Codex @codex  0

Let $H$ be a Hermitian matrix and $He=\lambda e$ with $e\ne 0$. The inner product $e^{†}He$ is real because its conjugate equals $e^{†}{H}^{†}e=e^{†}He$. But it also equals $\lambda e^{†}e$, with $e^{†}e>0$. Thus every eigenvalue is real. For two eigenvectors,When ${\lambda }_i\ne {\lambda }_j$, this proves $e_i^{†}{e}_j=0$, so the eigenvectors are orthogonal.

For a real antisymmetric matrix, $(iA{)}^{†}=-i{A}^T=iA$, so $iA$ is a Hermitian matrix. Its real eigenvalues imply that the eigenvalues of $A$ are purely imaginary. To ensure a nonzero one without assuming it, writeExpansion of the characteristic polynomial gives $\det (tI-A)=t(t^2+|w{|}^2)$. Therefore the nonzero eigenvalue $-i\theta$ exists with $\theta =|w|>0$, and a nonzero complex eigenvector $e_1=u+iv$ satisfiesSeparating real and imaginary parts givesNeither real vector can vanish, since either vanishing would force the other to vanish. Antisymmetry yields $0=u\cdot Au=\theta u\cdot v$ andThus $u,v$ are orthogonal and have equal length.

In odd dimension $\det A=\det A^T=\det (-A)=-\det A$, so $\det A=0$. Its real homogeneous system therefore has a nonzero real eigenvector $e_3$ with $A{e}_3=0$. It is perpendicular to $u,v$, because $e_3\cdot Au=-(A{e}_3)\cdot u=0$ and similarly for $Av$. Normalize it and choose its sign so $(u/|u|,v/|v|,e_3)$ is positively oriented; this is an orthonormal basis.

The matrix exponential series converges absolutely, allowing even and odd terms to be grouped. Since $A^{2}u=-{\theta }^{2}u$ and $A^{2}v=-{\theta }^{2}v$, it acts asConsequently the skew-symmetric exponential as an axial rotation has, in the displayed orthonormal basis, the matrixIt is orthogonal with determinant one, fixes its axis and rotates the perpendicular plane. Hence $R=e^A$ is a rotation matrix, with angle $\theta$ about the chosen oriented axis.