Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Orthogonal Transformations with Eigenvalue 1

  1. Nov 23, 2008 #1
    The problem statement, all variables and given/known data
    Prove that an orthogonal transformation T in Rm has 1 as an eigenvalue if the determinant of T equals 1 and m is odd. What can you say if m is even?

    The attempt at a solution
    I know that I can write Rm as the direct sum of irreducible invariant subspaces W1, W2, ..., Ws which are mutually orthogonal and have dimension 1 or 2. They can't all have dimension 2 for otherwise m would be even. Thus, there is at least one with dimension 1, say W1. For any nonzero w in W1, T(w) = w or T(w) = -w. In the former case, we're done. The latter case is giving me problems though. If T has -1 as an eigenvalue, I imagine that the determinant of T will not be 1 anymore (since I haven't used this fact), but I don't understand how. Any tips?

    Oh, and if m is even I imagine that T may not necessarily have eigenvalues since it's minimal polynomial may be multiples of powers of irreducible quadratics.
     
  2. jcsd
  3. Nov 23, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi e(ho0n3! :smile:

    Hint: you haven't yet use the fact that the determinant is the product of the individual determinants. :wink:
     
  4. Nov 23, 2008 #3
    And where would I use that fact exactly? I'm not calculating the determinant of a product as far as I know.
     
  5. Nov 23, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Perhaps I'm misunderstanding the problem …

    but what can you say about the determinant of one irreducible subspace? :smile:
     
  6. Nov 23, 2008 #5
    Nothing. Determinants are defined for matrices and linear transformations, not subspaces. Hmm...but now you've made me think of the following:

    Since we're told that the determinant of T is 1, does that mean the determinant of the restriction of T to each irreducible subspace is 1 as well? I know that for an irreducible subspace of dimension 2, T has determinant equal to 1. For an irreducible subspace of dimension 1, then it could be 1 or -1 depending on the eigenvalue of T on this subspace. Right?
     
  7. Nov 23, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Exactly! :smile:

    And there's an odd number of dimension-1 subspaces, so … ? :wink:
     
  8. Nov 23, 2008 #7
    So, if the determinant of T on each of the dimension-1 subspaces is -1, then the determinant of T on the direct sum of these one-dimensional subspaces is -1, right? (And also analogously for the two-dimensional subspaces.) Let's write Rm = U + V where U is the direct sum of the one-dimensional subspaces and V is the direct sum of the two-dimensional subspaces. Since the determinant of T on U is -1 and the determinant of T on V is 1, does that mean the determinant of T on U + V is -1?
     
  9. Nov 23, 2008 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes …

    so how many of the individual 1-dimensional subspaces can have negative determinant? :smile:
     
  10. Nov 23, 2008 #9
    An even number of them. So we have at least one 1-dimensional subspace where T has determinant 1 and so T has eigenvalue 1. Right?
     
  11. Nov 23, 2008 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    :biggrin: Woohoo! :biggrin:
     
  12. Nov 23, 2008 #11
    Thanks a lot for your help.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook