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Math Help - Proving the existence of a unitary operator...

  1. #1
    Junior Member gusztav's Avatar
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    Proving the existence of a unitary operator...

    Let A be a linear operator on a finite-dimensional unitary space V such that, for every x,y \in V the following implication holds:
    (x \perp y) \Rightarrow (Ax \perp Ay).
    Prove that then there exists a scalar \alpha and a unitary operator U (where U:V \rightarrow V) such that A= \alpha U.



    ******

    I'm aware of the following facts:
    if an operator U is unitary, and U:V \rightarrow V, then for every x, y \in V we have <Ux, Uy>=<x,y>

    Also, on a unitary space V the following is true for every x, y \in V: x \perp y \Leftrightarrow <x,y>=0

    And finally, if U is a unitary operator, then if \{ e_1,...e_n\} is an orthonormal basis for V, then \{ U e_1,...U e_n\} is also an orthonormal basis for V.

    ******

    But so far, I've had no luck in employing these to solve this problem, so would greatly appreciate your help!
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  2. #2
    MHF Contributor

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    Quote Originally Posted by gusztav View Post
    Let A be a linear operator on a finite-dimensional unitary space V such that, for every x,y \in V the following implication holds:
    (x \perp y) \Rightarrow (Ax \perp Ay).
    Prove that then there exists a scalar \alpha and a unitary operator U (where U:V \rightarrow V) such that A= \alpha U.



    ******

    I'm aware of the following facts:
    if an operator U is unitary, and U:V \rightarrow V, then for every x, y \in V we have <Ux, Uy>=<x,y>

    Also, on a unitary space V the following is true for every x, y \in V: x \perp y \Leftrightarrow <x,y>=0

    And finally, if U is a unitary operator, then if \{ e_1,...e_n\} is an orthonormal basis for V, then \{ U e_1,...U e_n\} is also an orthonormal basis for V.

    ******

    But so far, I've had no luck in employing these to solve this problem, so would greatly appreciate your help!
    i googled "unitary vector space" to see what exactly it is and then i found out that it's just an old name for a much more familiar name Hilbert space. anyway, first it's clear that we may assume

    that A \neq 0. let \{e_1, e_2, \cdots , e_n \} be an orthonormal basis for V. i claim that ||Ae_1||=||Ae_2|| = \cdots = ||Ae_n|| : we have <e_i+e_j,e_i-e_j>=0, for all i \neq j. thus, by the problem hypothesis, we

    must also have ||Ae_i||^2-||Ae_j||^2=<A(e_i+e_j),A(e_i-e_j)>=0, which proves the claim. let \alpha=||Ae_1|| and U=\frac{1}{\alpha}A. it's easy to see that <Ux,Uy>=<x,y>, for all x,y \in V. Q.E.D.
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