$\phi(A) =(I_n-A)(I_n+A)^{-1}$ is antisymmetric?

Question

Let $U=\{A\in SO_n(\mathbb R)/ \;\det(I_n+A)\neq0\}$. Let $\phi :U\to M_n(\mathbb R)$ the application defined by : $$\phi(A)=(I_n-A)(I_n+A)^{-1}$$ How can I show that $\phi(A)$ is antisymmetric !?

Any help is highlty appreciated!

Answer 1

Since $A$ is orthogonal, $AA^T = I$. Set $\phi = (I-A)(I+A)^{-1}$, so $$ \begin{split} \phi^T &= ((I+A)^{-1})^T (I-A)^T \\ &= (I+A^T)^{-1} (I-A^T) \\ &= (I+A^{-1})^{-1} (I-A^{-1}) \\ &= (A^{-1}(A+I))^{-1} (A^{-1}(A-I)) \\ &= (A+I)^{-1}AA^{-1}(A-I). \end{split} $$ Here, all the factors commute with each other because they just involve $A$ and $I$. From this, you can finish the steps to show that $\phi^T = -\phi$. So $\phi$ is antisymmetric.

Answer 2

You are writing down an inverse Cayley transform which is its own inverse: an involution. You may check all inverses below are nonsingular. Your matrices A are orthogonal (rotation) matrices by definition. The Cayley transform maps antisymmetric to orthogonal matrices, and hence orthogonal to antisymmetric ones. You might be interested in this question.

Since the left and right factor of what you write commute because both terms in each do, you can write $$ \phi= \frac{I-A}{I+A}, ~~~~~~~~~~~~~A A^T=I, \leadsto \\ A= \frac{I-\phi}{\phi +I}, ~~~~~~~~~~~~~ \frac{I-\phi}{I+\phi} \frac{I-\phi^T}{I+\phi^T}=I \leadsto \\ \frac{I-\phi}{I+\phi}= \frac{I+\phi^T}{I-\phi^T} ~. $$ Cayley-transforming both sides again, you conclude that $\phi^T= -\phi$. This is why you had to show the transform is an involution.

Reassure yourself no denominator vanishes. Try a simple example, e.g. $$ A= \tfrac{1}{2}\begin{bmatrix}\sqrt 3 & -1\\1& \sqrt 3 \end{bmatrix}, ~~~~~~~I-A= \tfrac{1}{2}\begin{bmatrix}2-\sqrt 3 & 1\\-1&2- \sqrt 3 \end{bmatrix},\\ (I+A)^{-1}= \frac{1}{4+2\sqrt3}\begin{bmatrix}2+\sqrt 3 & 1\\-1&2+ \sqrt 3 \end{bmatrix}\\ \implies \phi= \frac{1}{2+\sqrt 3}\begin{bmatrix}0 & 1\\-1&0 \end{bmatrix} $$