The following problem was left as an exercise in my class of operator theory and I am not able to prove it.
Let $H$ be a Hilbert Space and let $V\in L(H)$ be an isometry. Show that $V $ always has a unitary extension.
Let $T\in L(H_1,H_2)$. Then T is called an isometry if $||Tx||_2 =||x||_1 \forall x\in H_1$.
T is called a unitary operator if $T$ is a bijective isometry.
I have to show the existence of a Hilbert Space K containing H and a unitary operator $U\in L(K)$ such that $U|_{H} =V$.
I have to show that U is invertible and the isometry on V can be extended to U. But I am not getting any ideas on how can I prove these two assertions?
Do you mind helping me?
Best Answer
Let $$H':=\operatorname{im}(V).$$ The corestriction $W:H\to H'$ of $V$ is an isometric isomorphism.
Define the Hilbert space$$K:=H\times H'.$$ Its subspace $H\times\{0\}$ being identified to $H,$ the following is a unitary extension of $V:$ $$\begin {matrix}U:&K&\to&K,\\&(x,y)&\mapsto&(V(x)+(I-p)(W^{-1}(y)),p(W^{-1}(y)))\end{matrix}$$ where $p\in L(H)$ is the orthogonal projection onto $H'.$