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Homework Help: Linear Algebra - Normal and Unitary Matrices

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose V is a unitary space [over C] and T: V -> V is a normal transformation that satisfies T-1=-T. Prove that T is unitary transformation.

    2. Relevant equations
    I know that T is unitary if and only if it is normal and the absolute value of its eigenvalues is 1. [*2]

    3. The attempt at a solution
    T-1=-T so T2=-I, now suppose a is an eigenvalue of T so T2v=a2v=-Iv what in turn means [itex]a=\pm 1[/itex].
    So from [*2] we can conclude that T is unitary.

    Have I missed something?
  2. jcsd
  3. Oct 28, 2011 #2
    Did you prove normality? Also, can you be more specific about V? Is it just a vector space? A Banach Space? Hilbert Space? Is it finite or infinite dimension? What definition of unitary are you using?
  4. Oct 28, 2011 #3
    Thanks for posting!

    Normality is given.
    V is a general unitary space. [it means V is a general space over the complex field]
    Forgot to mention: V is finite dimension space.
    Unitary definition:[itex]T \cdot T^*=T^* \cdot T=I[/itex]
  5. Oct 28, 2011 #4
    Ah yes, I see now that it was given. It looks fine to me, though I have to say I've never heard of a "general unitary space." Perhaps this is okay in this instance since [itex] \mathbb C[/itex] is algebraically self-dual, but normally one requires at the very least a normed vector space.
  6. Oct 28, 2011 #5
    So I guess the prove is ok? [This question was formulated by myself so it is totally possible that I didn't define the question itself well]

    What is normed vector space?
  7. Oct 28, 2011 #6
    It looks fine so long as the result about being normal with modulus 1 is true. I think it's true, but haven't run through the proof myself. In any case, I would say you're good to go.

    A normed vector space is a vector space with a norm. :) Not sure I can be more explicit than that.
  8. Oct 28, 2011 #7

    I though every vector space has a norm, so I should not explicitly say it.
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