Show that rank of skew-symetric is even number $$A = -A^T$$
I assume that $A$ is not singular. 
So $$\det{A} \neq 0$$ Then $$ \det(A) = \det(-A^T) = \det(-I_{n} A^T) = (-1)^n\det(A^T) = (-1)^n\det(A)$$
So I get that $n$ must be even. 
But what about odd $n$? I know it has to be singular matrix. Hints?

Note: This answer is essentially based on this one by Jason DeVito. I have merely added some details.
I assume $A$ is real matrix. Note that it's rank as a real matrix equals its rank when considered as a complex matrix.
So from now on we consider $A$ as a complex matrix.
It is proved here that all the eigenvalues of $A$ are purely imaginary. Also, we know that for a real matrix, complex eigenvalues come in conjugate pairs. (Since the coefficients of the characteristic polynomial are real).
Since skew-symmetric matrices are digonalizable over $\mathbb{C}$, we get there is an even number of non-zero eigenvalues $\pm y_1 i,\pm y_2 i,...,\pm y_k i$ different from zero. Since the rank of a matrix is invariant under similarity, we get that $rank(A)$ equals the rank of it's diagonal form, which is trivially $2k$.
