Odd-dimensional complex skew-symmetric matrix has eigenvalue $0$ There is the standard proof using $$\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)$$ 
I would like a proof that avoids this. Specifically, there is the proof that for $A$ a $\bf{real} $ matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) $\lambda \|x\|^2 =\langle  Ax, x \rangle= \langle  x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}$, so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero.  This argument doesn't work for a general complex skew-symmetric matrix.  Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case?  Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)?
Thanks!
 A: Here is a "geometric" argument for the real case. The skew-symmetric condition is equivalent to $\langle Av, v\rangle =0$ for all vectors $v$. Geometrically, $Av$ is orthogonal to $v$. If $Av$ is never zero for $v\ne 0$, then we get a nonvanishing vector field on the unit sphere, contradicting the Hairy Ball theorem. 
A: The proof you wrote down works in any field of characteristic not equal to $2$. Here is a similar such proof which avoids determinants due to Ian Frenkel. 
Pass to the algebraic closure. Find $P$ such that $A = PDP^{-1}$ with $D$ upper triangular (for example using Jordan normal form, but you don't need to work this hard; it follows from the existence of eigenvectors). Then the diagonal entries of $D$ are the eigenvalues of $A$. Moreover,
$$A^T = (P^T)^{-1} D^T P^T$$
where $D^T$ is lower triangular, so the diagonal entries of $D$ are also the eigenvalues of $A^T$. Since $A^T = -A$ it follows that the multiset of eigenvalues of $A$ is closed under negation. Since there are an odd number of them there must be an eigenvalue which is its own negative, so it must be zero. 
