Even rank of a skew-adjoint linear map ($\mathrm{rank}A\in 2{\mathbb{N}}_0$)

On a finite-dimensional real inner product space, a skew-adjoint operator has even rank. In an orthonormal basis its matrix is skew-symmetric, so an invertible such matrix must have even dimension by $\det A=\det (-A)=(-1{)}^d\det A$. Moreover $\ker A=(\mathrm{im}A{)}^{\perp }$, so the image and kernel intersect trivially. The restriction of $A$ to its image is an invertible skew-adjoint map, and its dimension, the rank, is therefore even.