If `S` is a real skew-symmetric matrix, then prove that `I-S` is non-singular and the matrix `A=(I+S)(I-S)^(-1)` is orthogonal.

Explanation

`because ` S is skew-symmetric matrix
`therefore S^(T) =-S " "…(i) `
Frist we will show that `I-S` is non-singular. The equality
`abs(I-S) = 0 rArr I` is a characteristic root of the matrix S but this is
not possile, for a real skew-symmetric matrix can have zero
or purely imaginary numbers as its characterixtic roots. Thus,
`abs(I-S) ne 0` i.e. `I-S` is non-singular.
We have,
`A^(T) = {(I+S)(I-S)^(-1)}^(T) = {(I-S)^(-1) (I+S)}^(T)`
` = ((I+S)^(-1))^(T)(I+S)^(T) = (I+S)^(T) {(I+S)^(-1)}^(T)`
` = ((I-S)^(T))^(-1)(I+S)^(T) = (I+S)^(T) {(I-S)^(T)}^(-1)`
` = (I^(T)-S^(T))^(-1)(I^(T)+S^(T)) = (I^(T)+S^(T)) (I^(T)-S^(T))^(-1)`
`= (I+S)^(-1) (I-S) = (I-S) (I+S)^(-1)` [from Eq. (i)]
`therefore A^(T)A= (I+S)^(-1) (I-S) (I+S) (I-S)^(-1)`
`= (I-S)(I+S)^(-1) (I-S)^(-1) (I+S)`
`= (I+S)^(-1)(I+S) (I-S) (I-S)^(-1) `
`= (I-S) (I-S)^(-1)(I+S)^(-1)(I+S) `
`=Icdot I = Icdot I=I=I`
Hence, A is orthogonal.