Linear Algebra 25: Exponential of a skew-symmetric matrix

[peterbrown]

Thanks for this. I clearly made an error. Change one of the -1's in my example to a 1 and then for any skew-symmetric A, det(e^A) = exp(trace(A)) = 1, but det Q is -1.

[zivssps]

You have a mistake. Consider the matrix A=(0, pi; -pi 0) which is a skew-symmetric matrix.
Take the exp of A to get Q=-I.
Your assumption was that if A is over R, the eigenvalues over R.
If so, e^lambda > 0, but it's not always the case and e^lambda can be less than 0 (in this example, e^(i*pi) = -1 < 0).
Now, the question is very simple:
If A is a skew-symmetric, the Q is orthogonal. By simple logic we can switch the direction of the sentence to get: If Q isn't orthogonal, then A isn't skew-symmetric.
So, to answer the question, just pick an an-orthogonal matrix for Q, and therefore A can not be skew-symmetric.
Anther way to get an answer is to consider the determinant of Q:

If |Q|=0, to get an Identity we must let tr(A) go to -inf witch can't happened, at least in finite matrixs.
Furthermore, note that if |Q|=0, automatically Q isn't orthogonal.
So, either way, all we need is an an-orthogonal matrix Q and there is no skew-symmetric A s.t Q=e^A.
Personally, I would take Q to be the zero matrix. Than |Q|=0 (or Q isn't orthogonal) and there is no A.