If `A` is a skew symmetric matrix, then `B=(I-A)(I+A)^(-1)` is (where `I` is an identity matrix of same order as of `A`)

Explanation

We have `A=(I+S)(I-S)^(-1)`
`:. A^(T)=[(I-S)^(-1)]^(T) (I+S)^(T)=[(I-S)^(T)]^(-1) (I+S)^(T)`
But `(I-S)^(T)=I^(T)-S^(T)=I +S" "("as "S^(T)=-S)`
and `(I+S)^(T)=I^(T)+S^(T)=I-S`
`:. A^(T)=(I+S)^(-1) (I-S)`
`:. A^(T)A=(I+S)^(-1) (I-S) (I+S) (I-S)^(-1)`
`=(I+S)^(-1) (I+S) (I-S) (I-S)^(-1)`
`=I`
Thus, a is orthogonal.