TITLE: For any skew-symmetric matrix $K$, if $(I-K)$ is invertible, show that $B=(I+K)(I-K)^{-1}$ satisfies $B^tB=I=BB^t$.
QUESTION [1 upvotes]: $Def:$ A matrix $K$ is skew-symmetric if $K^t=-K$.
When manipulating the $B$ and $B^{-1}$ expressions, I don't know how to deal with the inverses. How can one derive the inverse of the sum of matrices?

REPLY [2 votes]: We can check it directly. Recall that $(AB)^t = B^tA^t$ and $(A^t)^{-1} = (A^{-1})^t$.
Note that $(I+K)(I-K) = I - K^2 = (I-K)(I+K)$.
$$B^tB = ((I+K)(I-K)^{-1})^t (I+K)(I-K)^{-1} = ((I-K)^{-1})^t (I+K)^t(I+K)(I-K)^{-1} = ((I-K)^t)^{-1} (I-K)(I+K)(I-K)^{-1} = (I+K)^{-1} (I-K)(I+K)(I-K)^{-1} = (I+K)^{-1} (I+K)(I-K)(I-K)^{-1} = I.$$
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