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The positive-definite symmetric form $b$ is a real inner product. Choose a $b$-orthonormal basis by the Gram-Schmidt process. The stated identity means that the matrix $K$ of $\psi$ satisfies $K^T=-K$. If $d=\dim U$, thenAn invertible $K$ has nonzero determinant, so $d$ is even.

To cover a possibly singular map, first show $\ker \psi =(\mathrm{im}\psi {)}^{\perp }$. If $x\in \ker \psi$, then $b(x,\psi y)=-b(\psi x,y)=0$ for every $y$, giving one inclusion. Conversely orthogonality to the image gives $b(\psi x,y)=0$ for all $y$, so nondegeneracy of $b$ gives $\psi x=0$. Positivity then implies $\mathrm{im}\psi \cap \ker \psi =\{0\}$.

The image is invariant under $\psi$, and its restricted map $\psi {|}_{\mathrm{im}\psi }$ is injective by that trivial intersection. In finite dimension it is consequently invertible. The restricted inner product is positive definite and the same skew-adjoint identity holds, so the first argument applied to this restriction makes $\dim \mathrm{im}\psi$ even. Thusincluding rank zero. This is the even rank of a skew-adjoint linear map.