TITLE: Proof that the rank of a skew-symmetric matrix is at least $2$
QUESTION [20 upvotes]: Is there a succinct proof for the fact that the rank of a non-zero skew-symmetric matrix ($A = -A^T$)  is at least 2? I can think of a proof by contradiction: Assume rank is 1. Then you express all other rows as multiple of the first row. Using skew-symmetric property, this matrix has to be a zero matrix. 
Why does such a matrix have at least 2 non-zero eigenvalues?

REPLY [12 votes]: The following answers the first part of the OP's question, without using the concept of eigenvalues. It works on all fields (including $\mathbb{R}$) with characteristic $\ne2$.*
Every rank-$1$ matrix can be written as $A=uv^\top$ for some nonzero vectors $u$ and $v$ (so that every row of $A$ is a scalar multiple of $v^\top$). If $A$ is skew-symmetric, we have $A=-A^\top=-vu^\top$. Hence every row of $A$ is also a scalar multiple of $u^\top$. It follows that $v=ku$ for some nonzero scalar $k$. But then $vu^\top=-uv^\top$ implies that $kuu^\top=-kuu^\top$ or $2kuu^\top=0$, which is impossible because both $k$ and $u$ are nonzero and the characteristic of the field is not $2$. Therefore, skew-symmetric matrices cannot be rank-1 matrices, and vice versa.
When the underlying field has characteristic 2, the notions of symmetric matrices and skew-symmetric matrices coincide. Hence every nonzero matrix of the form $uu^\top$ with nonzero vector $u$ is a rank-1 skew-symmetric matrix.
Remark. In most modern textbooks, a matrix $A$ in a field of characteristic $2$ is said to be skew-symmetric if $A$ has a zero diagonal and $A^T=-A$. This modern definition is better because the discrepancy between skew-symmetric matrix and alternating bilinear form now vanishes. With this definition, symmetric matrices and skew-symmetric matrices are different notions and a matrix of the form $uu^\top$ cannot be skew-symmetric unless it is zero.