Hilbert Space

**problem** · October 13th 2009, 07:32 PM

Show that an operator $T$ on a Hilbert Space is unitary iff $T(\{e_i\})$ is a complete orthonormal set whenever $\{e_i\}$ is.

**Jose27** · October 13th 2009, 07:37 PM

What is your definition of unitary operator?

**problem** · October 13th 2009, 09:17 PM

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

**Jose27** · October 14th 2009, 01:39 PM

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 $\mathbb{R} ^2$ and $T(e_1)=e_1$ and $T(e_2)=e_1+e_2)$ .

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

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

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

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

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

Thread: Hilbert Space

LinkBack

Thread Tools

Display

Hilbert Space

Similar Math Help Forum Discussions

Banach space & Hilbert space

Hilbert Space

Hilbert Space

Hilbert Space

Inverse of Mapping from Hilbert Space to Hilbert Space exists

Search Tags