# Hilbert Space

• Oct 13th 2009, 07:32 PM
problem
Hilbert Space
Show that an operator $\displaystyle T$ on a Hilbert Space is unitary iff $\displaystyle T(\{e_i\})$ is a complete orthonormal set whenever $\displaystyle \{e_i\}$ is.
• Oct 13th 2009, 07:37 PM
Jose27
What is your definition of unitary operator?
• Oct 13th 2009, 09:17 PM
problem
Unitary here means an isomorphism in Hilbert space(linear,one to one and onto)
• Oct 14th 2009, 01:39 PM
Jose27
Quote:

Originally Posted by problem
Unitary here means an isomorphism in Hilbert space(linear,one to one and onto)

I'm going to assume that it's also an isometry (otherwise take $\displaystyle \mathbb{R} ^2$ and $\displaystyle T(e_1)=e_1$ and $\displaystyle T(e_2)=e_1+e_2)$.

$\displaystyle \Rightarrow$ ) Let $\displaystyle T:H_1 \rightarrow H_2$ be an isometric isomorphism then (by the polarization identity) $\displaystyle T$ preserves the interior product, and so if $\displaystyle (e_a)_{a \in A}$ is an orthonormal basis for $\displaystyle H_1$ then $\displaystyle (T(e_a))_{a\in A}$ is also an orthonormal set in $\displaystyle H_2$. Take $\displaystyle y=T(x)$ since $\displaystyle (e_a)$ is a basis we have $\displaystyle (x_n)_{n \in \mathbb{N} }$$\displaystyle \subset$$\displaystyle span((e_a)_{a \in A})$ such that $\displaystyle x_n \rightarrow x$ then $\displaystyle (T(x_n))_{n\in \mathbb{N} } \subset span((T(e_a))_{a\in A})$ and $\displaystyle T(x_n) \rightarrow T(x)=y$ (Since T, being an isometry, belongs to $\displaystyle \mathcal{B} (H_1,H_2)$). Since $\displaystyle T$ is onto, we have that $\displaystyle (T(e_a))_{a\in A}$ is a basis for $\displaystyle H_2$.

$\displaystyle \Leftarrow$ ) Let $\displaystyle (e_a)_{a\in A}$ be an orthonormal basis in $\displaystyle H_1$ and $\displaystyle (T(e_a)){a\in A}$ one for $\displaystyle H_2$. Let $\displaystyle L: H_1 \rightarrow H_2$ be such that $\displaystyle L(e_a)=T(e_a)=b_a$ for all $\displaystyle a \in A$ and if $\displaystyle x=\sum_{a\in A} \ <x,e_a>e_a \in H_1$ $\displaystyle L(x):= \sum_{a\in A} \ <x,e_a>b_a$. It follows from Bessel's inequality that $\displaystyle L$ is well defined and it's clearly linear and one-one. If $\displaystyle y= \sum_{a\in A} \ <y,b_a>b_a \in H_2$ we define $\displaystyle x= \sum_{a\in A} \ <y,b_a>e_a \in H_1$ (again by Bessel) and by definition $\displaystyle L(x)=y$ and so $\displaystyle L$ is onto.

If $\displaystyle y=L(x)$ then $\displaystyle \sum_{a\in A} \ <y,b_a>b_a = \sum_{a\in A} \ <x,e_a>b_a$ then $\displaystyle <y,b_a>=<x,e_a>$ for all $\displaystyle a\in A$ and we have (Note that this previous argument proves injectivity):

$\displaystyle \Vert L(x) \Vert ^2 = \Vert y \Vert ^2 = \sum_{a\in A} \ \vert<y,b_a> \vert ^2 = \sum_{a\in A} \ \vert <x,e_a> \vert ^2 =\Vert x \Vert ^2$ and so $\displaystyle L$ is an isometry.

Since $\displaystyle L$ and $\displaystyle T$ coincide on a l.i set, they are equal on the whole linear span and (assuming $\displaystyle T$$\displaystyle \in \mathcal{B} (H_1,H_2)$ which I think is a necessary hypothesis) hence they agree on the closure of the linear span which is $\displaystyle H_2$. So $\displaystyle L \equiv T$ and this finishes the proof.