Let $U$ be a bounded operator on a Hilbert space. Show that the following are equivalent:

I. $U$ is surjective and $\|Uv\|=\|v\|$ for all $v\in H$;

II. $U$ is surjective and $\langle Uv,Uw\rangle=\langle v,w\rangle$ for all $v,w\in H$;

III. $UU^*=U^*U=I$, where $I$ is the identity operator.

An operator satisfying the conditions above is called a unitary operator Examples include translation operators on $L^2(X)$ that come from measure-preserving bijections of $X$.

Let $H$ be a Hilbert space with a fixed orthonormal basis $\{e_n\}_{n\in \mathbb{N}}$, and let $T$ be a compact self-adjoint operator on $H$. Show that there exists a unitary $U$ such that the matrix of $UTU^*$ is diagonal with real entries.

Hint: define $U$ by stipulating that it take a basis of eigenvectors for $T$ (which exists by the spectral theorem) to the given basis.


The Arveson form of the Polarization Identity for sesquilinear forms is $$ b(x,y) = \frac{1}{4}\sum_{n=0}^{3}b(x+i^ny,x+i^n y). $$ Therefore, if $b$ and $c$ are sesquilinear forms, then $b(x,x)=c(x,x)$ for all $x$ iff $b(x,y)=c(x,y)$ for all $x,y$. Examples of sesquilinear forms include $\langle x,y\rangle$, $\langle Ux,Uy\rangle$ and $\langle U^{\star}Ux,y\rangle$. Therefore $\langle x,y\rangle=\langle Ux,Uy\rangle$ for all $x,y$ iff $\|x\|=\|Ux\|$ for all $x$. And $I=U^{\star}U$ iff $$ \langle x,x\rangle = \langle U^{\star}Ux,x\rangle,\;\;\; x\in X. $$ Equivalently, $\|x\|=\|Ux\|$ for all $x\in X$. If $U^{\star}U=I$, then applying $U$ to the left of both sides gives $UU^{\star}Ux=Ux$ for all $x$, or $UU^{\star}y=y$ for all $y\in\mathcal{R}(U)$. Hence, $UU^{\star}=I$ if $U^{\star}U=I$ and if $U$ is surjective.

For the last part, let $\{e_n \}$ be an orthonormal basis of $H$, and let $\{ f_n\}$ be an orthonormal basis of eigenvectors of $T$. Define $$ U\sum_{n}\alpha_n e_n = \sum_{n}\alpha_n f_n $$ By Parseval's equality, $U$ is an isometry, because $$ \|\sum_n \alpha_n e_n\|^2=\sum_n|\alpha_n|^2 = \|\sum_n \alpha_n f_n\|^2 = \|U\sum_n\alpha_n e_n\|^2. $$ And $U$ is surjective because $\{ f_n\}$ is a basis. Therefore, $$ U^{\star}\sum_n\alpha_n f_n =U^{-1}\sum_n\alpha_n f_n = \sum_n \alpha_n e_n \\ UTU^{\star}\sum_n \alpha_n e_n = UT\sum_n\alpha_n f_n = U\sum_n\lambda_n\alpha_n f_n = \sum_n\lambda_n\alpha_n e_n $$


Better reduce your theorem..

Theorem (Isometry)
Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.
Let R be a bounded operator from $\mathcal{H}$ into $\mathcal{K}$. TFAE:

  1. $R$ has a left inverse by its adjoint, i.e. $R^*R=1$
  2. $R$ is an isometry, i.e. $\|R\varphi\|=\|\varphi\|$ for all $\varphi\in\mathcal{H}$.
  3. $R$ preserves scalar products, i.e. $\langle R\varphi,R\psi\rangle=\langle\varphi,\psi\rangle$ for all $\varphi,\psi\in\mathcal{H}$.

Proof (Polarization)

Proving by circular chain:

(1=>2): Scalar product induces norm: $$\|R\varphi\|^2=\langle R\varphi,R\varphi\rangle=\langle R^*R\varphi,\varphi\rangle=\langle\varphi,\varphi\rangle=\|\varphi\|^2$$ (2=>3): Norm generates scalar product: $$\langle R\varphi,R\psi\rangle=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|(R\varphi)+i^\alpha (R\psi)\|^2=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|(\varphi)+i^\alpha (\psi)\|^2=\langle\varphi,\psi\rangle$$

(3=>1): Scalar product defines adjoint: $$\langle R^*R\varphi,\chi\rangle=\langle R\varphi,R\psi\rangle=\langle\varphi,\psi\rangle=\langle 1\varphi,\psi\rangle$$ (3=>1): Scalar product is non-degenerate: $$\langle R^*R\varphi,\psi\rangle\equiv\langle 1\varphi,\psi\rangle\implies R^*R=1$$

Concluding the first lemma.

Lemma (Inverses)
Given plain spaces $X$ and $Y$.
Let $F$ be a function from $X$ into $Y$. TFAE:

  • $F$ is injective resp. surjective.
  • $F$ has a left resp. right inverse, i.e. $LF=1_X$ resp. $FR=1_Y$.

Remark (Uniqueness)
Left resp. right inverses are not necessarily unique!

Lemma (Left vs. Right)
Given plain spaces $X$ and $Y$.
Let $F$ be a function from $X$ into $Y$. Then: $$LF=1_X\quad FR=1_Y\implies L=R$$ That is left and right inverse become unique and agree.

Proof (Identity)

Identity function is a unit: $$L=L1_Y=L(FR)=(LF)R=1_XR=R$$

Concluding the third lemma.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.