Results 1 to 4 of 4

Math Help - Hilbert Space

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    78

    Hilbert Space

    Show that an operator T on a Hilbert Space is unitary iff T(\{e_i\}) is a complete orthonormal set whenever \{e_i\} is.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    What is your definition of unitary operator?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    78
    Unitary here means an isomorphism in Hilbert space(linear,one to one and onto)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by problem View Post
    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.
    Last edited by Jose27; October 14th 2009 at 02:00 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Banach space & Hilbert space
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 24th 2011, 01:06 PM
  2. Hilbert Space
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 19th 2009, 09:42 AM
  3. Hilbert Space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 10th 2009, 09:00 AM
  4. Hilbert Space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 25th 2009, 09:18 AM
  5. Inverse of Mapping from Hilbert Space to Hilbert Space exists
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: June 2nd 2009, 08:15 PM

Search Tags


/mathhelpforum @mathhelpforum