I have been given this exercise in my Operator theory class dealing with operators on Hilbert spaces, which reads as follows:
Let H be a Hilbert space. We are to prove, in two distinct ways, that if $ T \in B(H) $ is a contraction (Contraction: An operator T satisfying $ ||T|| \leq 1 $) which is not unitary, and if V is an isometric dilation of T, then the larger Hilbert space K, satisfying $ H \subset K $, is necessarily infinite dimensional.
Clarification: By isometric dilation I simply mean that the extended operator is itself an isometry.
Edit: the context for this is the Sz.-Nagy's dilation theorem as seen here
https://en.wikipedia.org/wiki/Sz.-Nagy%27s_dilation_theorem
And also from the text book by Nagy:
Here is an arxiv link to this result on page 1 the Nagy theorem:
http://arxiv.org/pdf/1012.4514.pdf
I unfortunately have to say that I truly have no idea on this one even after working hard at this, I don't have one way to prove this claim even in one method I thought assuming to get contradiction that I can assume K is finite dimensional and then work with matrices but I got nothing. I am stuck and truly desperate.
If $T \in \mathcal{B}(H)$ is a contraction, then $T$ has a unitary dilation on a Hilbert space $K$, meaning that there exists an isometry $V : H\rightarrow K$ and a unitary $U$ on $H$ such that $T^{n} = V^{\star}U^{n} V$ for all $n \ge 0$. Automatically $(T^{\star})^{n}=V^{\star}(U^{\star})^{n}V$.
Proof 1: Suppose $K$ is finite dimensional. Then $U$ is isometric and, hence, injective, which makes $U$ surjective; so $U$ must be unitary in the case of finite-dimensional $K$. If $m$ is the minimal polynomial for $U$, then $m(0) \ne 0$, which leads to a polynomial $q$ such that $q(U)=U^{-1}$. $T$ is normal because $$ T^{\star} = V^{\star}U^{\star}V = V^{\star}U^{-1}V=V^{\star}q(U)V = q(T) $$ And $T$ must be unitary because \begin{align} T^{\star}T & = (V^{\star}U^{\star}V)(V^{\star}UV) \\ & =(V^{\star}q(U)V)(V^{\star}UV) \\ & = V^{\star}q(U)UV= V^{\star}V = I. \end{align}
Proof 2: It appears you found a second proof in the comments.