# Linear ALgebra: Showing negative definteness

1. Mar 21, 2014

### N00813

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. Mar 21, 2014

### jbunniii

No need whatsoever for determinants or eigenvalues.

Hint: $x^T M^2 x = (x^T M) (M x) = ?$

3. Mar 21, 2014

### Mugged

1. Are you sure you didn't mistype that determinant formula?
2. anti-symmetry implies M^T = -M

4. Mar 22, 2014

### N00813

= $(M^T x)^T (Mx) = -(Mx)^T(Mx) = -|Mx|^2 <=0$

Thanks.