If $V\in L(H)$ is an isometry , then show that $V$ has unitary extension. 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?
 A: This is a very abstract problem because it occurs exclusively in infinite dimensions (due to every isometry on a finite-dimensional space automatically being unitary). Thus to get a better feeling for it let us attempt to solve the problem just for the right-shift operator
\begin{align*}
V:\ell^2(\mathbb C)&\to\ell^2(\mathbb C)\\
(x_1,x_2,\ldots)&\mapsto (0,x_1,x_2,\ldots)
\end{align*}
which is the prime example of an isometry which is not surjective (and hence not unitary). Extending this operator to something unitary means that we, in particular, have to "enlarge" the domain of $V$ somehow. A good first guess is to just take two copies, that is, $ K:= H\times H$ which is indeed a Hilbert space. Note that this is not the only choice for $ K$ as Anne's answer shows.
If we choose to embed $ H\hookrightarrow  K$ via $x\mapsto\scriptstyle\begin{pmatrix}x\\0\end{pmatrix}$, then $U: K\to K$ is called an extension of $V$ if
$$
U\begin{pmatrix}x\\0\end{pmatrix}=U\begin{pmatrix}(x_1,x_2,\ldots)\\0\end{pmatrix}=\begin{pmatrix}V(x_1,x_2,\ldots)\\0\end{pmatrix}=\begin{pmatrix}(0,x_1,x_2,\ldots)\\0\end{pmatrix}
$$
for all $x\in\ell^2(\mathbb C)$.
Hence the action of $U$ on the first component is already determined; our job now is to define on the second component such that we get something bijective and isometric. One possibility (of infinitely many) is
$$
U\begin{pmatrix}x\\y\end{pmatrix}=U\begin{pmatrix}(x_1,x_2,\ldots)\\(y_1,y_2,\ldots)\end{pmatrix}=\begin{pmatrix}(y_1,x_1,x_2,\ldots)\\(y_2,y_3,y_4,\ldots)\end{pmatrix},
$$
that is, we "fill the gap" caused by $V$ with the first element of the second component. It is obvious that $U$ is bijective and isometric.
Coming back to the general problem, i.e. to arbitrary $V$, let us express the above map $U$ using $V$:
$$
U\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}Vx+(y_1,0,0,\ldots)\\V^*y\end{pmatrix}=\begin{pmatrix}Vx+y-VV^*y\\V^*y\end{pmatrix}=\begin{pmatrix}Vx+(I-VV^*)y\\V^*y\end{pmatrix}\,.
$$
Now if (we can show that) the map
\begin{align*}
U: H\times H&\to  H\times H\\
\begin{pmatrix}x\\y\end{pmatrix}&\mapsto\begin{pmatrix}Vx+(I-VV^*)y\\V^*y\end{pmatrix}
\end{align*}
is a bijective linear isometry which satisfies $U(x,0)=(Vx,0)$ for all $x\in H$, then we are done. Indeed, proving this is straightforward, cf., e.g., Appendix 4 of "Functional Analysis" by F. Riesz and B. Sz.-Nagy (1990) or p. 6 in "Completely Bounded Maps and Operator Algebras" by V. Paulsen (2002).
A: 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 $H'.$
