If s is a real skew-symmetric matrix, the show that `I-S` is non-singular and matrix `A= (I+S) (I-S) ^(-1) = (I-S) ^(-1) (I+S) ` is orthogonal.A.B.C.D.

If s is a real skew-symmetric matrix, the show that `I-S` is non-singu

1.	If s is a real skew-symmetric matrix, the show that `I-S` is non-singular and matrix `A= (I+S) (I-S) ^(-1) = (I-S) ^(-1) (I+S) ` is orthogonal.A.B.C.D.
Answer» `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.

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If s is a real skew-symmetric matrix, the show that `I-S` is non-singular and matrix `A= (I+S) (I-S) ^(-1) = (I-S) ^(-1) (I+S) ` is orthogonal.A.B.C.D.

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