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?
 A: 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 $2k$.
A: Here's another proof, avoiding use of eigenvalues/eigenvectors:
Let $A$ be a skew-symmetric matrix, and consider the alternating bilinear form $$B(v,w) := \langle Av, w \rangle = -\langle v, Aw \rangle$$ where $\langle \cdot, \cdot \rangle$ is the standard inner product on $\mathbb{R}^n$. Let $W := \text{Im}(A)$ and let $P: \mathbb{R}^n \to W$ denote the orthogonal projection. 
For any nonzero $w = Au \in W$, note that $$B(Pu, w) = B(u,w) = \langle w, w \rangle > 0$$ where we have used that $B(u-Pu, w) = 0$ since $u-Pu \in W^{\perp}$. Thus, the restriction of $B$ to $W$ is nondegenerate. It follows that $\text{rank}(A) = \dim(W)$ is even, since the matrix of $B\vert_{W}$ is skew-symmetric with nonzero determinant (and any odd-dimensional skew-symmetric matrix has determinant zero).
