$(I-B)^{-1} (I+B)$ always orthogonal if $B$ anti-symmetric

Question

I saw this claim in Strang Intro to linear algebra 5th edition, page 336, and am finding it surprisingly difficult to prove.

The claim is that $A = (I-B)^{-1} (I+B)$ is orthogonal if $B^T = -B$.

We have $A^T = (I + B^T) (I - B^T)^{-1} = (I - B)(I + B)^{-1}$. Multiplying, we have

$$ A^T A = (I - B)(I + B)^{-1} (I-B)^{-1} (I+B).$$

Since the claim is that $A$ is orthogonal, this is supposed to be $I$, but it is unclear how to proceed from here. Do $(I + B)^{-1}$ and $(I-B)^{-1}$ commute? If yes, then the result follows trivially. However, I don't see any reason why they would commute.

It's certainly easy to show that $\det A = 1$, since $\det (I - B) = \det (I + B^T) = \det (I + B)$ and thus $\det A = \det (I + B) / \det (I - B) = 1$. However, I think that's necessary but not sufficient for $A$ to be orthogonal.

Answer 1

Observe that

$$(I-B)(I+B)=(I-B^2)=(I+B)(I-B)$$ and $$(I-B)^{-1} (I+B)(I - B)(I + B)^{-1}= (I-B)^{-1} (I-B)(I+B)(I + B)^{-1}.$$

Answer 2

$(I+B)^{-1}$ and $(I-B)^{-1}$ indeed commute (and thus the result follows): we have

$$(I+B)^{-1} (I-B)^{-1} = [(I-B)(I+B)]^{-1} = (I - B^2)^{-1}.$$

The result is clearly invariant under sign flip.