Proof that the rank of a skew-symmetric matrix is at least $2$

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?

Solution 1:

For a skew symmetric (real) matrix, the eigenvalues are all purely imaginary. This is because if $Av = \lambda v$, then we have $\lambda \langle v,v\rangle = \langle \lambda v, v\rangle = \langle Av,v\rangle = \langle v, -Av \rangle = \langle v, -\lambda v\rangle = -\overline{\lambda} \langle v,v\rangle$, so we conclude that $\lambda = -\overline{\lambda}$, i.e., that $\lambda$ is purely imaginary. Here, we're using an Hermitian inner product.

For a real matrix, complex eigenvalues come in conjugate pairs, so the rank must be even.

Solution 2:

There exists an invertible matrix $P$ such that $^t P A P$ is diagonal with blocks equal to $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ or $0$ (it is a simple exercise in bilinear forms), so that the rank of $A$ is necessarily even.