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?