I'm dealing recently with Nagy and Stinespring theorems and as idea behind them I was reading the following:
If we consider $V$ an isometry on $\mathcal{H}$ Hilbert space and let $P = I_{\mathcal{H}}-VV^*$ be the projection onto the orthocomplement of the range of $V$. If we define $U$ on $\mathcal{H} \oplus \mathcal{H} = \mathcal{K}$ as
\begin{equation} U = \begin{pmatrix} V & P \\ 0 & V^* \end{pmatrix} \end{equation}
It's easy to see that $U$ is unitary on $\mathcal{K}$. Until now everything's fine, but then if we identify $\mathcal{H}$ with $\mathcal{H} \oplus 0$ then
$$V^n = P_{\mathcal{H}}{U^n}|_{\mathcal{H}} \hspace{5mm} \forall n \ge 0 $$
First what is the mean of writing "$\mathcal{H} \oplus 0$", and why we do that?
how can I prove that this holds for the powers of both operators?
Also, which is the role of $P_{\mathcal{H}}$ if I'm already considering the restriction of $U$ to $\mathcal{H}$? Maybe cause we need to consider two things having the same dimension? (cause $U$ lives in $\mathcal{K} = \mathcal{H} \oplus \mathcal{H}$). Thank you!
What it means is this: $\mathcal H \oplus \mathcal H$ consists of pairs of vectors $(u,v)$ where $u$ and $v$ are in $\mathcal H$. We identify $\mathcal H$ as the subspace of $\mathcal H \oplus \mathcal H$ consisting of pairs of the form $(u,0)$. For an operator on $\mathcal H \oplus \mathcal H$ written in block form as $$ A = \pmatrix{ B & C\cr D & E\cr}$$ $\left. P_{\mathcal H} A\right|_{\mathcal H} = B$. What that means is that if you take a vector $u \in \mathcal H$ and identify it as $(u,0) \in \mathcal H \oplus \mathcal H$, $A (u,0) = (Bu, Du)$, and the projection on $\mathcal H$ (identified as the first block) of this is $Bu$.
If you take the $n$'th power of an operator where the lower left block is $0$, you get one where the lower left block is still $0$ and the diagonal blocks are the $n$'th powers of those blocks in the original operator (prove by induction). So in the case of $U$, this is saying that the top left block of $U^n$ is $V^n$.