If $(I-A)(I+A)^{-1}$ is orthogonal then prove that A is skew symmetric.

Question

If $(I-A)(I+A)^{-1}$ is orthogonal then prove that A is skew symmetric.

I can't solve this question.

Answer 1

You know that $$\left((I-A)(I+A)^{-1} \right)^T(I-A)(I+A)^{-1} =I \Rightarrow \\ ((I+A)^{-1})^{T} (I-A)^T(I-A)(I+A)^{-1} =I \Rightarrow \\ (I-A)^T(I-A) =(I+A)^{T} (I+A)\Rightarrow \\ (I-A^T)(I-A) =(I+A^T)(I+A)\Rightarrow \\ I-A-A^T+AA^T =I+A+A^T+AA^T\Rightarrow \\ A+A^T=0$$

Answer 2

Note that if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, it is also an eigenvector of $(I-A)(I+A)^{-1}$ with eigenvalue $\displaystyle\frac{1-\lambda}{1+\lambda}$. Noting that the eigenvalues of orthogonal matrices are either 1 or -1, we have that $\lambda$ must be 0. Since $(I-A)(I+A)^{-1}$ is orthogonal, it's diagonalizable and its eigenvectors form a basis, so $A$ is diagonalizable as well. So $A=0$ and is skew symmetric.