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Let $\phi, \psi \in \mathcal{H}$ be some element from a hilbert space $\mathcal{H}$ and $U$ a linear operator $U: \mathcal{H} \rightarrow \mathcal{H}$. Does

$$ \forall \phi: \| U \phi \|^2 = \| \phi \|^2 = (\phi, \phi) $$

imply that $U$ is unitary? I suspect not, and I think I have to require the more general relation:

$$ \forall \phi, \psi: (U \phi , U \psi) = (\phi, \psi)$$

However, I cant see why this is more general then the condition above. In other words, is there a $U$ that is not unitary and fulfills the first equation but not the second?

Michael Hardy
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cschwan
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1 Answers1

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Both above conditions are equivalent and define an isometry. A unitary operator satisfies $$ U^*U = UU^* = I,$$ while the isometry satisfies only $V^*V=I$. So for a unitary operator apart from the condition which you wrote we also have it for its adjoint, that is, $$ \left<U^*x, U^*y\right> = \left<x, y\right>.$$

Example of a map which is an isometry, but not unitary:

Let $V \colon \ell^2 \rightarrow \ell^2$ be given by $V(x_n)=(y_n)$, where $y_1=0$ and $y_{n+1}=x_n$ for every $n \in \mathbb{N}$. Then $V$ is an isometry, that is $$ \left<Vx, Vy\right>=\left<x,y\right>,$$ but it fails to be a unitary operator (it is not onto).

Proof that both conditions which you wrote are equivalent:

One direction is obvious.

Let us assume that the following holds $$ \|Vx\| = \|x\|$$ we would like to show that for any $x$, $y \in \mathcal{H}$ we have also $$ \left<Vx, Vy\right> = \left<x, y\right>.$$

Let $x$, $y \in \mathcal{H}$ and let $r \in \mathbb{C}$, then $$ LHS=\| x+ry\|^2 = \|V(x+ry)\|^2 = \| Vx + rVy\|^2=RHS.$$ Now note that $$LHS = \|x\|^2 + 2 \mathrm{Re} \overline{r}\left<x,y\right> + \|y\|^2$$ and $$RHS = \|Vx\|^2 + 2 \mathrm{Re} \overline{r}\left<Vx,Vy\right> + \|Vy\|^2= \|x\|^2 + 2 \mathrm{Re} \overline{r}\left<Vx,Vy\right> + \|y\|^2. $$ Therefore we obtain that $$\mathrm{Re} \overline{r}\left<Vx,Vy\right>=\mathrm{Re} \overline{r}\left<x,y\right>.$$ We end the proof by first taking $r=1$ and then taking $r=i$, note that if your Hilbert space was over $\mathbb{R}$, then you would just take $r=1$.

m_gnacik
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  • Thanks for your proof and particular nice example. Is $V^* (x_n)=(y_n), y_n=x_{n+1}$? Then I also see why $V V^=1$, but $V^ V \neq 1$! – cschwan Nov 05 '14 at 18:32
  • Yes, this is correct. – m_gnacik Nov 05 '14 at 18:38
  • Hello sir, may I ask you a question: I'm reading Friedberg linear algebra 4th ed. page 379, the definition of unitary operator is divided into two cases: finite/infinite dimensional, and in finite case the def is $||T(x)||=||x||$, but this is not the real def of unitary right? it's just that $||T(x)||=||x|| \iff T^T=TT^=I$ in finite case and from your answer the def of unitary should be $T^T=TT^=I$? Appreciate your reading. – linear_combinatori_probabi Sep 30 '18 at 06:41
  • In infinite-dimensional spaces, unitary operators are bijective with the inverse being equal to their adjoint. For a finite-dimensional vector space $X$ and every linear operator $V \colon X \rightarrow X$, injectivity (of $V$) is equivalent to its surjectivity. So (in finite-dimension) the condition $|Vx| = |x|$, which yields that $V$ is injective, implies also its surjectivity. This would not be that case if $X$ is an infinite dimensional Hilbert space (look at the counterexample in my answer above). I hope that it is clear, now. – m_gnacik Sep 30 '18 at 16:25