An \(n \times n\) matrix \(A\) over a field \(F\) is skew-symmetric if \(A^t = -A\). If \(A\) is a skew-symmetric \(n \times n\) matrix with complex entries and \(n\) is odd, prove that \(\det A = 0\).
Solution
Since \(A^t=-A\), \[ \det A=\det(A^t)=\det(-A)=(-1)^n\det A=-\det A \] because \(n\) is odd. Over \(\mathbb C\) this implies \(2\det A=0\), hence \(\det A=0\).