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in my research on dilations of contractions on Hilbert spaces and manifolds I have come across this nice publication concerning the classic Sz-Nagy theorem on the Arxiv by Levy and Shalit which states the classical Sz-Nagy theorem on the existence of unitary and isometric dilations for contractions and their minimality:

http://arxiv.org/pdf/1012.4514.pdf

Right below quoting the famous theorem they mentioned that if the operator is not a unitary contraction then we must have that the larger Hilbert space K is infinite dimensional. Please forgive my ignorance but I cannot seem to understand why and it seems like a nice result I wish to understand. If the contraction to be dilated is not a unitary operator then why is the larger space embedding H is necessarily infinite dimensional? I am sorry if it was trivial and I missed it but I think I need to understand this. I thank all helpers

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There are probably many ways to see this, but here is a quick one. Suppose $T$ is a contractive matrix which dilates to a unitary matrix $U$, we wish to show that $T$ is already unitary. Now, to say that the matrix $U$ dilates $T$ means there is an isometric matrix $V$ so that $$ T^n = V^* U^n V $$ for all $n\geq 0$. From this it follows that the minimal polynomial $p$ for $U$ must annihilate $T$, but since all the roots of $p$ are unimodular, it follows that all the eigenvalues of $T$ are unimodular, and thus $T$ is unitary.

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  • $\begingroup$ Thank you Mike this is certainly a fresh approach for operator theory $\endgroup$ – kroner Mar 24 '16 at 14:41
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A unitary dilation $U\in B(K)$ of $T\in B(H)$ is said to be minimal if $$\overline{span}\{U^nH : n\in \mathbb Z\} = K.$$ It is a well known result that the minimal unitary dilation is unique up to unitary equivalence.

For any contraction $T\in B(H)$, Schaeffer found a constructive version of Sz.-Nagy's dilation theorem. Specifically, $$ U = \left[\begin{array}{cccccc} \ddots \\ \ddots & 0\\ & 1& 0 \\ && D_{T^*} & T \\ && -T^* & D_T & 0 \\ &&&&1 & 0 \\ &&&&&\ddots&\ddots \end{array}\right] $$ is a unitary dilation of $T$, where $D_X = (1-X^*X)^{1/2}$.

Hence, let $K' = \overline{span}\{U^nH : n\in \mathbb Z\} \subset K$. $K'$ is a reducing subspace of $U$ and $U|_{K'}$ is the minimal unitary dilation of $T$. If $T$ is not a unitary already then either it isn't an isometry ($D_T \neq 0$), a co-isometry ($D_{T^*} \neq 0$) or both. Either way it is direct to see that $K'$ is infinite dimensional from the form of the Schaeffer dilation. Therefore, by minimality, any unitary dilation of $T$ must be on an infinite dimensional Hilbert space.

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