Prove the following for real \(n\times n\) matrices.
If \(A\) is orthogonal and has no eigenvalue \(-1\), then \(I+A\) is nonsingular and \(S=(I-A)(I+A)^{-1}\) is skew-symmetric.
If \(S\) is skew-symmetric, then \(A=(I-S)(I+S)^{-1}\) is orthogonal and has no eigenvalue \(-1\).
The correspondence between \(A\) and \(S\) is one-to-one.
Solution
1. Prove part (a).
Assume \(A\) is orthogonal and has no eigenvalue \(-1\). Then \[ \ker(I+A)=\{0\}, \] so \(I+A\) is nonsingular.
Define \[ S=(I-A)(I+A)^{-1}. \] Since \(A^t=A^{-1}\), \[ \begin{aligned} S^t &=(I+A^t)^{-1}(I-A^t)\\ &=(I+A^{-1})^{-1}(I-A^{-1}). \end{aligned} \] Now \[ I+A^{-1}=A^{-1}(I+A), \] and \[ I-A^{-1}=-A^{-1}(I-A). \] Therefore \[ S^t =-(I+A)^{-1}(I-A). \] The matrices \(I-A\) and \(I+A\) commute, so they also commute with \((I+A)^{-1}\). Hence \[ S^t=-(I-A)(I+A)^{-1}=-S. \] Thus \(S\) is skew-symmetric.
2. For skew-symmetric \(S\), both \(I+S\) and \(I-S\) are nonsingular.
Suppose \[ (I+S)x=0. \] Then \(Sx=-x\). Since \(S^t=-S\) and the matrices are real, \[ \langle Sx,x\rangle =-\langle x,Sx\rangle =-\langle Sx,x\rangle, \] so \[ \langle Sx,x\rangle=0. \] But \(Sx=-x\) would give \[ \langle Sx,x\rangle=-\|x\|^2, \] hence \(x=0\). Thus \(I+S\) is nonsingular. The same argument with \(Sx=x\) shows that \(I-S\) is nonsingular.
3. Prove that \(A=(I-S)(I+S)^{-1}\) is orthogonal and has no eigenvalue \(-1\).
Let \(S^t=-S\) and define \[ A=(I-S)(I+S)^{-1}. \] By step 2 this is well-defined. Taking transposes, \[ \begin{aligned} A^t &=(I+S^t)^{-1}(I-S^t)\\ &=(I-S)^{-1}(I+S). \end{aligned} \] Since \(I-S\) and \(I+S\) commute, \[ A^{-1}=(I+S)(I-S)^{-1}=(I-S)^{-1}(I+S)=A^t. \] Thus \(A\) is orthogonal.
Moreover \[ I+A =\bigl((I+S)+(I-S)\bigr)(I+S)^{-1} =2(I+S)^{-1}, \] which is nonsingular. Therefore \(-1\) is not an eigenvalue of \(A\).
4. The two formulas are mutual inverses.
Starting from \[ S=(I-A)(I+A)^{-1}, \] we obtain \[ I-S=2A(I+A)^{-1}, \] and \[ I+S=2(I+A)^{-1}. \] Hence \[ (I-S)(I+S)^{-1}=A. \]
Conversely, starting from \[ A=(I-S)(I+S)^{-1}, \] one similarly gets \[ I-A=2S(I+S)^{-1}, \qquad I+A=2(I+S)^{-1}, \] and therefore \[ (I-A)(I+A)^{-1}=S. \] Thus the two constructions undo each other. In particular the correspondence between orthogonal matrices without eigenvalue \(-1\) and skew-symmetric matrices is one-to-one.