A operator is unitary if and only if it is a surjective isometry

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Solution 1

Let $U$ be a unitary, then $UU^*=U^*U=1$.

$U^*U=1$ implies that $U$ is an isometry. Also $UU^*=1 $ implies that $U$ is onto, because ($UU^*H=H$).

Conversely, Suppose $U$ is an onto isometry, so $U^*U=1$. It's just necessary to show that $UU^* = 1$. If $\eta\in H$, then there is $\xi\in H$ such that $U\xi=\eta$ ($U $ is onto). Also $U$ is an isometry, so $ \xi=U^*U\xi = U^*\eta$ .Now clearly $UU^*\eta = U\xi =\eta$ for $\eta \in H$. Therefore $UU^*=1$.

Solution 2

Below I will prove the equivalence in a little more general settings.

The equivalence to be proven:

If $\mathcal{H}_1$ and $\mathcal{H}_2$ are Hilbert spaces over $\mathbb{C}$ with inner products $\langle \cdot,\cdot \rangle_1$ and $\langle \cdot,\cdot \rangle_2$, respectively, and $U:\mathcal{H}_1\rightarrow\mathcal{H}_2$ is a bounded linear operator between them, then $U$ is unitary $\Longleftrightarrow$ $U$ is surjective and an isometry.

To avoid any misunderstanding, below is the definition of a unitary operator:

Given the above settings, a unitary operator from $\mathcal{H}_1$ to $\mathcal{H}_2$ is an invertible linear operator $U$ that preserves inner products: $\langle Ux,Uy\rangle_2=\langle x,y\rangle_1$ for $\forall x,y\in\mathcal{H}_1$.

Note that this definition is equivalent to the normal form that $UU^*=U^*U=I$.

  • Proof of $\Longrightarrow$:

Since the definition of unitary operator requires that it be bijective, so $U$ is surjective. Isometry can be established by equalities $\|Ux\|_2^2=\langle Ux,Ux\rangle_2=\langle x,x \rangle_1=\|x\|_1^2$ for $\forall x\in \mathcal{H}_1$, which are just applications of the definition.

  • Proof of $\Longleftarrow$:

Before we proceed, let's rewrite the polarization identity as follows:

Re$\langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2),$

Im$\langle x,y\rangle=\frac{1}{4}(\|x+iy\|^2-\|x-iy\|^2)$.

Supposing any $x,y\in\mathcal{H}_1$ satisfying $Ux=Uy$, we have $0=\|Ux-Uy\|_2=\|U(x-y)\|_2=\|x-y\|_1$ due to isometry. So $x=y$ by positive definiteness of norm. This proves injectivity of $U$. Together with the surjectivity, we know $U$ is bijective and therefore invertible. Boundedness is clear from isometry.

To show $\langle Ux,Uy\rangle_2=\langle x,y\rangle_1$, we need to show that this equality is true for both the real and imaginary part (note that the inner product is a complex number). But this is straightforward in what follows:

$\begin{eqnarray} \mathop{\rm Re}\langle Ux,Uy\rangle_2&=&\frac{1}{4}(\|Ux+Uy\|_2^2-\|Ux-Uy\|_2^2)\\ &=&\frac{1}{4}(\|U(x+y)\|_2^2-\|U(x-y)\|_2^2)\\ &=&\frac{1}{4}(\|x+y\|_1^2-\|x-y\|_1^2)\\ &=&\mathop{\rm Re}\langle x,y\rangle_1\enspace, \end{eqnarray}$

$\begin{eqnarray} \mathop{\rm Im}\langle Ux,Uy\rangle_2&=&\frac{1}{4}(\|Ux+iUy\|_2^2-\|Ux-iUy\|_2^2)\\ &=&\frac{1}{4}(\|U(x+iy)\|_2^2-\|U(x-iy)\|_2^2)\\ &=&\frac{1}{4}(\|x+iy\|_1^2-\|x-iy\|_1^2)\\ &=&\mathop{\rm Im}\langle x,y\rangle_1\enspace. \end{eqnarray}$

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Updated on September 08, 2020

Comments

  • Slash_
    Slash_ about 3 years

    I'm trying to prove the following result.

    Let U be an operator of a Hilbert space H, then $U$ is an unitary operator $\iff$ $U$ is an isometry and $R_u = H$ ($U$ is onto and isometry)

    I tried to use the fact that $U$ is a linear operator and the fact that $U^*U=I$ defines an isometry but I couldn't proceed.

  • Slash_
    Slash_ almost 9 years
    I have yet to prove that $U$ is an unitary operator $\iff$ $R_u=H$, right?
  • niki
    niki almost 9 years
    @mathbbguy: No, U is unitary if and only if U is an isometry and $R_u=H$
  • Slash_
    Slash_ almost 9 years
    Oh Okay, I thought it was: 1. $U$ is an unitary operator $\iff$ $U$ is an isometry. 2. $U$ is an unitary operator $\iff$ $R_u = H$.
  • niki
    niki almost 9 years
    @mathbbguy: there are several examples of isometry operators that are not unitary, as shifts.
  • Slash_
    Slash_ almost 9 years
    Thanks you so much for your time, @niki. Have a nice day.
  • niki
    niki almost 9 years
    @mathbbguy: You are welcome.
  • user5280911
    user5280911 about 3 years
    "Also U is an isometry, so $\xi=U^*U\xi$" -- how did you get this? This is true only if you assume $U^*U=I$ which is exactly the definition of unitary mapping you proposed to prove. You have to use polarization identity to relate isometry with inner product.