If A is a symmetric matrix, B is a skew-symmetric matrix, `A+B` is nonsingular and `C=(A+B)^(-1) (A-B)`, then prove that
(i) `C^(T) (A+B) C=A+B` (ii) `C^(T) (A-B)C=A-B`
(iii) `C^(T)AC=A`

Explanation

(i) `(A+B)C=(A+B) (A+B)^(-1) (A-B)`
`implies (A+B)C=A-B` (1)
`C^(T)=(A-B)^(T) ((A+B)^(-1))^(T)`
`=(A+B)((A+B)^(T))^(-1)" "( :' A^(T)=A, B^(T)=-B)`
`{"as "|A+B| ne 0 implies |(A+B)^(T)| ne 0 implies |A-B| ne 0}`
`=(A+B) (A-B)^(-1)` (2)
From (1) and (2), we get
`C^(T) (A+B)C=(A+B)(A-B)^(-1) (A-B)`
`=(A+B)` (3)
(ii) Taking transpose in (3), we get
`C^(T) (A+B)^(T) (C^(T))^(T)=(A+B)^(T)`
`C^(T)(A-B) C=A-B` (4)
(iii) Adding (3) and (4), we get
`C^(T) [A+B+A-B]C=2A`
`C^(T) AC=A`