How do I prove that the determinant of an nxn, skew-symmetric matrix is always zero if n is odd?
Basic facts:
(*) det(A) = det(A^T)
(**) det(kA) = k^n det(A) where A is nxn matric and k is scalar
Let n be odd and let A be a skew-symmetric nxn matrix. Then A^T = -A. Then det(A^T) = det(-A). Note that by (**) where k = -1, we see that det(-A) = (-1)^n det(A). Then since n is odd, (-1)^n = -1. Then det(-A) = -det(A). Then by (*) we see that det(A) = det(A^T) = det(-A) = -det(A). Thus, det(A) = -det(A). Then 2*det(A) = 0. Then det(A) = 0.