1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook