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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.

Sebastiano
  • 8,290
m.c
  • 17

1 Answers1

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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, $C^TC=I$.

fengbiqian
  • 331
  • The thing I was missing is that I been able to commute (change the order). Now I understand that I was able to do it, Thanks ! – m.c Nov 22 '20 at 14:02