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

$Def:$ 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?

Answer 1

We can check it directly. Recall that $(AB)^t = B^tA^t$ and $(A^t)^{-1} = (A^{-1})^t$.

Note that $(I+K)(I-K) = I - K^2 = (I-K)(I+K)$.

$$B^tB = ((I+K)(I-K)^{-1})^t (I+K)(I-K)^{-1} = ((I-K)^{-1})^t (I+K)^t(I+K)(I-K)^{-1} = ((I-K)^t)^{-1} (I-K)(I+K)(I-K)^{-1} = (I+K)^{-1} (I-K)(I+K)(I-K)^{-1} = (I+K)^{-1} (I+K)(I-K)(I-K)^{-1} = I.$$