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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?

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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).

Frederik vom Ende
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  • Unrelated side note: had we not required domain and co-domain of $U$ to be the same, we could have extended $V$ "way smaller" via $U:H×(\operatorname{im}V)^\perp\to H$, $(x,y)↦Vx+y$ – Frederik vom Ende Dec 29 '22 at 00:02
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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 $H'.$

Anne Bauval
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  • Something seems to not add up here? You are defining $U:\mathcal H\times\operatorname{im}V\to\operatorname{im}V\times\mathcal H$ via $U(x,y):=(Vx,V^y)$ so domain and co-domain are not equal. But making them equal via $\operatorname{im}V\times\mathcal H\simeq\mathcal H\times\operatorname{im}V$ (as you do in the co-domain of $U$, where $=$ is rather $\simeq$) leads to the operator $(x,y)\mapsto(V^y,Vx)$ which is not an extension of $V$ anymore. – Frederik vom Ende Dec 28 '22 at 18:48
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    I don't "make them equal" via the isomorphism you wrote. They are equal by the sequence of true equalities (and not isomorphisms) in my answer. And my $U$ is not what you wrote. $(x,0)$ is really sent to the subspace $H'\oplus0$ of $H\oplus0\subset H\oplus H'=K.$ – Anne Bauval Dec 28 '22 at 18:59
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    I rewrote all this more formally, hopefully not damaging to much the readability – Anne Bauval Dec 28 '22 at 22:45
  • Looks great! So your $U$ operates as $(x,Vz)\mapsto(Vx+(I-VV^)z,VV^z)$ for all $x,z\in H$ which also gives a nice connection to the "larger" extension from my answer – Frederik vom Ende Dec 28 '22 at 23:58
  • If you say so... But it is a reformulation of my initial answer, which I personnaly found more understandable. – Anne Bauval Dec 29 '22 at 05:50
  • @AnneBauval Thank you very much for the answer. Can you please tell how you proved that $W $ is an isomorphism? –  Dec 31 '22 at 14:46
  • There is nothing to prove; $W$ is an isomorphism by definition, since $V$ is an isometry and $W$ is the corestriction to its image. – Anne Bauval Dec 31 '22 at 15:29
  • @AnneBauval But you didn't showed that extension U is unitary. –  Jan 04 '23 at 09:08
  • I didn't indeed. I leave it to you: either you find it (like me) obvious by construction (it was more obvious in my initial formulation), or you compute $U^$ and check that $UU^=U^*U=I,$ which is an easy but tedious exercise. – Anne Bauval Jan 04 '23 at 11:45