Decomposition property of a Hilbert space isometry I'm reading the book Foundations of Ergodic Theory (portuguese version) and I would be thankful if someone could shed some light on the solution to this exercise: 
Let $U:H\to H$ be an isometry of a Hilbert space. Show that there exist closed subspaces $V$ and $W$ such that $U(V)=V$, iteration of $W$ by $U$ produces a sequence of orthogonal subspaces which are also orthogonal to $V$, and 
$$H=V\oplus \bigoplus _{n=0}^{\infty }U^n\left(W\right)$$
So we conclude $U$ is an isomorphism if and only if $W={0}$. 
The book says that I should let $W=U(H)^\perp$ and $V=(\bigoplus^\infty_{n=0} U^n(W))^\perp$. From this, the last property follows easily. But how can I show all the iterations of $W$ are orthogonal?
I believe proving this exercise would be helpful in building intution for the differences between isometries and unitary operators.
 A: It is clear that $V$ is orthogonal to $U^n(W)$ for each $n$ by the definition of $V$ so to check the orthogonality condition, we need only check that for $n > m \geq 0$ we have that
$$U^n(W) \perp U^m(W).$$
For this, let $x,y \in W$ and consider $\langle U^n x, U^m y \rangle$. Since $U$ is an isometry, we have that $\langle U^n x, U^m y \rangle = \langle U^{n-m} x, y \rangle$. Since $y \in W \perp U(H)$ and $U^{n-m}x \in U(H)$ this tell us that $\langle U^n x, U^m y \rangle = 0$ so that $U^n(W) \perp U^m(W)$ as desired.
The only thing that remains is to see that $U(V) = V$.
First I check that $U(V) \subseteq V$. For this, we need to check that for $v \in V$, $w \in W$ and $n \geq 0$ we have that
$$\langle U v, U^n w \rangle = 0$$
In the case $n = 0$, this follows by definition of $W$. If $n > 0$ then
$$\langle U v, U^n w \rangle = \langle v, U^{n-1}w \rangle = 0$$
since $v \in V$. Therefore $U(V) \subseteq V$.
Now for the remaining inclusion we want to show that if $v \in V$ then $v \in U(V)$. For this, first note that since $v \perp W$, $v \in U(H)^{\perp \perp} = \overline{U(H)} = U(H)$ where the last equality follows since $U$ is an isometry so that $U(H)$ is closed. Therefore there is a $w \in H$ such that $Uw = v$. We want to check that $w \in V$. To do this take $x \in W$ and notice that
$$\langle w, U^n x \rangle = \langle v, U^{n+1}x \rangle = 0$$
where the first equality is by the isometry property. Therefore $w \in V$ and so $v \in U(V)$. Hence $U(V) = V$ as desired.
