= Even rank of a skew-adjoint linear map
{title2=$\operatorname{rank}A\in2\mathbb N_0$}
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 $\ker A=(\operatorname{im}A)^\perp$, 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.
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