the orthonormal basis $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ of $(TH)^{\perp}$.
$T$ is unitary on $\bigcap_{\alpha\in\Lambda}H_{e_{\alpha}}$, unless the intersection is $\{0\}$.