Let \(S\) be a real skew-symmetric matrix or a complex skew-Hermitian matrix. Define the Cayley transform \[ U=(S+I)(S-I)^{-1}. \] Show that \(U\) is respectively orthogonal or unitary, and solve for \(S\) in terms of \(U\).
Solution
It suffices to treat the skew-Hermitian case; the real skew-symmetric case is identical with transpose in place of conjugate transpose.
1. The matrix \(S-I\) is invertible.
If \((S-I)v=0\), then \(Sv=v\). But every eigenvalue of a skew-Hermitian matrix is purely imaginary, while \(1\) is real and nonzero. Hence \(v=0\).
2. The Cayley transform is unitary.
Because \(S^*=-S\), \[ U^*=\bigl((S-I)^{-1}\bigr)^*(S+I)^* =(-S-I)^{-1}(-S+I). \] Thus \[ U^*=(S+I)^{-1}(S-I). \] All factors are rational functions of \(S\), hence commute. Therefore \[ U^*U =(S+I)^{-1}(S-I)(S+I)(S-I)^{-1}=I. \] So \(U\) is unitary. In the real case the same computation gives \(U^TU=I\).
3. Recover \(S\) from \(U\).
From \[ U(S-I)=S+I \] we obtain \[ (U-I)S=U+I. \] Moreover \[ U-I=2(S-I)^{-1}, \] so \(U-I\) is invertible over \(\RR\) or \(\CC\). Hence \[ S=(U-I)^{-1}(U+I). \] Since \(U\) commutes with polynomials in \(U\), this may equivalently be written \((U+I)(U-I)^{-1}\).