Orthogonal normal form of a finite real skew matrix

Regard $B=iA$ as a complex matrix on ${\mathbb{C}}^{2N}$. Since $A^{\mathsf{T}}=-A$ and $A$ is real, $B^{\ast }=(-i)A^{\mathsf{T}}=iA=B$. By [F1], $B$ has an orthonormal eigenbasis with real eigenvalues. Complex conjugation sends an eigenvector of eigenvalue $ϵ>0$ to an eigenvector of eigenvalue $-ϵ$, because conjugating $iAz=ϵz$ gives $-iA\bar{z}=ϵ\bar{z}$. Thus nonzero eigenvalues occur in opposite pairs.

Fix a unit eigenvector $z=u+iv$ for $ϵ>0$, with real vectors $u,v$. The vectors $z$ and $\bar{z}$ belong to distinct eigenspaces of the Hermitian matrix $B$, so $⟨z,\bar{z}⟩=0$. Expanding this equality gives $\mathrm{\parallel }u{\mathrm{\parallel }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2$ and $u\cdot v=0$; $\mathrm{\parallel }z\mathrm{\parallel }=1$ then gives $\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }v\mathrm{\parallel }=1\mathrm{/}\sqrt{2}$. From $Az=-iϵz$, comparison of real and imaginary parts gives $Au=ϵv$ and $Av=-ϵu$. Therefore $e_1=\sqrt{2}u$ and $e_2=-\sqrt{2}v$ are orthonormal and $A{e}_1=-ϵe_2$, $A{e}_2=ϵe_1$, which is the displayed block in this basis.

Choose an orthonormal basis in each positive eigenspace of $B$; [F1] supplies these finite-dimensional bases. Together with their complex conjugates, the chosen vectors are orthonormal: conjugation preserves orthonormality, and every positive eigenspace is orthogonal to every negative eigenspace of the Hermitian matrix $B$. Hence the real planes from step 2.1 are mutually orthogonal. On the zero eigenspace, $Az=0$, so the real and imaginary parts of each vector lie in the real kernel of $A$. The real and complex kernels have equal dimension because row reduction of the real matrix $A$ has the same pivots over $\mathbb{R}$ and $\mathbb{C}$; this dimension is $2N$ minus twice the total number of positive eigenvectors, hence even. Choose a real basis of $\ker A$ and successively subtract projections onto earlier vectors and normalize each nonzero residual; this gives an orthonormal real basis. Pairing consecutive vectors gives zero blocks. The resulting real orthonormal basis gives the required $O$.

Since $O$ is orthogonal, $(OA{O}^{\mathsf{T}}{)}^{\mathsf{T}}(OA{O}^{\mathsf{T}})=O{A}^{\mathsf{T}}A{O}^{\mathsf{T}}={⨁}_{r=1}^N{ϵ}_r^2{I}_2$. Thus the singular-value multiset from the Statement has two entries ${ϵ}_r$ for each block $r$, including its two zero entries when ${ϵ}_r=0$. This identifies every positive block value once per real two-plane and completes the multiplicity assertion.