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The Cayley transform is the rational matrix function $C(X)=(I+X)(I-X{)}^{-1}$, defined when $I-X$ is invertible. For the orthogonal group, unitary group, and analogous groups preserving a nondegenerate bilinear form or Hermitian form, it gives a local parametrization near the identity by their Lie algebras. If $X^{†}J+JX=0$, expanding both sides givesso $C(X{)}^{†}JC(X)=J$. The inverse chart is $X=(Q-I)(Q+I{)}^{-1}$ wherever $Q+I$ is invertible. The Cayley transform of a real skew-symmetric matrix is the positive-definite real orthogonal example.

The formula is not a parametrization of an arbitrary Matrix Lie group. For example, $X=\mathrm{diag}(1/4,1/4,-1/2)$ belongs to the special linear Lie algebra, but $\det C(X)=25/27$, so its transform is not in the special linear group. In indefinite form-preserving groups, the Cayley chart's full domain can also meet components outside the identity component of a Lie group; its local behavior near zero alone does not imply a global parametrization.

Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are polynomials in $B$. Also $P$ has a matrix inverse since it is the transpose of the invertible $Q$. Thus the Cayley transform of a real skew-symmetric matrix, with its parameter replaced by $-B$, satisfiesHence $C$ is an orthogonal matrix. Its inverse is $Q{P}^{-1}$; this lets us distinguish the symmetric and skew-symmetric Cayley transforms below.

Commuting inverse. Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication givesMultiply on both sides by $(I-M{)}^{-1}$ to obtainMore generally, a matrix commuting with an invertible matrix also commutes with its inverse.

Orthogonality. For a real skew-symmetric matrix $A$, $x^{T}Ax=0$. If $(I-A)x=0$ for a real vector $x$, multiply by $x^T$ to obtain $\Vert x{\Vert }^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the Cayley transform of a real skew-symmetric matrix,The factors commute by the first result, soTherefore $\boxed{N\text{ is an orthogonal matrix}.}$

Eigenvector mapping. If $Av=av$, then $(I-A{)}^{-1}v=(1-a{)}^{-1}v$. ConsequentlyThe denominator cannot vanish. In fact the eigenvalues of a real skew-symmetric matrix are purely imaginary: $v^{†}Av$ is the negative of its complex conjugate, so $a=-\overline{a}$. For $a=i\beta$ the corresponding eigenvalue of $N$ is $(1+i\beta )/(1-i\beta )=e^{2i\arctan \beta }$, consistent with orthogonality.

Spectrum in three dimensions. Let $Q$ be any real three-by-three orthogonal matrix. For a complex eigenvector $v$, $\Vert Qv\Vert =\Vert v\Vert$ implies $|\lambda |=1$. A real cubic characteristic polynomial has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the eigenvalues can be writtenTheir product is $\epsilon =\det Q$.

Determinant and fixed axis for the Cayley transform. Since $A^T=-A$,so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an eigenvector with $Nx=x$. Set $z=(I-A{)}^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. HenceThis is the axis fixed by the three-dimensional rotation. Equivalently, a real odd-dimensional skew-symmetric matrix has zero determinant, because $\det A=\det A^T=\det (-A)=-\det A$.

First, $I-A$ is invertible. If $(I-A)x=0$ for a real vector, then $Ax=x$, whereas the skew-symmetric matrix identity gives $x^{T}Ax=0$; hence $x^{T}x=0$ and $x=0$. Thus its real kernel is trivial, which suffices for invertibility. The same argument applies to $I+A$.

Using $A^T=-A$ and the rule for the transpose of a product,The factors $I+A$ and $I-A$ commute, as do the required inverses. ConsequentlyThus $\boxed{N\text{ is orthogonal}}$. This is the Cayley transform of a real skew-symmetric matrix.

For a real skew-symmetric matrix $B$, its Cayley transform of a real skew-symmetric matrix with the opposite sign convention has inverse $C^{-1}=(I+B)(I-B{)}^{-1}$. Since $C$ is an orthogonal matrix, symmetry means $C=C^{-1}$, or $(I-B{)}^2=(I+B{)}^2$. This forces $B=0$. Skew-symmetry means $C=-C^{-1}$, or $(I-B{)}^2=-(I+B{)}^2$, which is equivalent to $B^2=-I$. In that case $C=-B$. Such examples exist in even dimension by taking two-dimensional right-angle rotation matrices in diagonal blocks; in odd real dimension the determinant equation $(\det B{)}^2=(-1{)}^n$ rules them out.