Given that $A$ is antisymmetric matrix, and $(I + A)$ it is invertible.
I had to prove that $(I - A)$ is also invertible, and I managed to do so, But now given that $$C = (I - A)(I + A)^{-1}$$ I have to prove that $$C^\intercal C = I$$
- Without determinants.
I tried many different ways, First tried to prove that C is invertible and to find it's inverse matrix, also tried to find the inverse of the transposed matrix, I have no clue how to continue...
Thanks.
Lemma: If $AB=I$, then $BA=I$. see here
The antisymmetric matrix $A$ means $A^T=-A$.
We first check the value of $CC^T$: \begin{equation} \begin{aligned} CC^T&=(I-A)(I+A)^{-1}(I+A^T)^{-1}(I-A^T)\\ &=(I-A)(I+A)^{-1}(I-A)^{-1}(I+A)\\ &=(I-A)(I-A^2)^{-1}(I+A) \end{aligned} \end{equation}
It follows from $(I+A)(I-A)(I-A^2)^{-1}=(I-A^2)(I-A^2)^{-1}=I$ and the above lemma that$$(I-A)(I-A^2)^{-1}(I+A)=I.$$
Therefore, $CC^T=I$. Furthermore, $C^TC=I$.