Linear ALgebra: Showing negative definteness

[N00813]

Homework Statement

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.

Homework Equations

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

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.

 

[jbunniii]

No need whatsoever for determinants or eigenvalues.

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

 

[Mugged]

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

 

[N00813]

No need whatsoever for determinants or eigenvalues.

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

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

Thanks.