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A²=Id => A diagonalizable?

  1. Dec 23, 2011 #1


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    Hello matrix gurus,

    Is is true that if A is real with A²=I (eigenvalues ±1), it is diagonalizable over R?

    What if I add that A is in O(m,m), where O(m,m) is the split indefinite orthogonal group of 2m x 2m matrices M such that [itex]M^TI_{m,m}M=I_{m,m}[/itex], where [itex]I_{m,m}[/itex] is the block diagonal matrix diag(I_m,-I_m) for I_m the m x m identity matrix?

    Last edited: Dec 23, 2011
  2. jcsd
  3. Dec 23, 2011 #2
    Yes, an involution is always diagonalizable over the reals.

    We use the following result:

    See http://en.wikipedia.org/wiki/Diagonalizable_matrix A proof can be found in Hoffman & Kunze theorem 6 page 204.

    The minimal polynomial of an involution is [itex]x^2-1=(x+1)(x-1)[/itex] and satisfies our theorem.
    Last edited: Dec 23, 2011
  4. Dec 24, 2011 #3


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    A.v - v and A.v + v are both eigen vectors of A. They can not both be zero.
    Use this inductively to find a basis of the vector space made up of eigen vectors of A.

    If A.v is not a multiple of v then both A.v - v and A.v + v are non-zero eigen vectors. Choose any other vector that is not in the span of v and A.v. and continue.

    One certainly does not need all of the reals to diagonalize A. The rationals will work. I think all you really need to be able do is divide by 2.

    Over the integers A can not always be diagonalized. The map (1,0) -> (1,-1) (0,1) -> (0,-1) is an example. The problem of inequivalent integer representations of groups is daunting.
    Last edited: Dec 24, 2011
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