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Diagonalizing a 3x3 matrix

  1. Nov 28, 2007 #1
    [SOLVED] Diagonalizing a 3x3 matrix

    1. The problem statement, all variables and given/known data

    I want to show that a real 3x3 matrix, A, whose square is the identity is diagonalizable by a real matrix P and that A has (real) eigenvalues of modulus 1.

    2. Relevant equations


    3. The attempt at a solution

    Since any matrix is diagonalizable over the complex numbers, I deduced that since there exists a complex matrix P such that PAP^{-1} = diag{x,y,z} (so x,y,z the eigenvalues of A), then diag{x^2,y^2,z^2} = (PAP^{-1})^2 = Id, hence the eigenvalues are square roots of 1 therefore must be real of modulus 1 as required.

    I'm not totally sure my reasoning is sound. Even if it is, there is still the problem that P may be complex.
  2. jcsd
  3. Nov 28, 2007 #2


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    How do you figure that every matrix is diagonalizable over the complex numbers?? That's not true. Try diagonalizing [1 1,0 1]. On the other hand A^2=I can be factored into (A+I)(A-I)=0. Try thinking about the kernel and range of (A+I) and (A-I).
  4. Nov 30, 2007 #3
    OK at last I think I see how to do this. I consider A+I and A-I as endomorphisms of a 3-dimensional vector space. Then use the rank-nullity theorem to deduce that

    (i) dim(K+) + dim(R+) = 3


    (ii) dim(K-) + dim(R-) = 3

    (I hope the notation is obvious.)

    Then I note that the kernels (resp. ranges) have intersection {0}, hence giving

    (iii) dim(K+) + dim(K-) \le 3


    (iv) dim(R+) + dim(R-) \le 3

    Then (i) - (iii) combined with (iv) - (ii) gives say dim(R+) = dim(K-) whence we deduce that

    dim(K+) + dim(K-) = 3

    Finally note that these are the dimensions of the eigenspaces for 1 and -1 and so A is diagonalizable and once we know that the condition on the eigenvalues follows.

    I'm not 100% sure that the matrix that we would use to actually DO the diagonalizing is real though.
  5. Nov 30, 2007 #4


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    That's basically it. Once you know K has three independent eigenvectors with eigenvalues +/-1 you are done. You know P can be taken to be real because your eigenvectors are real. It's just the change of basis from the standard basis to your eigenvectors.
  6. Dec 7, 2007 #5
    Thank you so much!
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