Results 1 to 8 of 8

Math Help - Proofs, linear transformation is unitary

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Proofs, linear transformation is unitary

    Let T: V \to V be the linear transformation given by T(\vec x)=c \vec x where c\in \mathbb{C}. Demonstrate that T is unitary if and only if |c|=1.
    If V is a unidimensional vector space, demonstrate that the only unitary linear transformations are the one given previously. In particular, if V is a unidimensional vector space, there exist only 2 orthogonal transformations: T(x)=x and T(x)=-x.
    Attempt: Not much. I want to prove \Rightarrow first.
    If T is unitary, then T^{-1}(\vec x)=T^{\dagger}(\vec x). I don't know how to go further, I must show that this implies that |c|=1.
    Hmm... T^{-1} (\vec x)=\frac{\vec x}{c}=x^{\dagger} c^{\dagger}.
    Stuck here.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You know that unitary transformations preserve length, don't you? Can't you use that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Ackbeet View Post
    You know that unitary transformations preserve length, don't you? Can't you use that?
    Well I wasn't aware of that. My freshman course in L.A. wasn't made for physicists so we didn't cover unitary transformations in particular.
    So it means that |\vec x|=|c \vec x|=|c||\vec x|. Assuming \vec x \neq 0, we get |c|=1 as required.
    For \Leftarrow ) I don't really know where to start or better said: what property to use.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    If you can use "a linear transformation is unitary if and only if it preserves lengths (||Ax||= ||x||)" then look at ||Ax||= |c|||x|.

    If not the what is your definition of unitary?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by HallsofIvy View Post
    If you can use "a linear transformation is unitary if and only if it preserves lengths (||Ax||= ||x||)" then look at ||Ax||= |c|||x|.

    If not the what is your definition of unitary?
    I think the definition we're using is "T is unitary if and only if its matrix representative is unitary". Our definition of a unitary matrix is A^{-1}=A^{\dagger}.
    But I'll try to follow your tip anyway. If I have any problem I'll repost.
    Edit: Ok so if I understand well, A would be the matrix representative of T. We have ||Ax||=|c||x|=|x|, therefore T preserves lengths and so is unitary. Is this a valid proof?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Ok, so suppose matrix A is unitary. That is, assume A^{-1}=A^{\dagger}. Let x be a nonzero vector. Then

    \|Ax\|^{2}=\langle Ax|Ax\rangle=\langle x|A^{\dagger}Ax\rangle<br />
=\langle x|A^{-1}Ax\rangle=\langle x|Ix\rangle=\langle x|x\rangle=\|x\|^{2}.

    Therefore, since \|Ax\|\ge 0 and \|x\|>0, we may conclude that \|Ax\|=\|x\| for all nonzero x. If x=0, the result is also true.

    This proof you can use as an intermediate step in the proof you're ultimately interested in.

    Does this help?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Yes it helped Ackbeet. If I'm not wrong, you're reasoning is for the \Rightarrow ) part of the proof, right? What you've showed is equivalent to say \|Ax\|= \|cx\|= |c|  \|x\|=\|x\| \Rightarrow |c|=1, for all x. Which completes the proof for the \Rightarrow ) part. I hope I'm not wrong on this.
    For part \Leftarrow ), is my "proof" of post #5 valid?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Right. I think you've correctly finished the forward direction, if you include the lemma I proved in your proof.

    The reverse direction I'm not so sure of. You have to show that, assuming there exists c\in\mathbb{C} such that |c|=1, such that T\vec{x}=c\vec{x} for all vectors \vec{x}, that T^{-1}=T^{\dagger}.

    So, assume the hypotheses with all the labels the same. I think you might also be able to assume that T^{-1} exists. You must show that T^{\dagger}T=TT^{\dagger}=I.

    Can you compute T^{-1} and T^{\dagger} directly? I think you might be able to do that. Then you simply show that they are the same.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 1st 2011, 10:00 PM
  2. 2 proofs for unitary matrix/transformation prob.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 18th 2011, 02:20 AM
  3. Transformation Proofs Help Please!!
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 6th 2010, 04:42 PM
  4. Linear Alegebra linear equations proofs
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 13th 2010, 09:47 AM
  5. Linear Algebra.Linear Transformation.Help
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 5th 2009, 01:14 PM

Search Tags


/mathhelpforum @mathhelpforum