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Linear ALgebra: Showing negative definteness

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data

    If M is a real anti-symmetric n x n matrix, M^2 is a real symmetric matrix. Show that M^2 is a non-positive matrix, i.e. x(transposed) M^2 x <= 0, for all vectors x.


    2. Relevant equations

    det(M) = (-1)^n det (M)

    3. The attempt at a solution

    I attempted to use the relevant equation above to find the determinant of M^2, and found that it is >=0. Diagonalising M^2 gives me the matrix of diagonal eigenvalues, which shares the same determinant as M^2. Thus, in the eigenvalue basis, I proved y(trans) D y >=0, which is the opposite of what the question wants.
     
  2. jcsd
  3. Mar 21, 2014 #2

    jbunniii

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    No need whatsoever for determinants or eigenvalues.

    Hint: ##x^T M^2 x = (x^T M) (M x) = ?##
     
  4. Mar 21, 2014 #3
    1. Are you sure you didn't mistype that determinant formula?
    2. anti-symmetry implies M^T = -M
     
  5. Mar 22, 2014 #4
    = ##(M^T x)^T (Mx) = -(Mx)^T(Mx) = -|Mx|^2 <=0 ##

    Thanks.
     
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