Dilation of an operator

Question

Let $T\in B(\mathcal{H})$. Based on https://en.wikipedia.org/wiki/Dilation_(operator_theory) dilation of the operator $T$ is an operator on a larger Hilbert space $K$, whose restriction to $\mathcal{H}$ composed with the orthogonal projection $P$ onto $\mathcal{H}$ is $T$.

I have several questions about this definition. Is there another version of this definition? Is the operator $P$ always projection? I mean could one find a larger Hilbert space $K$ and operators $V, U$ on $K$, when $U$ is not an orthogonal projection, such that $UV|_{\mathcal{H}}= T$?

Answer 1

Your candidate for a definition is exactly the same definition of dilation you quoted. If $UV|_H=T$, then since $PT=T$ you have $PUV|_H=T$, and so $UV$ is a dilation of $T$ in the sense of your first paragraph.

A "dilation" has to be a "dilation". The spirit is that you have $T$ (typically, $T$ is a contraction), and you get a dilation $$\tilde T=\begin{bmatrix} T&A\\ B& C\end{bmatrix}$$ that is better behaved, say $\tilde T$ is a unitary, or a projection.

Answer 2

Another version:

Theorem: Let $A$ be a bounded linear operator on a Hilbert space $\mathcal{H}$ with $\|A\|\le 1$. There there exists a Hilbert space $\mathcal{K}$, an isometry $V : \mathcal{H}\rightarrow\mathcal{K}$, and a unitary operator $U : \mathcal{K}\rightarrow\mathcal{K}$ such that $$ A^n = V^*U^n V \mbox{ for all } n \ge 0. $$